Abstract Algebra — Chapter 4 Advanced group theory

Chapter 4
Advanced group theory

4.1. Isomorphism theorems

We have already proven the first isomorphism theorem. In this section, we shall use it in deriving two others.

Definition 4.1.1. §

Let $H$ and $K$ be subgroups of a group $G$ . We define the join $H \lor K$ of $H$ and $K$ to be the subgroup of $G$ generated by $H \cup K$ .

Remark 4.1.2. §

Note that the join $H \lor K$ contains (and is generated by) the set

𝐻𝐾 = {h𝑘 ∣ h \in H, k \in K}

and $H \lor K = 𝐻𝐾$ and only if $𝐻𝐾 \leq G$ .

Example 4.1.3. §

Take $H = ⟨ (1 2) ⟩$ and $K = ⟨ (1 3) ⟩$ as subgroups of $S_{3}$ . We have

𝐻𝐾 = {e, (1 2), (1 3), (1 3 2)},

which is not a subgroup of $S_{3}$ , while $H \lor K = S_{3}$ .

Lemma 4.1.4. §

We have that $𝐻𝐾 \leq G$ if and only if $𝐻𝐾 = 𝐾𝐻$ .

Proof.

Suppose first that $𝐻𝐾 \leq G$ . Let $h \in H$ and $k \in K$ . Since $h, k \in 𝐻𝐾$ , we have $𝑘h \in 𝐻𝐾$ , as $𝐻𝐾 \leq G$ . Thus $𝐾𝐻 \subseteq 𝐻𝐾$ . On the other hand, we have ${(h𝑘)}^{- 1} = k^{- 1} h^{- 1} \in 𝐾𝐻$ , so the inverse of every element of $𝐻𝐾$ is contained in $𝐾𝐻$ . But every element of $𝐻𝐾$ is the inverse of some element of $𝐻𝐾$ since $𝐻𝐾 \leq G$ , so $𝐻𝐾 \subseteq 𝐾𝐻$ as well. Thus, we have $𝐻𝐾 = 𝐾𝐻$ .

Now suppose that $𝐻𝐾 = 𝐾𝐻$ . We always have $e = e \cdot e \in 𝐻𝐾$ . Moreover, if $h, h^{'} \in H$ and $k, k^{'} \in K$ , then

h𝑘 \cdot h^{'} k^{'} = h (k h^{'}) k^{'},

and since $𝐾𝐻 = 𝐻𝐾$ , there exists $h^{″} \in H$ , $k^{″} \in K$ such that $k h^{'} = h^{″} k^{″}$ , so

h (k h^{'}) k^{'} = h (h^{″} k^{″}) k^{'} = h h^{″} \cdot k^{″} k^{'} \in 𝐻𝐾 .

Moreover, we have

{(h𝑘)}^{- 1} = k^{- 1} h^{- 1} \in 𝐾𝐻,

but $𝐾𝐻 = 𝐻𝐾$ , so ${(h𝑘)}^{- 1} \in 𝐻𝐾$ . Thus, we have that $𝐻𝐾 \leq G$ . □

Corollary 4.1.5. §

Suppose that $H$ and $N$ are subgroups of $G$ with $N$ normal. Then we have $𝐻𝑁 \leq G$ . If $H$ is normal in $G$ as well, then we have $𝐻𝑁 ⊴ G$ .

Proof.

By Lemma 4.1.4, it suffices to show that $𝐻𝑁 = 𝑁𝐻$ . But $N ⊴ G$ , so $h𝑁 = 𝑁h$ for all $h \in H$ , which means that

𝐻𝑁 = ⋃_{h \in H} h𝑁 = ⋃_{h \in H} 𝑁h = 𝑁𝐻 .

Moreover, if $H ⊴ G$ , then for any $g \in G$ , we have

𝑔𝐻𝑁 g^{- 1} = 𝑔𝐻 g^{- 1} \cdot 𝑔𝑁 g^{- 1} = 𝐻𝑁 .

□

Theorem 4.1.6 (Second Isomorphism Theorem). §

Let $H$ be a subgroup of a group $G$ , and let $N$ be a normal subgroup of $G$ . Then we have an isomorphism

H ∕ (H \cap N) \to_{}^{\sim} 𝐻𝑁 ∕ N .

Proof.

Define

ϕ : H \to 𝐻𝑁 ∕ N, ϕ (h) = h𝑁 .

Then

\ker ϕ = {h \in H ∣ h \in N} = H \cap N .

Moreover, if $h \in H$ and $n \in N$ , then $h𝑛𝑁 = h𝑁 = ϕ (h)$ , so $ϕ$ is surjective. The result therefore follows by the first isomorphism theorem. □

Example 4.1.7. §

Consider the subgroups $H = ℤ \times ℤ \times {0}$ and $N = {0} \times ℤ \times ℤ$ of $G = ℤ^{3}$ . We have $𝐻𝑁 = G$ , and $H \cap N = {0} \times ℤ \times {0}$ . Note that

H ∕ (H \cap N) = \frac{ℤ \times ℤ \times {0}}{{0} \times ℤ \times {0}} \to_{}^{\sim} ℤ

via the map that takes $(a, b, 0) (H \cap N)$ to $a$ . On the other hand, we have

𝐻𝑁 ∕ N = \frac{ℤ \times ℤ \times ℤ}{{0} \times ℤ \times ℤ} \to_{}^{\sim} ℤ

via the map that takes $(a, b, c) N$ to $a$ .

Remark 4.1.8. §

Suppose that $H$ and $K$ are subgroups of a group $G$ with $K \leq H$ . If $K$ and $H$ are both normal subgroups of $G$ , then $K ⊴ H$ . On the other hand, the property of being a normal subgroup is not transitive. One may have $K ⊴ H$ and $H ⊴ G$ but $𝐾⋬𝐺$ !

Example 4.1.9. §

Take $G = A_{4}$ ,

H = ⟨ (1 2) (3 4), (1 3) (2 4) ⟩,

and $K = ⟨ (1 2) (3 4) ⟩$ . Since

(1 2) (3 4) \cdot (1 3) (2 4) = (1 4) (2 3) = (1 3) (2 4) \cdot (1 2) (3 4),

the group $H$ is an abelian group of order $4$ (isomorphic to the Klein four-group) consisting of the three cycles of cycle type $(2, 2)$ and the identity. Now $K ◃ H$ since $H$ is abelian, and $H ◃ G$ since conjugation preserves cycle type. On the other hand, $K$ is not a normal subgroup of $G$ since

(1 2 3) \cdot (1 2) (3 4) \cdot {(1 2 3)}^{- 1} = (1 4) (2 3) \notin K .

Theorem 4.1.10 (Third Isomorphism Theorem). §

Let $H$ and $K$ be normal subgroups of a group $G$ with $K \leq H$ . Then we have an isomorphism

G ∕ H \to_{}^{\sim} (G ∕ K) ∕ (H ∕ K) .

Proof.

We first remark that $H ∕ K \leq G ∕ K$ since it is a subset of $G ∕ K$ that is a group under the operation on $G ∕ K$ . Moreover, $H ∕ K ⊴ G ∕ K$ since if $h \in H$ and $a \in G$ , then $𝑎h a^{- 1} = h^{'}$ for some $h^{'} \in H$ , so

𝑎𝐾 \cdot h𝐾 \cdot a^{- 1} K = h^{'} K \in H ∕ K .

We may now define

𝜃 : G \to (G ∕ K) ∕ (H ∕ K)

𝜃 (a) = (𝑎𝐾) (H ∕ K) .

By the group laws on $G ∕ K$ and $(G ∕ K) ∕ (H ∕ K)$ , we have

𝜃 (𝑎𝑏) = (𝑎𝑏𝐾) (H ∕ K) = (𝑎𝐾 \cdot 𝑏𝐾) (H ∕ K) = (𝑎𝐾) (H ∕ K) \cdot (𝑏𝑘) (H ∕ K) = 𝜃 (a) 𝜃 (b) .

Then $𝜃 (a) = H ∕ K$ if and only if $𝑎𝐾 = h𝐾$ for some $h \in H$ , so if and only if $a \in H$ . Thus $\ker 𝜃 = H$ . On the other hand, $𝜃$ is surjective by definition. The result now follows from the first isomorphism theorem. □

Remark 4.1.11. §

For $H$ , $K$ and $G$ as in the third isomorphism theorem, the composite map

G \to G ∕ K \to (G ∕ K) ∕ (H ∕ K) \to G ∕ H,

where the first and second maps are quotient maps and the third is the inverse of the isomorphism in the third isomorphism theorem, is exactly the quotient map $G \to G ∕ H$ .

Example 4.1.12. §

Let $G = ℤ$ , $H = 𝑚ℤ$ and $K = 𝑛ℤ$ , where $m, n \geq 1$ and $m$ divides $n$ , so that $K ⟨ H$ . Then $G ∕ H = ℤ ∕ 𝑚ℤ$ , $G ∕ K = ℤ ∕ 𝑛ℤ$ , and $H ∕ K = 𝑚ℤ ∕ 𝑛ℤ$ . We note that

(G ∕ K) ∕ (H ∕ K) = \frac{ℤ ∕ 𝑛ℤ}{𝑚ℤ ∕ 𝑛ℤ} = \frac{ℤ ∕ 𝑛ℤ}{⟨ m ⟩} \to_{}^{\sim} ℤ ∕ 𝑚ℤ = G ∕ H,

the map in the last isomorphism being induced by the natural reduction-modulo- $m$ map from $ℤ ∕ 𝑛ℤ$ to $ℤ ∕ 𝑚ℤ$ and the first isomorphism theorem.

Example 4.1.13. §

Let $G = ℤ^{3}$ , $H = ℤ \times ℤ \times {0}$ and $K = ℤ \times {0} \times {0}$ . Then $G ∕ 𝐻≅ℤ$ via the map that takes $(a, b, c) H$ to $c$ , while $G ∕ 𝐾≅ℤ \times ℤ$ via the map that takes $(a, b, c) K$ to $(b, c)$ , and $H ∕ K$ has image $ℤ \times {0}$ under this map. Then

(G ∕ K) ∕ (H ∕ K) \to_{}^{\sim} \frac{ℤ \times ℤ}{ℤ \times {0}} \to_{}^{\sim} ℤ,

where the latter map takes $(b, c) (ℤ \times ℤ)$ to $c$ .

We also have the following, known as the butterfly (or Zassenhaus) lemma, which we state without proof.

Theorem 4.1.14 (Butterfly lemma). §

Let $H$ , $K$ , $A$ , $B$ be subgroups of a group $G$ with $A ⊴ H$ and $B ⊴ K$ . Then there is a canonical isomorphism

\frac{A (H \cap K)}{(A (H \cap B)} ≅ \frac{B (H \cap K)}{B (A \cap K)} .

4.2. Commutators and simple groups

Definition 4.2.1. §

Let $G$ be a group and $a, b \in G$ . The commutator of $a$ and $b$ is

[a, b] = 𝑎𝑏 a^{- 1} b^{- 1} .

Definition 4.2.2. §

The commutator subgroup $[G, G]$ of a group $G$ is the subgroup of $G$ generated by its commutators, which is to say

[G, G] = ⟨ [a, b] ∣ a, b \in G ⟩ .

Remark 4.2.3. §

If $G$ is an abelian group, then $[G, G] = {e}$ .

Example 4.2.4. §

In $D_{n}$ , we have

\begin{array}{rcl} [r^{i}, r^{j} s] = r^{i} (r^{j} s) (r^{- i}) (s r^{- j}) = r^{2 i}, \\ [r^{i} s, r^{j} s] = (r^{i} s) (r^{j} s) (s r^{- i}) (s r^{- j}) = r^{i - j} s r^{j - i} s = r^{2 (i - j)} . \end{array}

Therefore, we have that

[D_{n}, D_{n}] = ⟨ r^{2} ⟩,

which has index $2$ and $4$ in $D_{n}$ in the cases that $n$ is odd and even, respectively.

Example 4.2.5. §

We have

[{GL}_{n} (ℝ), {GL}_{n} (ℝ)] \subseteq {SL}_{n} (ℝ) .

since $\det (𝐴𝐵 A^{- 1} B^{- 1}) = 1$ for any $A, B \in {GL}_{n} (ℝ)$ . The opposite equality also holds, but we shall not prove it here.

Lemma 4.2.6. §

The commutator subgroup of $G$ is a normal subgroup of $G$ .

Proof.

Let $a, b, g \in G$ . We have

g [a, b] g^{- 1} = 𝑔𝑎𝑏 a^{- 1} b^{- 1} g^{- 1} = (𝑔𝑎) b {(𝑔𝑎)}^{- 1} b^{- 1} \cdot 𝑏𝑔 b^{- 1} g^{- 1} = [𝑔𝑎, b] [b, g] \in [G, G] .

Since every element of $[G, G]$ is a product of powers of elements of the form $[a, b]$ with $a, b \in G$ and every such element is sent to another element of $[G, G]$ by the conjugation homomorphism $γ_{g}$ , the image of $γ_{g}$ is contained in $[G, G]$ . Since this holds for all $g \in G$ , we have that $[G, G]$ is normal. □

Example 4.2.7. §

Let $n \geq 3$ . We claim that $[S_{n}, S_{n}] = A_{n}$ . Note that

[σ, (a b)] = σ (a b) σ^{- 1} (a b) = (σ (a) σ (b)) (a b)

for $σ \in S_{n}$ and $a, b \in X_{n}$ with $a < b$ . It follows that $[G, G]$ contains all products of two transposiitons in $S_{n}$ . Moreover, these generate $A_{n}$ by definition, so $A_{n} \subseteq [S_{n}, S_{n}]$ . Furthermore, every element of $[S_{n}, S_{n}]$ is even as every such element is a product of elements of the form $[σ, τ]$ with $σ, τ \in S_{n}$ , and these satisfy

sign ([σ, τ]) = sign (σ) sign (τ) sign {(σ)}^{- 1} sign {(τ)}^{- 1} = 1 .

Theorem 4.2.8. §

Let $N$ be a normal subgroup of $G$ . Then $G ∕ N$ is an abelian group if and only if $[G, G] \leq N$ .

Proof.

Let $a, b \in G$ . We have $𝑎𝑏𝑁 = 𝑏𝑎𝑁$ if and only if $a^{- 1} b^{- 1} 𝑎𝑏 \in N$ , so if and only if $[a^{- 1}, b^{- 1}] \in N$ . But $[G, G]$ is the smallest subgroup of $G$ containing $[c, d]$ for every $c, d \in G$ , so $N$ is normal if and only if $[G, G]$ is contained in $N$ . □

Definition 4.2.9. §

The maximal abelian quotient, or abelianization, $G^{ab}$ of a group $G$ is the quotient group

G^{ab} = G ∕ [G, G] .

We have the following consequence of Theorem 4.2.8.

Corollary 4.2.10. §

Let $G$ be a group and $H$ be an abelian group, and suppose $ϕ : G \to H$ is a homomorphism. Then there exists a homomorphism $\bar{ϕ} : G^{ab} \to H$ with $\bar{ϕ} \circ π_{[G, G]} = ϕ$ , where $π_{[G, G]} : G \to G^{ab}$ is the quotient map.

Proof.

By the first isomorphism theorem, there exists a unique map $ψ : G ∕ \ker ϕ \to H$ with $ψ \circ π_{\ker ϕ} = ϕ$ . By Theorem 4.2.8, we have that $[G, G] \subseteq \ker ϕ$ . and now the third isomorphism theorem provides a composite map

β : G^{ab} = G ∕ [G, G] \to (G ∕ [G, G]) ∕ (\ker ϕ ∕ [G, G]) \to_{}^{\sim} G ∕ \ker ϕ

such that $π_{\ker ϕ} = β \circ π_{[G, G]}$ . Set $\bar{ϕ} = ψ \circ β$ . Then

\bar{ϕ} \circ π_{[G, G]} = ψ \circ β \circ π_{[G, G]} = ψ \circ π_{\ker ϕ} = ϕ,

as desired. □

Example 4.2.11. §

The abelianization of $D_{n}$ is $D_{n} ∕ ⟨ r^{2} ⟩$ , and if $n$ is even this group is isomorphic to $ℤ ∕ 2 ℤ \times ℤ ∕ 2 ℤ$ under the map $\bar{ϕ}$ induced by the homomorphism

ϕ : D_{n} \to ℤ ∕ 2 ℤ \times ℤ ∕ 2 ℤ

that takes $r^{i} s^{j}$ to $(i, j)$ . If $n$ is odd, then $D_{n} ∕ ⟨ r^{2} ⟩$ is isomorphic to $ℤ ∕ 2 ℤ$ via the map from $D_{n}$ that takes $r^{i} s^{j}$ to $j$ .

Example 4.2.12. §

We have $S_{n}^{ab} = S_{n} ∕ A_{n}$ , and so $S_{n}^{ab} ≅ℤ ∕ 2 ℤ$ via the map induced by the sign map.

Here is another nice class of normal subgroups.

Definition 4.2.13. §

The center of a group $G$ is the subgroup

Z (G) = {z \in G ∣ 𝑧𝑎 = 𝑎𝑧 for all a \in G} .

Since $𝑎𝑧 = 𝑧𝑎$ for all $z \in Z (G)$ and $a \in G$ , we clearly have that $𝑎𝑍 (G) = Z (G) a$ for all $a \in G$ , and hence we have that $Z (G) ⊴ G$ . We leave the verification of the following examples to the reader.

Examples 4.2.14. §

a.: If $G$ is abelian, then $Z (G) = G$ .
b.: For $n \geq 3$ , we have $Z (S_{n}) = {e}$ .
c.: For $n \geq 3$ , we have $Z (D_{n}) = ⟨ r^{n ∕ 2} ⟩$ if $n$ is even and $Z (D_{n}) = {e}$ if $n$ is odd.
d.: For $n \geq 2$ , we have that $Z ({GL}_{n} (ℝ))$ is the subgroup of scalar matrices.

Definition 4.2.15. §

A nontrivial group $G$ is called simple if it has no nontrivial, improper normal subgroups.

Example 4.2.16. §

An abelian group $G$ is simple if and only if it is cyclic of prime order, since otherwise it will have a nontrivial, improper subgroup, which is automatically normal since $G$ is abelian.

Examples 4.2.17. §

The groups $S_{n}$ and $D_{n}$ for $n \geq 3$ are not simple, since they contain improper, nontrivial normal subgroups. Moreover, $A_{4}$ is not simple, as it contains the normal subgroup $⟨ (1 2) (3 4), (1 3) (2 4) ⟩$ of order $8$ , as seen in Example 4.1.9.

We have the following easy lemma.

Lemma 4.2.18. §

If $G$ is simple and nonabelian, then $Z (G) = {e}$ .

Proof.

If $G$ is nonabelian, then $Z (G) \neq G$ , and if $G$ is also simple, then since $Z (G)$ is normal, we must have $Z (G) = {e}$ . □

We note that if a group is not simple, we can find a nontrivial quotient of it by a nontrivial normal subgroup that is.

Definition 4.2.19. §

We say that a normal subgroup $M$ of a group $G$ is a maximal normal subgroup if it is not contained in any larger proper normal subgroup of $G$ .

Example 4.2.20. §

Any subgroup of index $2$ in a group is a maximal normal subgroup, since such a subgroup is normal and is not contained in a larger proper normal subgroup, being that its index would have to be smaller than $2$ , but greater than $1$ .

Proposition 4.2.21. §

Let $N$ be a normal subgroup of $G$ . Then $G ∕ N$ is simple if and only if $N$ is maximal.

Proof.

This is an immediate consequence of Proposition 2.13.10, since $G ∕ N$ is simple if and only if it has no proper normal subgroups, which are in bijection with the proper normal subgroups of $G$ containing $N$ . □

Example 4.2.22. §

Since $A_{n}$ is a maximal normal subgroup of $S_{n}$ , the quotient $S_{n} ∕ A_{n}$ is simple (which we already knew since it is isomorphic to $ℤ ∕ 2 ℤ$ ).

Finite simple groups are considered the building blocks of all finite groups. Their classification was the major project in group theory during the 20th century, and it was finally finished at the beginning of the 21st. Many examples of nonabelian finite simple groups are simple enough to give, though proving they are simple is another matter.

Example 4.2.23. §

The groups $A_{n}$ are simple for all $n \geq 5$ . In fact, $A_{5}$ is a finite simple group with the smallest possible order, which is $60$ , and it is the unique such group up to isomorphism. We defer the proofs of these facts until later.

We mention one more broad class of examples of finite simple groups.

Example 4.2.24. §

We remark that $ℤ ∕ 𝑝ℤ$ has two binary operations of addition and multiplication, and these satsify the distributive property. Hence we may consider the set ${GL}_{n} (ℤ ∕ 𝑝ℤ)$ of invertible $n$ by $n$ matrices with entries in $ℤ ∕ 𝑝ℤ$ , and this forms a group under multiplication. We also have its subgroup ${SL}_{n} (ℤ ∕ 𝑝ℤ)$ of matrices with determinant $1$ . It is not necessarily simple, as it is possible that it can have nontrivial center: the group of scalar matrices with determinant $1$ . I.e., $𝑎𝐼 \in {SL}_{n} (ℤ ∕ 𝑝ℤ)$ if and only if $a^{n} = 1$ in $ℤ ∕ 𝑝ℤ$ . The quotient of ${SL}_{n} (ℤ ∕ 𝑝ℤ)$ by its center is called ${PSL}_{n} (ℤ ∕ 𝑝ℤ)$ . It turns out that ${PSL}_{n} (ℤ ∕ 𝑝ℤ)$ is simple for all primes $p$ for all $n \geq 3$ and for all primes $p \geq 5$ and $n = 2$ .

4.3. Automorphism groups

Definition 4.3.1. §

An automorphism of a group $G$ is an isomorphism $ϕ : G \to G$ .

The subgroup test shows quickly that the set of automorphisms of $G$ forms a subgroup of $G$ under composition.

Proposition 4.3.2. §

The set of automorphisms $Aut (G)$ of a group $G$ forms a group under composition.

Proof.

Since composition of functions is associative, to check that $Aut (G)$ is a group, we need only check that it contains an identity element, which it clearly does, and that it contains inverses, which is does since the inverse of an isomorphism is an isomorphism. □

Definition 4.3.3. §

The automorphism group $Aut (G)$ of a group $G$ is the group of automorphisms of $G$ under conjugation.

Examples 4.3.4. §

a.: We have $Aut (ℤ ∕ 𝑛ℤ) ≅ {(ℤ ∕ 𝑛ℤ)}^{\times}$ for $n \geq 1$ via the map that takes $ϕ \in Aut (ℤ ∕ 𝑛ℤ)$ to $ϕ (1)$ . In fact, we have $ϕ (a) = 𝑎𝜙 (1)$ for any $a \in ℤ ∕ 𝑛ℤ$ , and so $ϕ$ is multiplication by $ϕ (1)$ . This can be invertible if and only if $a$ is a unit in $ℤ ∕ 𝑛ℤ$ .
b.: The same discussion as in part a tells us that $Aut (ℤ) = ℤ^{\times} = ⟨ - 1 ⟩$ .
c.: We have $Aut (ℤ^{n}) ≅ {GL}_{n} (ℤ)$ . That is, if $ϕ \in Aut (ℤ^{n})$ and $e_{i}$ is the $i$ th element in the standard basis of $ℤ^{n}$ (with a $1$ in its $i$ th entry and $0$ in the others), then $ϕ (e_{i})$ determines the $i$ th column of a matrix in ${GL}_{n} (ℤ)$ . The inverse map is given by allowing ${GL}_{n} (ℤ)$ to act on $ℤ^{n}$ by left multiplication, viewing an element of $ℤ^{n}$ as a column vector.

We give the example of the automorphisms of the dihedral group as a proposition.

Proposition 4.3.5. §

For $n \geq 3$ , the group $Aut (D_{n})$ is isomorphic to the subgroup $Aff (ℤ ∕ 𝑛ℤ)$ of ${GL}_{2} (ℤ ∕ 𝑛ℤ)$ given by

{(\begin{matrix} a & b \\ 0 & 1 \end{matrix}) | a \in {(ℤ ∕ 𝑛ℤ)}^{\times}, b \in ℤ ∕ 𝑛ℤ} .

Proof.

The isomorphism $f : Aff (ℤ ∕ 𝑛ℤ) \to Aut (D_{n})$ is given by

f : (\begin{matrix} a & b \\ 0 & 1 \end{matrix}) \mapsto ϕ_{a, b},

where $ϕ_{a, b} (r) = r^{a}$ and $ϕ_{a, b} (s) = r^{b} s$ . Since $D_{n}$ is generated by $r$ and $s$ , there exists at most one element of $Aut (D_{n})$ taking these values on $r$ and $s$ . Since $F_{r, s}$ is free, we can define $Φ : F_{r, s} \to D_{n}$ by $Φ (r) = r^{a}$ and $Φ (s) = r^{b} s$ for $a$ and $b$ as above. Note that $Φ (r^{n}) = r^{𝑎𝑛} = e$ , $Φ (s^{2}) = {(r^{b} s)}^{2} = e$ , and $Φ ({(𝑟𝑠)}^{2}) = {(r^{a + b} s)}^{2} = e$ , so by the presentation $D_{n} ≅ ⟨ r, s ∣ r^{n}, s^{2}, 𝑟𝑠𝑟𝑠 ⟩$ , we have the existence of $ϕ_{a, b}$ . In that $a$ is invertible modulo $n$ , we have

⟨ r^{a}, r^{b} s ⟩ = ⟨ r, r^{b} s ⟩ = ⟨ r, s ⟩ = D_{n},

so $ϕ_{a, b}$ is onto and hence in $Aut (D_{n})$ as $D_{n}$ is finite.

Now, any $ϕ \in Aut (D_{n})$ must send $r$ to another element of order $n$ , so $r^{a}$ with $a \in ℤ$ prime to $n$ . It must also send $s$ to an element of order $2$ that cannot be in the subgroup $⟨ ϕ (r) ⟩ = ⟨ r ⟩$ , since $ϕ$ is surjective. Thus, $ϕ (s) = r^{b} s$ for some $b \in ℤ$ . Thus, $f$ is onto, and it is one-to-one by definition. To see it is a homomorphism, note that

(\begin{matrix} a & b \\ 0 & 1 \end{matrix}) (\begin{matrix} a^{'} & b^{'} \\ 0 & 1 \end{matrix}) = (\begin{matrix} a a^{'} & b + a b^{'} \\ 0 & 1 \end{matrix}),

while

\begin{array}{rcl} ϕ_{a, b} (ϕ_{a^{'}, b^{'}} (r)) = ϕ_{a, b} (r^{a^{'}}) = r^{a a^{'}} & and & ϕ_{a, b} (ϕ_{a^{'}, b^{'}} (s)) = ϕ_{a, b} (r^{b^{'}} s) = r^{a b^{'} + b} s . \end{array}

□

Definition 4.3.6. §

An automorphism of $G$ is called an inner automorphism (or inner) if it is equal to a conjugation map $γ_{a} : G \to G$ for some $a \in G$ .

Lemma 4.3.7. §

The set $Inn (G)$ of inner automorphisms of $G$ is a subgroup of $Aut (G)$ under composition.

Proof.

That the inner automorphisms form a subgroup amounts to the facts that $γ_{e} = {id}_{G}$ , that $γ_{𝑎𝑏} = γ_{a} γ_{b}$ , and that $γ_{a^{- 1}} = γ_{a}^{- 1}$ for $a \in G$ . □

Definition 4.3.8. §

The inner automorphism group $Inn (G)$ is the subgroup of $G$ consisting of inner automorphisms.

Lemma 4.3.9. §

For a group $G$ , the inner automorphism group $Inn (G)$ is a normal subgroup of $G$ .

Proof.

For $ϕ \in Aut (G)$ and $g, x \in G$ , we have

(ϕ \circ γ_{g} \circ ϕ^{- 1}) (x) = ϕ (g ϕ^{- 1} (x) g^{- 1}) = ϕ (g) 𝑥𝜙 {(g)}^{- 1} = γ_{ϕ (g)} (x),

so $ϕ γ_{g} ϕ^{- 1} = γ_{ϕ (g)}$ lies in $Inn (G)$ . □

Definition 4.3.10. §

The outer automorphism group of $G$ is the quotient group $Out (G) = Aut (G) ∕ Inn (G)$ .

Remark 4.3.11. §

An automorphism is sometimes called outer if it is not inner. However, the outer automorphism group is not a group of automorphisms, but rather cosets thereof.

Remark 4.3.12. §

If $G$ is an abelian group, then every inner automorphism of $G$ is trivial, so $Out (G) ≅ Aut (G)$ .

Example 4.3.13. §

The group $Inn (D_{n})$ for $n \geq 3$ is generated by the images $γ_{r}$ and $γ_{s}$ of $r$ and $s$ under $γ : D_{n} \to Aut (D_{n})$ . We have $γ_{r} (s) = r^{2} s$ and $γ_{s} (r) = r^{- 1}$ , and of course $γ_{r} (r) = r$ and $γ_{s} (s) = s$ . Using the isomorphism of Proposition 4.3.5, we that $Inn (G)$ is isomorphic to the subgroup of ${GL}_{2} (ℤ ∕ 𝑛ℤ)$ given by

K = {(\begin{matrix} a & b \\ 0 & 1 \end{matrix}) | a = \pm 1, b \in 2 ℤ ∕ 𝑛ℤ} .

The quotient group $Out (D_{n})$ is then in bijection with pairs $(i, j) \in {(ℤ ∕ 𝑛ℤ)}^{\times} ∕ ⟨ - 1 ⟩ \times ℤ ∕ (n, 2) ℤ$ . We leave it to the reader to check that

Out (D_{n}) ≅ {\begin{matrix} {(ℤ ∕ 𝑛ℤ)}^{\times} ∕ ⟨ - 1 ⟩ & if n is odd, \\ {(ℤ ∕ 𝑛ℤ)}^{\times} ∕ ⟨ - 1 ⟩ \times ℤ ∕ 2 ℤ & if n is even \end{matrix}

using the fact that $(\begin{matrix} 1 & n ∕ 2 \\ 0 & 1 \end{matrix})$ is in the center of the group $H$ of Proposition 4.3.5 if $n$ is even.

Definition 4.3.14. §

A subgroup $N$ of a group $G$ is characteristic if $ϕ (N) = N$ for every $ϕ \in Aut (G)$ .

Lemma 4.3.15. §

Let $G$ be a group.

a.: If $H$ is the unique subgroup of $G$ of a given order, then $H$ is characteristic.
b.: The center $Z (G)$ of a group $G$ is characteristic.
c.: The commutator subgroup $[G, G]$ of a group $G$ is characteristic.

Proof.

Let $ϕ \in Aut (G)$ . For part a, note that $ϕ (H)$ has the same order as $H$ . For part $b$ , note that $ϕ (a) ϕ (x) = ϕ (x) ϕ (a)$ for any $a \in Z (G)$ and $x \in G$ , but $ϕ$ is onto, so $ϕ (a)$ commutes with every element of $G$ . So, $ϕ : Z (G) \to Z (G)$ , and $ϕ^{- 1} \in Aut (G)$ has the same property, so $ϕ (Z (G)) = Z (G)$ . For part c, note that $[G, G]$ is generated by commutators $[a, b]$ with $a, b \in G$ , and $ϕ ([a, b]) = [ϕ (a), ϕ (b)] \in [G, G]$ . We can see that $ϕ ([G, G])$ actually equals $[G, G]$ by noting that $ϕ$ is onto. □

Lemma 4.3.16. §

If $K$ is a characteristic subgroup of a normal subgroup $N$ of a group $G$ , then $K ⊴ G$ . If, moreover, $N$ is characteristic in $G$ , then $K$ is characteristic in $G$ .

Proof.

Let $a \in G$ . Then the restriction of $γ_{a}$ to $N$ provides an element of $Aut (N)$ as $N$ is normal, and so $𝑎𝐾 a^{- 1} = γ_{a} (K) = K$ as $K$ is characteristic. Thus $K ⊴ G$ .

If $N$ is characteristic in $G$ and $ϕ \in Aut (G)$ , then the restriction of $ϕ$ to $N$ is an automorphism of $N$ as $N$ is characteristic in $G$ , and so $ϕ (K) = K$ as $K$ is characteristic in $N$ . Thus $K$ is characteristic in $G$ . □

4.4. Free abelian groups

The theory of free abelian groups is the analogue of the theory of vector spaces when the scalars are taken to be not real or complex numbers, but rather integers. In this section, we briefly explore this theory.

Definition 4.4.1. §

An abelian group $G$ (under addition) is said to be a free abelian group if it has a generating set $X$ of $G$ such that for any $n \geq 1$ , distinct $x_{1}, x_{2}, \dots, x_{n} \in X$ , and $c_{1}, c_{2}, \dots, c_{n} \in ℤ$ with

\sum_{i = 1}^{n} c_{i} x_{i} = 0,

one has $c_{1} = c_{2} = \dots = c_{n} = 0$ . Such a set $X$ is called a basis of $G$ , and $G$ is said to be free on $X$ .

Example 4.4.2. §

The group $ℤ^{n}$ is free on the set ${e_{1}, e_{2}, \dots, e_{n}}$ , where $e_{i} \in ℤ^{n}$ is the tuple that is $0$ in every coordinate but the $i$ th, where it is $1$ .

Example 4.4.3. §

The group $ℤ ∕ 𝑛ℤ$ is not free for $n \geq 1$ , since one has $𝑛𝑎 = 0$ for every $a \in ℤ ∕ 𝑛ℤ$ .

Remark 4.4.4. §

Much as in linear algebra, freeness of an abelian group $G$ on a set $X$ implies that there is a unique way to represent any nonzero element $a \in G$ as a sum

a = \sum_{i = 1}^{n} c_{i} x_{i}

for some $n \geq 1$ , distinct elements $x_{1}, x_{2}, \dots, x_{n}$ of $X$ , and nonzero elements $c_{1}, c_{2}, \dots, c_{n}$ of $ℤ$ .

Definition 4.4.5. §

If $x_{1}, x_{2}, \dots, x_{n} \in G$ , where $G$ is a free abelian group, then we refer to a sum

\sum_{i = 1}^{n} c_{i} x_{i}

with $c_{1}, c_{2}, \dots, c_{n} \in ℤ$ as an integral linear combination of elements of $G$ .

Let us begin with a very general construction of a direct sum of groups, which we will then specialize immediately to the case of interest that the groups are all $ℤ$ .

Definition 4.4.6. §

Let $I$ be an indexing set and ${G_{i} ∣ i \in I}$ a collection of abelian groups. Let

⨁_{i \in I} G_{i} = {{(a_{i})}_{i \in I} \in \prod_{i \in I} G_{i} ∣ a_{i} \in G_{i}, a_{i} = 0 for all but finitely many i \in I} .

Then $\oplus_{i \in I} G_{i}$ is a subgroup of $\prod_{i \in I} G_{i}$ known as the direct sum of the groups $G_{i}$ .

Remark 4.4.7. §

When $I$ is finite, we have $\oplus_{i \in I} G_{i} = \prod_{i \in I} G_{i}$ .

Notation 4.4.8. §

The symbol $δ_{i, j}$ (or $δ_{𝑖𝑗}$ ), for $i$ and $j$ in some set $I$ , is taken to mean

δ_{𝑖𝑗} = {\begin{matrix} 1 & if i = j \\ 0 & if i \neq j . \end{matrix}

Example 4.4.9. §

For any indexing set $I$ , the direct sum

⨁_{i \in I} ℤ = {{(a_{i})}_{i \in I} \in \prod_{i \in I} ℤ ∣ a_{i} \in ℤ, a_{i} = 0 for all but finitely many i \in I} .

is a free group under coordinate-wise addition with basis ${e_{j} = {(δ_{𝑖𝑗})}_{i \in I} ∣ j \in I}$ .

Definition 4.4.10. §

The basis ${e_{i} ∣ i \in I}$ in Example 4.4.9 is known as the standard basis of $\oplus_{i \in I} ℤ$ .

Free groups have the property that homomorphisms are defined uniquely by their values on a basis, as we now show.

Proposition 4.4.11. §

Let $G$ be an abelian group. Then $G$ is free on a subset $X$ if and only if, for every function $\bar{ϕ} : X \to G^{'}$ , where $G^{'}$ is an abelian group, there exists a unique homomorphism $ϕ : G \to G^{'}$ with $ϕ (x) = \bar{ϕ} (x)$ for all $x \in X$ .

Proof.

Suppose first that $G$ is free on a basis $X$ . Then for $c_{1}, c_{2}, \dots, c_{n} \in ℤ$ and distinct elements $x_{1}, x_{2}, \dots, x_{n} \in X$ , define

ϕ (\sum_{i = 1}^{n} c_{i} x_{i}) = \sum_{i = 1}^{n} c_{i} \bar{ϕ} (x_{i}) .

The map $ϕ$ is then a well-defined map on all of $G$ by Remark 4.4.4, and it is easy to check that it is a homomorphism. Moreover, if $ψ : G \to G^{'}$ is any homomorphism with $ψ (x) = \bar{ϕ} (x)$ for all $x \in X$ , then

ψ (\sum_{i = 1}^{n} c_{i} x_{i}) = \sum_{i = 1}^{n} c_{i} ψ (x_{i}) = \sum_{i = 1}^{n} c_{i} \bar{ϕ} (x_{i}) = ϕ (\sum_{i = 1}^{n} c_{i} x_{i}) .

Conversely, suppose that $G$ and $X$ have the property of the proposition. We claim that $G$ is free on $X$ . First, suppose that $x_{i} \in X$ and $c_{i} \in ℤ$ for $1 \leq i \leq n$ and some $n \geq 1$ are such that

a = \sum_{i = 1}^{n} c_{i} x_{i} = 0 .

Define $\bar{ϕ} : X \to ℤ^{n}$ by $\bar{ϕ} (x_{i}) = e_{i}$ . Then

0 = \bar{ϕ} (a) = \sum_{i = 1}^{n} c_{i} e_{i},

which forces $c_{i} = 0$ for all $i$ , as the $e_{i}$ form a basis of $ℤ^{n}$ .

Next, let $H$ be the subgroup of $G$ generated by $X$ . We define two homomorphisms $G \to G$ . One is given by ${id}_{G}$ , while is the composition of the map $π : G \to H$ uniquely determined by $π (x) = x$ for all $x \in X$ with the inclusion map $ι : H \to G$ . By assumption, then, we must have ${id}_{G} = ι \circ π$ , and as the latter map has image $H$ , we have $G = H$ . Thus, $G$ is free on $X$ . □

Remark 4.4.12. §

The existence of unique homomorphisms of a free abelian group $G$ with prescibed values on a basis $X$ , as found in Proposition 4.4.11, is often referred to as the universal property of $G$ .

Corollary 4.4.13. §

Suppose that $G$ is a free abelian group on a basis $X = {x_{i} ∣ i \in I}$ , where $I$ is an indexing set. Then there is a unique isomorphism

ψ : ⨁_{i \in I} ℤ \to_{}^{\sim} G

such that $ψ (e_{i}) = x_{i}$ for all $i \in I$ , where ${e_{i} ∣ i \in I}$ is the standard basis of $G$ .

Proof.

We can define $ψ$ as in the statement of the corollary by Proposition 4.4.11, which also implies the existence of a unique homomorphism $ϕ : G \to \oplus_{i \in I} ℤ$ such that $ϕ (x_{i}) = e_{i}$ for all $i \in I$ . Since $ϕ \circ ψ (e_{i}) = e_{i}$ and $ψ \circ ϕ (x_{i}) = x_{i}$ for all $i \in I$ , the same proposition implies that $ϕ \circ ψ$ and $ψ \circ ϕ$ are the identity homomorphisms. In particular, $ψ$ is an isomorphism. □

Definition 4.4.14. §

The general linear group ${GL}_{n} (ℤ)$ of degree $n$ is the group of $n$ -by- $n$ matrices with integer entries which have inverses with integer entries, with respect to the operation of matrix multiplication.

Remark 4.4.15. §

A $n$ -by- $n$ matrix $A$ with integer entries has an inverse with integer entries if and only if $\det (A) = \pm 1$ .

For two free abelian groups to be isomorphic, their bases must have the same cardinality. Equivalently, an abelian group cannot have bases of two different cardinalities. We prove this only in the special case of finitely generated abelian groups.

Theorem 4.4.16. §

Suppose that $G$ is a free abelian group with basis $X$ having $n$ elements. Then every basis of $G$ has $n$ elements.

Proof.

By Corollary 4.4.13, we have that $𝐺≅ \prod_{i = 1}^{n} ℤ$ . We then have that

G ∕ 2 𝐺≅ \prod_{i = 1}^{n} ℤ ∕ 2 ℤ,

and so has order $2^{n}$ . If $G$ had a different basis with a finite number of elements $m$ , then $G ∕ 2 G$ would have order $2^{m}$ , forcing $m = n$ . On the other hand, if $G$ had an infinite basis indexed by a set $I$ , then the same argument would tell us that

G ∕ 2 𝐺≅ ⨁_{i \in I} ℤ ∕ 2 ℤ,

which is infinite, so impossible. □

Definition 4.4.17. §

If $G$ is a finitely generated, free abelian group, we refer to the number of elements in any basis of it as its rank.

We have the following analogue of the change-of-basis theorem in linear algebra.

Proposition 4.4.18. §

Let $G$ be a free abelian group with basis $X = {x_{1}, x_{2}, \dots, x_{n}}$ . Then $X^{'} = {x_{1}^{'}, x_{2}^{'}, \dots, x_{n}^{'}}$ is also a basis of $G$ if and only if there exists a matrix $A = (a_{𝑖𝑗}) \in {GL}_{n} (ℤ)$ such that

x_{i}^{'} = \sum_{j = 1}^{n} a_{𝑖𝑗} x_{j}

for each $1 \leq i \leq n$ .

Proof.

Since $X$ generates $G$ , we may write each $x_{i}^{'}$ as

x_{i}^{'} = \sum_{j = 1}^{n} a_{𝑖𝑗} x_{j}

for some $a_{𝑖𝑗} \in ℤ$ and then form an $n$ -by- $n$ matrix $A = (a_{𝑖𝑗})$ . If $X^{'}$ also generates $G$ , then we may write

x_{i} = \sum_{j = 1}^{n} b_{𝑖𝑗} x_{j}^{'} = \sum_{j = 1}^{n} \sum_{k = 1}^{n} b_{𝑖𝑗} a_{𝑗𝑘} x_{k}

for some $b_{𝑖𝑗} \in ℤ$ and then form $B = (b_{𝑖𝑗})$ . Since $X$ is a basis, this tells us that $𝐵𝐴 = I$ , so $A \in {GL}_{n} (ℤ)$ .

Conversely, if there exists a $B$ with $𝐵𝐴 = I$ , then

x_{i} = \sum_{k = 1}^{n} (\sum_{j = 1}^{n} b_{𝑖𝑗} a_{𝑗𝑘}) x_{k} = \sum_{j = 1}^{n} b_{𝑖𝑗} x_{j}^{'},

so the $x_{j}^{'}$ generate $G$ , and moreover they form a basis as, if

\sum_{i = 1}^{n} c_{i} x_{i}^{'} = 0,

we then have

\sum_{i = 1}^{n} \sum_{j = 1}^{n} a_{𝑖𝑗} c_{i} x_{j} = 0,

\sum_{i = 1}^{n} a_{𝑖𝑗} c_{i} = 0

for each $i$ , or in other words the vector $c = (c_{1}, c_{2}, \dots, c_{n})$ satisfies $𝐴𝑐 = 0$ , which means $𝐵𝐴𝑐 = 0$ , or $c = 0$ . □

4.5. Finitely generated abelian groups

We begin with the following corollary of Theorem 4.4.18.

Lemma 4.5.1. §

Suppose that $X = {x_{1}, x_{2}, \dots, x_{n}}$ is a basis of a free abelian group $G$ , and let $c_{i} \in ℤ$ for $2 \leq i \leq n$ . Then $X^{'} = {x_{1}^{'}, x_{2}, \dots, x_{n}}$ with

x_{1}^{'} = x_{1} + c_{2} x_{2} + \dots + c_{n} x_{n}

is also a basis of $G$ .

Proof.

Take $A \in {GL}_{n} (ℤ)$ to be $A = I + \sum_{k = 2}^{n} c_{j} E_{1 j}$ , where $E_{𝑖𝑗}$ is the $n$ -by- $n$ matrix with exactly one nonzero entry, which is a $1$ in the $i$ th row and $j$ th column. It is easy to see that $\det (A) = 1$ , so $A \in {GL}_{n} (ℤ)$ . We then apply Proposition 4.4.18. □

We are now ready to prove the following result.

Lemma 4.5.2. §

Let $G$ be a finitely generated, free abelian group of rank $n$ , and let $H$ be a nontrivial subgroup. Then there is an isomorphism

φ : G \to ℤ \times G^{'},

where $G^{'}$ is a subgroup of $G$ that is free abelian of rank $n - 1$ , such that

φ (H) = 𝑑ℤ \times H^{'},

for some $d \geq 1$ , where $H^{'} = H \cap G^{'}$ .

Proof.

Consider the set $B$ of all bases of $G$ . Let $d \geq 1$ be minimal such that there exists $X^{'} = {x^{'}, x_{2}, \dots, x_{n}} \in B$ such that

y = d x^{'} + \sum_{i = 2}^{n} d_{j} x_{j} \in H

for some $d_{2}, \dots, d_{n} \in ℤ$ , and fix such an $X^{'}$ and $y$ . We may divide each $d_{i}$ for $2 \leq i \leq n$ by $d$ to obtain

d_{i} = q_{i} d + r_{i}

with $q_{i} \in ℤ$ and $0 \leq r_{i} < n$ . Then

y = d (x^{'} + q_{2} x_{2} + \dots + q_{n} x_{n}) + r_{2} x_{2} + \dots + r_{n} x_{n} .

Let $x_{1} = x^{'} + q_{2} x_{2} + \dots + q_{n} x_{n}$ . Then $X = {x_{1}, x_{2}, \dots, x_{n}}$ is a basis by Lemma 4.5.1. The minimality of $d$ now forces $r_{2} = \dots = r_{n} = 0$ . In other words, we have $y = d x_{1} \in H$ .

Let $G^{'} = ⟨ x_{2}, \dots, x_{n} ⟩$ , which is free abelian of rank $n - 1$ , and define a homomorphism

φ : G \to ℤ \times G^{'}, φ (a) = (c_{1}, a - c_{1} x_{1}),

for $a \in G$ , where $c_{1} \in ℤ$ is such that

a = \sum_{i = 1}^{n} c_{i} x_{i}

for some $c_{2}, \dots, c_{n} \in ℤ$ . We have that $φ (a) = 0$ if and only if $c_{1} = 0$ and $a - c_{1} x_{1} = 0$ by definition, which occurs exactly when $a = 0$ as $X$ is a basis. Therefore, $φ$ is injective. Moreover, for $a^{'} \in G^{'}$ and $c \in ℤ$ , we clearly have that

φ (c x_{1} + a^{'}) = (c, a^{'}),

so $φ$ is surjective. Therefore, $φ$ is an isomorphism.

Finally, we compute $φ (H)$ . Suppose $a \in H$ is written as above, and let $c_{1} = 𝑞𝑑 + r$ with $q \in ℤ$ and $0 \leq r < d$ . Then $b = a - 𝑞𝑑 x_{1} \in H$ , and

b = r x_{1} + \sum_{i = 2}^{n} c_{i} x_{i} .

By the minimality of $d$ , we must have $r = 0$ . In other words, we have $b \in H^{'}$ and

a = q (d x_{1}) + b,

so $φ (a) \in 𝑑ℤ \times H^{'}$ . Conversely, if $(m, b) \in 𝑑ℤ \times H^{'}$ , then

(m, b) = φ (m x_{1} + b),

and $m x_{1} \in H$ since $d$ divides $m$ , so $m x_{1} + b \in H$ . Therefore, $φ (H) = 𝑑ℤ \times H^{'}$ , as desired. □

We also note the following easy corollary of Theorem 2.5.17, obtained by applying it recursively.

Corollary 4.5.3. §

Let $m$ be a positive integer, and for some $k \geq 0$ , write

m = p_{1}^{r_{1}} p_{2}^{r_{2}} \dots p_{k}^{r_{k}}

for distinct prime numbers $p_{1}, p_{2}, \dots, p_{k}$ and $r_{1}, r_{2}, \dots, r_{k} \geq 2$ . Then

ℤ ∕ 𝑚ℤ≅ℤ ∕ p_{1}^{r_{1}} ℤ \times ℤ ∕ p_{2}^{r_{2}} ℤ \times \dots \times ℤ ∕ p_{k}^{r_{k}} ℤ .

We can now classify the finitely generated abelian groups up to isomorphism.

Theorem 4.5.4 (Structure theorem for finitely generated abelian groups). §

Let $G$ be a finitely generated abelian group. Then there exist $k, r \geq 0$ and positive integers $d_{1}, d_{2}, \dots, d_{k} \geq 2$ such that there is an isomorphism

𝐺≅ ℤ^{r} \times (ℤ ∕ d_{1} ℤ) \times \dots \times (ℤ ∕ d_{k} ℤ) .

In fact, the integers $d_{i}$ may be chosen so that $d_{i + 1}$ divides $d_{i}$ for each $1 \leq i \leq k - 1$ , and then these are the unique $r, k$ , and $d_{1}, d_{2}, \dots, d_{k}$ with those properties. Alternatively, we may choose the isomorphism so that each $d_{i}$ is a power of a prime number, in which case the decomposition is again unique up to reordering.

Proof.

We prove the result by induction on the number of elements $n$ in a finite generating set of $G$ , where we may consider the trivial group to be generated by the empty set. The case $n = 0$ is then just the case that $r = k = 0$ , and we have the result. Suppose we know the result for all abelian groups that can be generated by $n$ elements. Let $G$ be an abelian group for which $X = {x_{1}, x_{2}, \dots, x_{n + 1}}$ is a minimal set of generators. Then there exists a unique surjective homomorphism $ψ : ℤ^{n + 1} \to G$ such that $ψ (e_{i}) = x_{i}$ for $1 \leq i \leq n + 1$ . Let $H = \ker ψ \leq ℤ^{n + 1}$ .

By Lemma 4.5.2, we have an isomorphism

φ : ℤ^{n + 1} \to ℤ \times ℤ^{n}

such that $φ (H) = 𝑑ℤ \times H^{'}$ for some $H^{'} \leq ℤ^{n}$ and $d \geq 1$ . By the first isomorphism theorem, we have

𝐺≅ ℤ^{n + 1} ∕ 𝐻≅ \frac{ℤ \times ℤ^{n}}{𝑑ℤ \times H^{'}} ≅ℤ ∕ 𝑑ℤ \times ℤ^{n} ∕ H^{'} .

Now, since $ℤ^{n} ∕ H^{'}$ can be generated by $n$ elements, it may be written by induction as

ℤ^{n} ∕ H^{'} ≅ ℤ^{r} \times (ℤ ∕ d_{1} ℤ) \times \dots \times (ℤ ∕ d_{k - 1} ℤ)

for some $d_{1}, d_{2}, \dots, d_{k - 1} \geq 2$ with $r + k - 1 \leq n$ . Letting $d_{k} = d$ if $d \geq 2$ and noting that $𝐺≅ ℤ^{n} ∕ H^{'}$ if $d = 1$ , we therefore have the first statement of the theorem.

By Corollary 4.5.3, we may can decompose each $ℤ ∕ d_{i} ℤ$ into a finite direct product of groups of the form $ℤ ∕ p^{k} ℤ$ with $p$ prime and $k \geq 2$ , proving the last decomposition. On the other hand, suppose we have decomposed $G$ up to isomorphism as

𝐺≅ ℤ^{r} \times P_{1} \times P_{2} \dots \times P_{t},

where $p_{1}, p_{2}, \dots, p_{t}$ are distinct prime numbers and each $P_{i}$ for $1 \leq i \leq t$ is a finite abelian $p_{i}$ -group, which in turn we have written as

P_{i} ≅ℤ ∕ p_{i}^{m_{i 1}} ℤ \times ℤ ∕ p_{i}^{m_{i 2}} ℤ \times \dots ℤ ∕ p_{i}^{m_{i s_{i}}} ℤ

for some $s_{i} \geq 1$ and $m_{i 1} \geq m_{i 2} \geq \dots \geq m_{i s_{i}} \geq 1$ . Let $k = \max {s_{i} ∣ 1 \leq i \leq t}$ , set $m_{𝑖𝑗} = 0$ if $j > s_{i}$ , and let

d_{j} = \prod_{i = 1}^{t} p_{i}^{m_{𝑖𝑗}}

for each $1 \leq j \leq k$ . Then $d_{1}, d_{2}, \dots, d_{k} \geq 2$ , and $d_{i + 1}$ divides $d_{i}$ for each $1 \leq i \leq k - 1$ , as desired. Moreover, Theorem 2.5.17 implies that

ℤ ∕ d_{j} ℤ≅ (ℤ ∕ p_{1}^{m_{1 j}} ℤ) \times (ℤ ∕ p_{2}^{m_{2 j}} ℤ) \times \dots \times (ℤ ∕ p_{t}^{m_{𝑡𝑗}} ℤ),

which yields the desired decomposition of $G$ by gathering terms and applying these isomorphisms.

Finally, we address uniqueness of the latter two decompositions. First, we claim that in any decomposition of $G$ (without restriction on the $d_{i}$ ), we must have the same $r$ . For this, let

G_{tor} = {a \in G ∣ 𝑛𝑎 = 0 for some n \geq 1} .

If we have written

G = ℤ^{r} \times (ℤ ∕ d_{1} ℤ) \times (ℤ ∕ d_{2} ℤ) \times \dots \times (ℤ ∕ d_{k} ℤ)

for some $r, k \geq 0$ and $d_{1}, d_{2}, \dots, d_{k} \geq 2$ , then

G_{tor} = {0} \times (ℤ ∕ d_{1} ℤ) \times (ℤ ∕ d_{2} ℤ) \times \dots \times (ℤ ∕ d_{k} ℤ),

and $G ∕ G_{tor} ≅ ℤ^{r}$ . But $ℤ^{r} ≇ ℤ^{s}$ for $s \neq r$ , so the $r$ in the decomposition must be unique. Moreover, if $G_{tor} = {0}$ , which is to say that $| G_{tor} | = 1$ , then uniqueness of the decomposition is simply that $r$ is unique such that $𝐺≅ ℤ^{r}$ , which we have just proven.

Now, suppose $k \geq 1$ and we have chosen the $d_{i}$ either to be prime powers, which we list in descending order, or such that $d_{i + 1}$ divides $d_{i}$ for all $1 \leq i \leq k - 1$ (so also in descending order). In the former case, $d_{1}$ is the largest order of any element of prime power order in $G$ , and in the latter, $d_{1}$ is the exponent of $G$ . Therefore, if we have a second decomposition,

G = ℤ^{r} \times (ℤ ∕ d_{1}^{'} ℤ) \times (ℤ ∕ d_{2}^{'} ℤ) \times \dots (ℤ ∕ d_{k^{'}}^{'} ℤ),

written in the same form as the first, then we must have $d_{1} = d_{1}^{'}$ . Take the quotient, therefore, by the subgroup $ℤ^{r} \times ℤ ∕ d_{1} ℤ \times {0} \times {0}$ in each decomposition. Then we have

ℤ ∕ d_{2} ℤ \times \dots \times ℤ ∕ d_{k} ℤ≅ℤ ∕ d_{2}^{'} ℤ \times \dots \times ℤ ∕ d_{k^{'}}^{'} ℤ .

By induction on the order of $G_{tor}$ , assuming that we have proven the uniqueness for all $G^{'}$ with $| G_{tor}^{'} | < | G_{tor} |$ , we must have $k = k^{'}$ and $d_{i} = d_{i}^{'}$ for all $2 \leq i \leq k$ , proving uniqueness. □

In the course of the proof of , we considered a subgroup of an abelian group $G$ that we denote $G_{tor}$ . We give it a formal definition.

Definition 4.5.5. §

Let $G$ be an abelian group.

a.

The torsion subgroup of $G$ is the subgroup

G_{tor} = {a \in G ∣ 𝑛𝑎 = 0 for some n \geq 1}

of $G$ .

b.

An element of $G_{tor}$ is called a torsion element of $G$ .

Remark 4.5.6. §

If $G$ is a finite abelian group, then $G = G_{tor}$ .

We leave the proof of the following direct corollary of Theorem 4.5.4 to the reader.

Corollary 4.5.7. §

If $G$ is a finitely generated abelian group, then $𝐺≅ ℤ^{r} \times G_{tor}$ for some $r \geq 0$ . Moreover, $G_{tor}$ is a finite abelian group, and it is isomorphic to a direct product of cyclic groups.

4.6. Group actions on sets

Definition 4.6.1. §

An action of a group $G$ on a set $X$ is an operation

⋆ : G \times X \to X

satisfying the following properties

i.: $e ⋆ x = x$ for all $x \in X$ ,
ii.: $a ⋆ (b ⋆ x) = (𝑎𝑏) ⋆ x$ for all $a, b \in G$ and $x \in X$ .

We then say that $G$ acts on $X$ and that the operation $⋆$ is a $G$ -action, and we refer to $X$ as a $G$ -set.

Remark 4.6.2. §

As with binary operations, we typically denote “ $⋆$ ” more simply by “ $\cdot$ ”.

Examples 4.6.3. §

a.: The symmetric group $S_{X}$ acts on $X$ . In particular, $S_{n}$ acts on $X_{n} = {1, 2, \dots, n}$ .
b.: The group of isometries of $ℝ^{n}$ acts on $ℝ^{n}$ .
c.: The wallpaper group of a tiling of the plane acts on $ℝ^{2}$ .
d.: For $n \geq 3$ , the group $D_{n}$ acts on the set of vertices of the regular $n$ -gon of which it is the symmetry group, as well as the set of its edges.
e.: The group ${GL}_{n} (ℝ)$ acts on $ℝ^{n}$ by left multiplication of column vectors.

Here are a couple of more abstract examples.

Examples 4.6.4. §

a.: A group $G$ acts on itself by left multiplication: $a ⋆ x = 𝑎𝑥$ for $a, x \in G$ .
b.: A group $G$ acts on itself by conjugation: $a ⋆ x = 𝑎𝑥 a^{- 1}$ for $a, x \in G$ .

Remark 4.6.5. §

A group $G$ does not act on itself by right multiplication. If we defined $a ⋆ x = 𝑥𝑎$ , then

a ⋆ (b ⋆ x) = a ⋆ (𝑥𝑏) = (𝑥𝑏) a = x (𝑏𝑎),

while $(𝑎𝑏) ⋆ x = x (𝑎𝑏)$ . The action by right multiplication is an example of what is known as a right action (as opposed to a usual, or left, action).

Definition 4.6.6. §

We say that an action of a group $G$ on a set $X$ is transitive if for every $x, y \in X$ , there exists $a \in G$ with $𝑎𝑥 = y$ . We then say that $G$ acts transitively on $X$ .

Examples 4.6.7. §

a.

The group $S_{X}$ acts transitively on $X$ .

b.

The group $D_{n}$ acts transitively on the set of vertices of a regular $n$ -gon, as well as the set of edges.

c.

The group of isometries of $ℝ^{n}$ acts transitively on $ℝ^{n}$ .

d.

The group ${GL}_{n} (ℝ)$ does not act transitively on $ℝ^{n}$ , as an invertible matrix times a nonzero vector is always nonzero.

e.

Any group $G$ acts transitively on itself by left multiplication. This is simply the cancellation theorem: if $x, y \in G$ , then $a = y x^{- 1}$ satisfies $𝑎𝑥 = y$ .

f.

The action of $G$ on itself by conjugation is not transitive if $G$ is nontrivial. For example, the identity element is not a conjugate of any other element.

g.

The group $S_{X}$ acts on the power set of $X$ ,

σ ⋆ Y = {σ (y) ∣ y \in Y}

for $Y \subseteq X$ , but this action is not transitive if $X$ is nonempty. For instance, $σ ⋆ Y$ always has the same cardinality as $Y$ , so it cannot be the empty set if $Y$ is nonempty.

h.

The group $G$ acts on the set $G ∕ H$ of left cosets a subgroup $H$ of $G$ by left multiplication:

a \cdot 𝑏𝐻 = 𝑎𝑏𝐻,

and this is a transitive action.

Definition 4.6.8. §

Let $G$ be a group and $X$ be a $G$ -set. The orbit of $x \in X$ is the set

G \cdot x = {𝑔𝑥 ∣ g \in G} .

Remark 4.6.9. §

Recall that the orbit of $x \in X_{n}$ under $σ \in S_{n}$ was defined as

O_{σ} (x) = {σ^{i} x ∣ i \in ℤ},

and we can reinterpret this orbit as the orbit $⟨ σ ⟩ \cdot x$ .

The following lemma is nearly immediate.

Lemma 4.6.10. §

An action of a group $G$ on a set $X$ is transitive if and only if $G \cdot x = X$ for every (equivalently, some) $x \in X$ .

Examples 4.6.11. §

a.: Since $S_{X}$ acts transitively on $X$ , we have that $S_{X} \cdot x = X$ for every $x \in X$ .
b.: Consider the action of $S_{n}$ on the power set of $X_{n}$ . We have
$S_{n} \cdot {1, 2, \dots, k} = {Y \subset X_{n} ∣ | Y | = k} .$
c.: The orbit of $v \in ℝ^{n}$ under ${GL}_{n} (ℝ)$ is $ℝ^{n} - {0}$ if $v \neq 0$ and ${0}$ if $v = 0$ .
d.: The orbit of $x \in G$ under the action of $G$ on itself by conjugation is the conjugacy class $C_{x}$ of $x$ .
e.: The orbit of $H \leq G$ under the action of $G$ of its set of subgroups by conjugation is the set of all conjugate subgroups to $G$ :
${𝑎𝐻 a^{- 1} ∣ a \in G} .$

We remark that the property of being in the same orbit is an equivalence relation on a $G$ -set $X$ , and therefore we obtain a partition of $X$ as a disjoint union of its orbits.

Proposition 4.6.12. §

Let $X$ be a $G$ -set. The relation $x \sim_{G} y$ if and only if $G \cdot x = G \cdot y$ for $x, y \in X$ is an equivalence relation on $G$ , and the equivalence class of $x \in X$ under $\sim_{G}$ is the orbit $𝐺𝑥$ of $x$ . Therefore, $G$ is the disjoint union of its distinct orbits.

Proof.

That $\sim_{G}$ is an equivalence relation is checked immediately. We remark that for $x, y \in X$ , we have $G \cdot x = G \cdot y$ if and only if $y \in G \cdot x$ , since $𝑎𝑥 = 𝑏𝑦$ for some $a, b \in G$ if and only if $𝑐𝑥 = y$ for some $c \in G$ (that $c$ being $b^{- 1} a$ ). Therefore, the equivalence class of $x \in X$ is exactly the orbit $G \cdot x$ , and the final statement is just Lemma 1.2.10. □

Definition 4.6.13. §

Let $X$ be a $G$ -set for some group $G$ . Let $a \in G$ and $x \in X$ . We say that $a$ fixes $x$ if $𝑎𝑥 = x$ .

Definition 4.6.14. §

We say that an action of a group $G$ on a set $X$ is faithful if the only element $a \in G$ that fixes all $x \in X$ is the identity element. We then say that $G$ acts faithfully on $X$ , and $X$ is a faithful $G$ -set.

In other words, $G$ acts faithfully on $X$ if $𝑎𝑥 = x$ for all $x \in X$ implies $a = e$ .

Examples 4.6.15. §

a.: The group $S_{X}$ acts faithfully on $X$ , since a nontrivial permutation of $X$ does not fix every element of $X$ .
b.: For $n \geq 3$ , the group $D_{n}$ acts faithfully on the set of vertices of the regular polygon, as well as the set of edges.
c.: The group $G$ acts on itself faithfully by left multiplication, since if $𝑎𝑥 = x$ for any $x \in G$ , then $a = e$ .
d.: The action of a group $G$ on itself by conjugation is faithful if and only if the group has trivial center. To see this, note that
$Z (G) = {a \in G ∣ 𝑎𝑥 a^{- 1} = x for all x \in G} .$

Definition 4.6.16. §

Let $G$ be a group and $X$ be a $G$ -set. The stabilizer, or isotropy subgroup, of $G_{x}$ of an element $x \in X$ is the set of elements of $G$ that fix $x$ . That is, we have

G_{x} = {a \in G ∣ 𝑎𝑥 = x} .

Remark 4.6.17. §

The stabilizer $G_{x}$ is indeed a subgroup of $G$ , since $e \in G_{x}$ , and for $a, b \in G_{x}$ , we have $(𝑎𝑏) x = a (𝑏𝑥) = 𝑎𝑥 = x$ , so $𝑎𝑏 \in G_{x}$ , while

a^{- 1} x = a^{- 1} (𝑎𝑥) = 𝑒𝑥 = x,

so $a^{- 1} \in G_{x}$ .

Lemma 4.6.18. §

A group $G$ acts faithfully on a set $X$ if and only if

⋂_{x \in X} G_{x} = {e} .

Proof.

We have $a \in G_{x}$ if and only if $𝑎𝑥 = x$ . Thus $a \in G_{x}$ for all $x \in X$ if and only if $𝑎𝑥 = x$ for all $x \in X$ , and the action of $G$ on $X$ is not faithful if and only if the latter occurs for some $a \in G$ with $a \neq e$ . So, $⋂_{x \in X} G_{x}$ contains a non-identity element if and only if $G$ acts non-faithfully on $X$ . □

Examples 4.6.19. §

We give some examples of stabilizers.

a.: The stabilizer of $n$ under the action of $S_{n}$ is the image of $S_{n - 1}$ under the homomorphism $ι : S_{n - 1} \to S_{n}$ of Example 2.10.20.
b.: The stabilizer of a vertex under the action of $D_{n}$ on a regular $n$ -gon consists exactly of the subgroup of order $2$ generated by the unique reflection in $D_{n}$ for which the line of reflection passes through the vertex.
c.: The stabilizer of $4$ under the action of $⟨ σ ⟩ \leq S_{5}$ , where $σ = (1 2 3) (4 5)$ , is $⟨ σ^{2} ⟩$ .
d.: The stabilizer of $x \in G$ under the action of $G$ on itself by left multiplication is trivial:
$G_{x} = {a \in G ∣ 𝑎𝑥 = x} = {e}$

The following definition gives an interesting class of examples of stabilizers.

Definition 4.6.20. §

The stabilizer of $x \in G$ under the action of $G$ on itself by conjugation is the subgroup of elements in $G$ that commute with $x$ and is known as the centralizer $Z_{x}$ of $x$

Z_{x} = {a \in G ∣ 𝑎𝑥 = 𝑥𝑎} .

Example 4.6.21. §

The centralizer $Z_{(1 2 3)}$ in $S_{5}$ is

Z_{(1 2 3)} = ⟨ (1 2 3), (4 5) ⟩ .

We end by comparing orbits and stabilizers.

Theorem 4.6.22. §

Let $X$ be a $G$ -set, and let $x \in X$ . Then there is a bijection

ψ_{x} : G ∕ G_{x} \to G \cdot x,

given by $ψ_{x} (a G_{x}) = 𝑎𝑥$ for any $a \in G$ .

Proof.

First, we note that $ψ_{x}$ is well-defined, since if $b \in a G_{x}$ , then $b = 𝑎𝑔$ for some $g \in G_{x}$ , and

ψ_{x} (b G_{x}) = 𝑏𝑥 = 𝑎𝑔𝑥 = a (𝑔𝑥) = 𝑎𝑥 = ψ_{x} (a G_{x}) .

Moreover, it is one-to-one since, if $𝑎𝑥 = 𝑏𝑥$ , then $x = a^{- 1} 𝑏𝑥$ , so $a^{- 1} b \in G_{x}$ , and therefore $a G_{x} = b G_{x}$ . Finally, it is onto by definition. □

Corollary 4.6.23. §

If $G$ is a finite group, then every element $x \in X$ has a finite orbit, and

| G \cdot x | = [G : G_{x}] .

In particular, the number of elements in the orbit of $x$ divides $| G |$ .

Example 4.6.24. §

The centralizer of $(1 2 3)$ in $S_{5}$ has order $6$ , while the orbit of $(1 2 3)$ is the set of $3$ -cycles in $S_{5}$ , of which there are $20$ , and we note that $| S_{5} | = 20 \cdot 6$ .

We find an application in the class equation.

Proposition 4.6.25 (The class equation). §

Let $G$ be a finite group. Then

| G | = | Z (G) | + \sum_{x \in X} [G : Z_{x}],

where $X$ is set of representatives of the conjugacy classes in $G$ with more than one element.

Proof.

By Corollary 4.6.23, we have that $[G : Z_{x}] = | C_{x} |$ for $x \in X$ . Moreover, $| C_{x} | = 1$ if $x \in Z (G)$ . The equality we wish to prove is therefore reduced to the known fact that $| G |$ is the sum of the orders of its distinct conjugacy classes. □

Here is one application.

Proposition 4.6.26. §

The group $A_{5}$ is simple.

Proof.

Any normal subgroup of a group is a disjoint union of conjugacy classes in that group including the conjugacy class ${e}$ . Let us determine the conjugacy classes in $A_{5}$ . The conjugacy classes in $S_{5}$ of nontrivial elements in $A_{5}$ are the products of $2$ transpositions, the $3$ -cycles, and the $5$ -cycles. The centralizer of $(1 2 3)$ in $A_{5}$ is the group $⟨ (1 2 3) ⟩$ , so $| C_{(1 2 3)} | = \frac{60}{3} = 20$ , and $C_{(1 2 3)}$ is thus the set of $3$ -cycles. The centralizer of $(1 2) (3 4)$ is $⟨ (1 2) (3 4), (1 3) (2 4) ⟩$ , so $| C_{(1 2) (3 4)} | = 15$ , and $C_{(1 2) (3 4)}$ is thus the set of products of two transpositions. The centralizer of $(1 2 3 4 5)$ is $⟨ (1 2 3 4 5) ⟩$ , so $| C_{(1 2 3 4 5)} | = 12$ and so there are two conjugacy classes of $5$ cycles, each with $12$ elements. We then have that the distinct conjugacy classes of elements in $A_{5}$ have $1$ , $12$ , $12$ , $15$ , and $20$ elements. Aside from $1$ and $60$ , no sum of these numbers including $1$ divides $60$ , so $A_{5}$ has no nontrivial, improper normal subgroups. □

4.7. Permutation representations

In this brief section, we give a characterization of group actions on sets as permutation representations.

Theorem 4.7.1. §

Let $X$ be a $G$ -set, and define $σ_{a} : X \to X$ by $σ_{a} (x) = 𝑎𝑥$ for any $a \in G$ . Then $σ_{a} \in S_{X}$ , and the map

ρ : G \to S_{X}

such that $ρ (a) = σ_{a}$ for all $a \in G$ is a homomorphism. Conversely, if $ρ : G \to S_{X}$ is a homomorphism, then the operation defined by

a ⋆ x = ρ (a) (x)

on $a \in G$ and $x \in G$ is a $G$ -action.

Proof.

We have

(σ_{a^{- 1}} \circ σ_{a}) (x) = x = (σ_{a} \circ σ_{a^{- 1}}) (x),

so $σ_{a}$ is a bijection, which is to say $σ_{a} \in S_{X}$ . For $a, b \in G$ and $x \in X$ , we have

ρ (𝑎𝑏) (x) = σ_{𝑎𝑏} (x) = (𝑎𝑏) x = a (𝑏𝑥) = σ_{a} (σ_{b} (x)) = (σ_{a} \circ σ_{b}) (x) = (ρ (a) \circ ρ (b)) (x),

so $ρ$ is a homomorphism.

Conversely, given $ρ$ , we check that $e ⋆ x = ρ (e) (x) = {id}_{X} (x) = x$ , while

a ⋆ (b ⋆ x) = ρ (a) (ρ (b) (x)) = ρ (𝑎𝑏) (x) = (𝑎𝑏) ⋆ x,

so $⋆$ is a $G$ -action. □

Definition 4.7.2. §

If $X$ is a $G$ -set, then the homomorphism $ρ_{X}$ associated to $X$ by Theorem 4.7.1 is called its permutation representation.

Example 4.7.3. §

The action of $S_{X}$ on $X$ gives rise to a permutation representation $ρ : S_{X} \to S_{X}$ satisfying $ρ (σ) (x) = σ (x)$ for all $x \in X$ . In other words, we have $ρ = {id}_{S_{X}}$ .

Example 4.7.4. §

Consider the action of ${GL}_{n} (ℝ)$ on $ℝ^{n}$ . The permutation representation

ρ : {GL}_{n} (ℝ) \to S_{ℝ^{n}}

takes $A \in {GL}_{n} (ℝ)$ to a map $T_{A} : ℝ^{n} \to ℝ^{n}$ that satisfies $T_{A} (v) = 𝐴𝑣$ . In other words, the image of $ρ$ is the subgroup of $S_{ℝ^{n}}$ consisting of invertible linear transformations.

The following lemma is almost immediate.

Lemma 4.7.5. §

A group $G$ acts faithfully on a set $X$ if and only if $ρ_{X}$ is injective. In fact, the kernel of $ρ_{X}$ is the intersection of the stabilizers $G_{x}$ over all $x \in X$ .

Proof.

We have $ρ_{X} (a) = {id}_{X}$ if and only if $𝑎𝑥 = ρ_{X} (a) (x) = x$ for all $x \in X$ . □

Example 4.7.6. §

The permutation representation attached to $G$ acting on itself by conjugation is a homomorphism $γ : G \to S_{G}$ given by $a \mapsto γ_{a}$ , and its image is the inner automorphism group, a subgroup of $Aut (G) \leq S_{G}$ . The kernel of $γ$ is $Z (G)$ .

We now prove Cayley’s theorem, which tells us that every group is a subgroup of a symmetric group.

Theorem 4.7.7 (Cayley). §

Every group $G$ is isomorphic to a subgroup of $S_{G}$ .

Proof.

Consider the permutation representation $ρ_{G} : G \to S_{G}$ associated to the action of $G$ on itself by left multiplication. By Example 4.6.4, $ρ_{G}$ is injective, and therefore, $G$ is isomorphic to $im ρ_{G} \leq S_{G}$ . □

Remark 4.7.8. §

Note that if $G$ has order $n$ , Cayley’s theorem tells us that $G$ is isomorphic to a subgroup of $S_{n}$ . However, this is not always the smallest permutation group in which it is contained. For example, the action of $D_{n}$ on its vertices is faithful, and so a choice of numbering of these vertices identifies $D_{n}$ with a subgroup of $S_{n}$ , as opposed to $S_{2 n}$ . Even more simply, $S_{n}$ is obviously a subgroup of itself, and not just isomorphic to a subgroup of $S_{n!}$ .

4.8. Burnside’s formula

Notation 4.8.1. §

Let $X$ be a $G$ -set, and let $S$ be a subset of $G$ . Then we set

X^{S} = {x \in X ∣ 𝑎𝑥 = x for all a \in S} .

If $S = {a}$ for some $a \in X$ , we sometimes write $X^{a}$ for $X^{{a}}$ .

Remark 4.8.2. §

For $x \in X$ and $a \in G$ , where $X$ is a $G$ -set, the statement that $𝑎𝑥 = x$ is equivalent both to $a \in G_{x}$ and to $x \in X^{a}$ .

Examples 4.8.3. §

a.: Take $σ = (1 3) (2 5) \in S_{6}$ , and let $X = X_{6}$ . We have $X^{σ} = {4, 6}$ .
b.: Let $X$ be the $D_{n}$ -set that is the set of vertices of the regular $n$ -gon inscribed on the unit circle in $ℝ^{2}$ with a vertex at $(1, 0)$ . Then $X^{s} = {(1, 0)}$ if $n$ is odd and $X^{s} = {(- 1, 0), (0, 1)}$ if $n$ is even, while $X^{r} = \emptyset$ .
c.: Let $X = ℝ^{n}$ , and let $A \in {GL}_{n} (ℝ)$ . Then $X^{A} = {v \in ℝ^{n} ∣ 𝐴𝑣 = v}$ is the eigenspace of $A$ with eigenvalue $1$ (so ${0}$ if $1$ is not an eigenvalue). We have $X^{{GL}_{n} (ℝ)} = {0}$ .

We now state Burnside’s formula.

Theorem 4.8.4 (Burnside). §

Let $G$ be a finite group, and let $X$ be a finite $G$ -set. Let $r$ be the number of distinct orbits in $X$ under $G$ . Then

r = \frac{1}{| G |} \sum_{a \in G} | X^{a} | .

Proof.

We will count the set of pairs

S = {(a, x) ∣ a \in G, x \in X, 𝑎𝑥 = x} \subseteq G \times X

in two different ways. First, note that

S = \underset{a \in G}{∐} {(a, x) ∣ x \in X^{a}},

so we have

| S | = \sum_{a \in G} | X^{a} | .

On the other hand, note that

S = \underset{x \in X}{∐} {(a, x) ∣ a \in G_{x}},

so we have

| S | = \sum_{x \in X} | G_{x} | = \sum_{x \in X} \frac{| G |}{| G \cdot x |},

the latter statement being Corollary 4.6.23 and Lagrange’s theorem. If $𝒪$ is an orbit in $X$ , then it is the orbit of all $x \in 𝒪$ , so we have

\sum_{x \in 𝒪} \frac{| G |}{| G \cdot x |} = | G | \sum_{x \in 𝒪} \frac{1}{| 𝒪 |} = | G | .

Since $X$ is the disjoint union of its orbits $𝒪$ and there are $r$ orbits, we obtain $| S | = r | G |$ , and therefore we have

r | G | = \sum_{a \in G} | X^{a} |,

as desired. □

Burnside’s formula has an amusing use in certain problems involving counting.

Example 4.8.5. §

Suppose we have a table with which is a regular octagon, with one chair placed at each side. Let us consider two seatings of eight people at the table to be the equivalent if and only if every person has the same two neighbors under both seatings (though possibly on different sides). We can ask: how many equivalence classes of seatings are there?

Let $X$ denote the set of all seatings, so $| X | = 8!$ . Two seatings are equivalent if and only if there is an element of $D_{8}$ that takes the positions of the people under one seating to the their positions under the other. In other words, the seatings are in one-to-one correspondence with the orbits under the action of $D_{8}$ on $X$ . We note that $X^{e} = X$ , and $X^{a} = \emptyset$ if $a \neq e$ since any nontrivial element of $D_{8}$ will change the position of at least one person (in fact, at least six people). Applying Burnside’s formula, we have that the number of equivalence classes $r$ of seatings is $8! ∕ 16 = 2520$ .

Example 4.8.6. §

How many different ways are there to color the faces of a cube either red or blue that actually look different? (Here: two colorings are the same if one is a rotation of another.)

Let $X$ denote the set of all colorings, so $| X | = 2^{6} = 64$ . The group of rotations (orientation-preserving isometries) of a die has 5 types of elements: the identity, 6 rotations of order 4 through the centers of opposite faces, 3 rotations of order 2 of the same form, 6 rotations of order 2 through the centers of opposite edges, and 8 rotations of order 3 through the centers of opposite vertices. Respectively, these elements $a$ have $| X^{a} | = 64, 8, 16, 8$ , and $4$ . We then have

r = \frac{1}{24} (64 + 6 \cdot 8 + 3 \cdot 16 + 6 \cdot 8 + 8 \cdot 4) = \frac{240}{24} = 10 .

This means there are exactly 10 different-looking colorings. Note that, as is often the case with these sorts of problems, it would have been easier to simply count them directly.

4.9. $p$ -groups

Definition 4.9.1. §

A group $G$ is said to be a $p$ -group if every element of $G$ is finite of order a power of $p$ .

Note that we have already classified the finite abelian $p$ -groups up to isomorphism.

Example 4.9.2. §

The group $D_{4}$ is a nonabelian $2$ -group of order $8$ .

All finite groups of $p$ -power order are clearly $p$ -groups. We shall see that the converse is true as well. For this, we require the following useful lemma.

Lemma 4.9.3. §

Let $G$ be a finite group of $p$ -power order, and let $X$ be a $G$ -set. Then

| X | \equiv | X^{G} | mod p .

Proof.

By Corollary 4.6.23, every orbit in $X$ has order dividing $| G |$ , hence a power of $p$ . Note that the orbits of order $1$ are exactly the ${x}$ with $x \in X^{G}$ . On the other hand, the other orbits all have order divisible by $p$ , so if $Y$ is a set of representatives of the orbits of $G$ , then it contains $X^{G}$ , and we have

| X | = \sum_{y \in Y} | G \cdot y | \equiv \sum_{\begin{array}{c} x \in X^{G} \end{array}} | {x} | mod p,

□

In general, if the order of a finite group $G$ is $n$ , then while we know that every element of $G$ has order dividing $n$ , we do not have the converse (unless $G$ is cyclic). On the other hand, Cauchy’s theorem, which we now prove, tells us that $G$ contains elements of every prime order dividing $n$ .

Theorem 4.9.4 (Cauchy). §

Let $p$ be a prime number, and let $G$ be a finite group of order divisible by $p$ . Then $G$ contains an element of order $p$ .

Proof.

We consider the set

X = {(a_{1}, a_{2}, \dots, a_{p}) \in G^{p} ∣ a_{1} a_{2} \dots a_{p} = e} .

Note that if $(a_{1}, a_{2}, \dots, a_{p}) \in X$ , then $a_{1}, a_{2}, \dots, a_{p - 1} \in G$ can be chosen arbitrarily, and then $a_{p} = {(a_{1} a_{2} \dots a_{p - 1})}^{- 1}$ is determined by those $a_{i}$ . It follows that $| X | = | G |^{p - 1}$ . Let $τ = (1 2 \dots p) \in S_{p}$ . We let $τ$ , and hence $⟨ τ ⟩$ , act on $X$ by

τ \cdot (a_{1}, a_{2}, \dots, a_{p}) = (a_{2}, \dots, a_{p}, a_{1}) .

Note that this is an action, as

(a_{2} \dots a_{p}) a_{1} = e,

since $a_{1}$ being left inverse to $a_{2} \dots a_{p}$ implies that it is also right inverse to $a_{2} \dots a_{p}$ . Then

X^{⟨ τ ⟩} = X^{τ} = {(a, a, \dots, a) \in G^{p} ∣ a^{p} = e} .

By Lemma 4.9.3, we have

| X | \equiv | X^{⟨ τ ⟩} | mod p .

Since $p$ divides $| G |$ , it divides $| X |$ , and hence it divides the order of $X^{⟨ τ ⟩}$ . But $X^{⟨ τ ⟩}$ is in bijection with the set of elements of $G$ of order dividing $p$ , and $e$ is such an element. So, we must have at least $p$ distinct elements in $G$ of order dividing $p$ , hence at least $p - 1$ of order $p$ . □

Cauchy’s theorem has the above-mentioned corollary.

Corollary 4.9.5. §

Every finite $p$ -group has $p$ -power order.

Proof.

If $G$ is a finite group and $ℓ$ is a prime dividing $| G |$ , then $G$ has an element of order $ℓ$ by Cauchy’s theorem. So if $G$ is a $p$ -group, then by definition the only prime that can divide $| G |$ is $p$ . □

The following result is very useful in the study of $p$ -groups.

Proposition 4.9.6. §

The center $Z (G)$ of a nontrivial finite $p$ -group is nontrivial.

Proof.

Consider the action of $G$ on itself by conjugation. The set of elements of $G$ fixed by every element of $G$ under conjugation is exactly the center of $G$ . By Lemma 4.9.3, we therefore have that $| G | \equiv | Z (G) | mod p$ . Since $| G |$ is a nontrivial power of $p$ , this means that $Z (G)$ is not the trivial subgroup. □

We give an application of Proposition 4.9.6 to the study of the structure of $p$ -groups of order $p^{2}$ .

Theorem 4.9.7. §

Every group of order $p^{2}$ , where $p$ is a prime, is abelian.

Proof.

Let $G$ be a group of order $p^{2}$ . By Proposition 4.9.6, we have that $Z (G)$ is nontrivial, so has either order $p$ or $p^{2}$ . We must show that it is the latter, since $Z (G) = G$ if and only if $G$ is abelian. So, suppose by way of contradiction that $| Z (G) | = p$ , and let $b \in G$ be an element that is not in the center of $G$ . Then $H = Z (G) ⟨ b ⟩$ has order greater than $p$ , hence is all of $G$ . But $H$ is abelian, since $b$ commutes with every element of $Z (G)$ and certainly every element of $Z (G)$ commutes with itself. So, $G = H$ is abelian as well, contradicting $| Z (G) | = p$ . □

In particular, this tells us that there are only two isomorphism classes of groups of order $p^{2}$ , those of $ℤ ∕ p^{2} ℤ$ and $ℤ ∕ 𝑝ℤ \times ℤ ∕ 𝑝ℤ$ .

4.10. The Sylow theorems

Definition 4.10.1. §

Let $G$ be a group, and let $p$ be a prime number.

a.: A subgroup of $G$ is called a $p$ -subgroup if it is a $p$ -group.
b.: A $p$ -subgroup of $G$ is called a Sylow $p$ -subgroup of $G$ if it is not properly contained in any $p$ -subgroup of $G$ .

Remark 4.10.2. §

If $G$ is a finite group and $p^{n}$ is the largest power of $p$ dividing $| G |$ , then every $p$ -subgroup of $G$ has order dividing $p^{n}$ .

Examples 4.10.3. §

a.

In $S_{5}$ , the Sylow $5$ -subgroups are the subgroups generated by the $5$ -cycles, the Sylow $3$ -subgroups are the subgroups generated by the $3$ -cycles, and the Sylow $2$ -subgroups are the subgroups of order $8$ the form

⟨ (a b c d), (a c) ⟩

with $a, b, c, d$ distinct elements of $X_{5}$ .

b.

If $n \geq 3$ and $n = 2^{k} m$ with $m$ odd, then every Sylow $2$ -subgroup of $D_{n}$ has the form $D_{n} = ⟨ r^{i} s, r^{m} ⟩$ for some $0 \leq i < m$ .

The Sylow theorems, which we now state in one compact result that we refer to as Sylow’s theorem, constitute an extremely useful tool for the study of finite groups.

Theorem 4.10.4 (Sylow). §

Let $G$ be a finite group, let $p$ be a prime number, and let $n$ be exponent of the highest power of $p$ dividing $| G |$ . Then the following hold.

a.: Every Sylow $p$ -subgroup of $G$ has order $p^{n}$ .
b.: Every two Sylow $p$ -subgroups of $G$ are conjugate.
c.: The number of Sylow $p$ -subgroups divides $| G |$ and is congruent to $1$ modulo $p$ .

We defer the proof of the Sylow’s theorem to below.

Notation 4.10.5. §

Let $G$ be a finite group and $p$ a prime number. We let ${Syl}_{p} (G)$ denote the set of Sylow $p$ -subgroups of $G$ , and we let $n_{p} (G) = | {Syl}_{p} (G) |$ .

The fact that $n_{p} (G)$ both divides $| G |$ and is congruent to $1$ modulo $P$ can be very useful in determining the possible isomorphism classes of groups of a given order.

Example 4.10.6. §

It is easy to see from our description of the Sylow $p$ -subgroups of $S_{5}$ that every two Sylow $p$ -sugroups of $S_{5}$ are conjugate, as the elements generating such groups are conjugate. We have $n_{5} (S_{5}) = 6 \equiv 1 mod 5$ . We also have $n_{3} (S_{5}) = 10 \equiv 1 mod 3$ , and $n_{2} (S_{5}) = 15$ , which is odd.

To understand the second part of Sylow’s theorem, we introduce the concept of a normalizer.

Definition 4.10.7. §

Let $G$ be a group and $H$ be a subgroup. The normalizer $N_{G} (H)$ of $H$ in $G$ is the subgroup

N_{G} (H) = {a \in G ∣ 𝑎𝐻 a^{- 1} = H}

of $G$ .

Remarks 4.10.8. §

a.: By definition, $N_{G} (H)$ is the stabilizer of $H$ under the action of $G$ on its set of subgroups by conjugation, so in particular is a subgroup.
b.: We have $H ⊴ N_{G} (H)$ and $N_{G} (H)$ is the largest subgroup of $G$ in which $H$ is normal.
c.: We have $N_{G} (H) = G$ if and only if $H ⊴ G$ .

Examples 4.10.9. §

a.: We have $N_{D_{n}} (⟨ r ⟩) = D_{n}$ and $N_{D_{n}} (⟨ s ⟩) = ⟨ s ⟩$ if $n$ is odd and $⟨ s, r^{n ∕ 2} ⟩$ if $n$ is even.
b.: The normalizer of $⟨ (1 2 3 4) ⟩$ in $S_{5}$ is $⟨ (1 2 3 4), (1 3) ⟩$ .

The following lemma is crucial for proving the first part of Sylow’s theorem.

Lemma 4.10.10. §

Suppose that $G$ is a finite group, and let $H$ be a subgroup of $G$ of order a power of a prime $p$ . Then

[G : H] \equiv [N_{G} (H) : H] mod p .

Proof.

Let $L = G ∕ H$ . Then $H$ acts on $L$ by left multiplication: $h \cdot (𝑎𝐻) = h𝑎𝐻$ for $h \in H$ , $a \in G$ . We have that $𝑎𝐻 \in L^{H}$ if and only if $h (𝑎𝐻) = 𝑎𝐻$ for every $h \in H$ , which is to say that $𝑎h a^{- 1} \in H$ for every $h \in H$ , which means exactly that $𝑎𝐻 a^{- 1} = H$ . In other words,

L^{H} = {𝑎𝐻 ∣ a \in N_{G} (H)} = N_{G} (H) ∕ H .

By Lemma 4.9.3, we have that

| L | \equiv | L^{H} | mod p,

which is exactly the statement of the lemma. □

Let us fix a prime $p$ throughout the rest of this section. We prove a strengthening of the first part of Sylow’s theorem.

Theorem 4.10.11 (First Sylow theorem). §

Let $G$ be a group, and let $n$ be the exponent of the highest power of $p$ dividing $G$ . Every subgroup of $G$ of order $p^{k}$ with $k < n$ is a normal subgroup of a subgroup of $G$ of order $p^{k + 1}$ .

Proof.

Suppose that $H < G$ has order $p^{k}$ . By Lemma 4.10.10, its index in its normalizer is congruent to $[G : H]$ modulo $p$ , so is divisible by $p$ . But then $N_{G} (H) ∕ H$ has order divisible by $p$ , and so by Cayley’s theorem there exists a subgroup of it of order $p$ . By Proposition 2.13.10, there then exists a subgroup $K$ of $N_{G} (H)$ in which $H$ is normal and such that $| K ∕ H | = p$ . Lagrange’s theorem then implies that $| K | = p^{k + 1}$ , as desired. □

Theorem 4.10.11 tells us, in particular, that every $p$ -subgroup of $G$ of order less than $p^{n}$ is not maximal, so part a of Sylow’s theorem holds. In fact, recursion tells us that:

Corollary 4.10.12. §

Every $p$ -subgroup of a finite group $G$ is contained in a Sylow $p$ -subgroup of order $p^{n}$ , where $n$ is the exponent of the highest power of $p$ dividing $G$ . In particular, every Sylow $p$ -subgroup has order $p^{n}$ .

The first Sylow theorem also has the following simple corollary.

Corollary 4.10.13. §

Let $G$ be a group, and let $n$ be the exponent of the highest power of $p$ dividing $G$ . The $G$ has subgroups of order $p^{k}$ for every $1 \leq k \leq n$ .

Proof.

Suppose without loss of generality that $p$ divides $| G |$ . By Cauchy’s theorem, $G$ has an element of order $p$ , so it has a subgroup of order $p$ . By recursion, the first Sylow theorem then tells us that $G$ has subgroups of every $p$ -power order dividing $G$ . □

We next prove the second part of Sylow’s theorem, which we state as a separate result.

Theorem 4.10.14 (Second Sylow theorem). §

If $P$ and $Q$ are Sylow $p$ -subgroups of a finite group $G$ for some prime $p$ , then $P$ and $Q$ are conjugate subgroups of $G$ .

Proof.

We consider the action of $Q$ on the set of left cosets $G ∕ P$ via $h \cdot 𝑎𝑃 = (h𝑎) P$ for $h \in Q$ and $a \in G$ . By Lemma 4.9.3, we have that $| {(G ∕ P)}^{Q} | \equiv | G ∕ P | mod p$ . Since $p$ does not divide $| G ∕ P |$ , we therefore have that $p$ does not divide $| {(G ∕ P)}^{Q} |$ . In particular, there exists an element $𝑏𝑃 \in {(G ∕ P)}^{Q}$ . Since $h𝑏𝑃 = 𝑏𝑃$ for all $h \in Q$ , we have $b^{- 1} h𝑏 \in P$ for all $h \in Q$ , so $b^{- 1} 𝑄𝑏 \leq P$ . Since $P$ and $Q$ have the same order, we therefore have that $Q = 𝑏𝑃 b^{- 1}$ is a conjugate of $P$ in $G$ . □

Corollary 4.10.15. §

Suppose that $G$ is a finite group, and let $P$ be a Sylow $p$ -subgroup of $G$ . Then $P$ is normal in $G$ if and only if $n_{p} (G) = 1$ .

Proof.

We know that every conjugate of a Sylow $p$ -subgroup is a Sylow $p$ -subgroup, as it has the same order, so it is an immediate corollary of Theorem 4.10.14 that $P$ is normal in $G$ if and only if $n_{p} (G) = 1$ . □

The second Sylow theorem also has, after a short argument, the following consequence.

Proposition 4.10.16. §

Let $G$ be a finite group, let $p$ be a prime number, and let $P$ be a Sylow $p$ -subgroup. Then $n_{p} (G) = [G : N_{G} (P)]$ . In particular, the number of Sylow $p$ -subgroups of $G$ divides $| G |$ .

Proof.

Consider the action of $G$ on the set ${Syl}_{p} (G)$ of Sylow $p$ -subgroups of $G$ by conjugation By Theorem 4.10.4b, we have that ${Syl}_{p} (G)$ has just one orbit under this action, which is all of ${Syl}_{p} (G)$ . Since the stabilizer of $P$ is $N_{G} (P)$ , the result follows from Corollary 4.6.23. □

We will require a special case of the following lemma.

Lemma 4.10.17. §

Let $Q$ be a Sylow $p$ -subgroup and $P$ be a $p$ -subgroup of a finite group $G$ . Then $P \cap N_{G} (Q) = P \cap Q$ .

Proof.

Let $H = P \cap N_{G} (Q)$ . We need only show that $H \leq Q$ . Since $H \leq N_{G} (Q)$ , we have that $h𝑄 h^{- 1} = Q$ for all $h \in H$ , so $𝐻𝑄 = 𝑄𝐻$ , and therefore $𝐻𝑄$ is a subgroup of $G$ with $Q$ as a normal subgroup. The second isomorphism theorem implies that

| 𝐻𝑄 | = \frac{| H | | Q |}{| H \cap Q |},

and yields in particular that $𝐻𝑄$ is a $p$ -subgroup of $G$ . On the other hand, $Q$ is a Sylow $p$ -subgroup of $G$ , so $𝐻𝑄$ cannot be larger, and therefore must equal $Q$ . Thus, we have the required containment $H \leq Q$ . □

Finally, we prove the third part of Sylow’s theorem.

Theorem 4.10.18 (Third Sylow theorem). §

The number $n_{p} (G)$ of Sylow $p$ -subgroups of a finite group $G$ divides $| G |$ and is congruent to $1$ modulo $p$ .

Proof.

The first part is just Proposition 4.10.16 and Lagrange’s theorem. For the second part, we assume that $p$ divides $| G |$ , as the result is otherwise trivial. Let $P$ be a Sylow $p$ -subgroup of $G$ , and let $P$ act on ${Syl}_{p} (G)$ by conjugation: if $Q \in {Syl}_{p} (G)$ and $a \in P$ , then $a$ takes $Q$ to $𝑎𝑄 a^{- 1}$ . By Lemma 4.9.3, we then have

n_{p} (G) \equiv | {Syl}_{p} {(G)}^{P} | mod p .

Let $Q \in {Syl}_{p} {(G)}^{P}$ , which tells us that $P \leq N_{G} (Q)$ . By Lemma 4.10.17, we then have that

P = P \cap N_{G} (Q) \leq Q,

which forces $P = Q$ as $P$ and $Q$ have the same order by Corollary 4.10.12. Thus, we have that $| {Syl}_{p} {(G)}^{P} | = 1$ , and so $n_{p} (G) \equiv 1 mod p$ . □

4.11. Applications of Sylow theory

We can use Sylow’s theorem to classify, or simply to give information on, the structure of groups of a given order. For instance, Sylow’s theorem can be used to show that there are no simple groups of certain small orders or of orders with certain sorts of prime factorizations, as we see in the following examples.

Example 4.11.1. §

There are no simple groups of order $42$ . If $G$ is a group of order $42$ , then $n_{7} (G)$ divides $42$ and is $1$ modulo $7$ , which forces $n_{7} (G) = 1$ . By Corollary 4.10.15, we have that the unique subgroup of $G$ of order $7$ is normal, so $G$ is not simple.

Example 4.11.2. §

Let $G$ be a group of order $30$ . Suppose that $n_{3} (G) > 1$ and $n_{5} (G) > 1$ . Then the third part of Sylow’s theorem tells us that $n_{5} (G) = 6$ and $n_{3} (G) = 10$ . Now, any Sylow $5$ -subgroup has order $5$ , hence is cyclic with $4$ elements of order $5$ , and any two distinct Sylow $5$ -subgroups have trivial intersection. Therefore, $G$ contains $24$ elements of order $5$ . On the other hand, the same argument with $3$ replacing $5$ tells us that $G$ contains $20$ elements of order $3$ . This is clearly impossible. In particular, there are no simple groups of order $30$ .

Example 4.11.3. §

There are no simple groups of order $p^{n}$ , where $p$ is a prime and $n \geq 2$ . This follows from Corollary 4.10.13, which tells us that such a group has a subgroup of order $p^{n - 1}$ and the first Sylow theorem, which tells us that the subgroup is normal in a subgroup of order $p^{n}$ , which is necessarily the whole group.

We can also study groups with orders having a particularly nice form. The following result is useful for that.

Proposition 4.11.4. §

Let $G$ be a group. Suppose that $H$ and $K$ are normal subgroups of $G$ with $𝐻𝐾 = G$ and $H \cap K = {e}$ . Then the function $ψ : H \times K \to G$ given by $ψ (h, k) = h𝑘$ for $h \in K$ and $k \in K$ is an isomorphism.

Proof.

Let $h \in H$ and $k \in K$ . Then $[h, k]$ equals both $(h k^{- 1} h^{- 1}) k^{- 1}$ , from which it is seen to be an element of $K$ , as $K ◃ G$ , and $h (k h^{- 1} k^{- 1})$ , which is similarly seen to be an element of $H$ . As $H \cap K = {e}$ , we therefore have $[h, k] = e$ , and therefore elements of $H$ commute with elements of $K$ . It follows that $ψ$ as defined is a homomorphism. It is onto as $G = 𝐻𝐾$ and one-to-one as $h𝑘 = e$ implies $h, k \in H \cap K$ , so $(h, k) = (e, e)$ . □

Proposition 4.11.4 has the following application in conjunction with Sylow’s theorems.

Theorem 4.11.5. §

Suppose that $p$ and $q$ are prime numbers with $p < q$ . Then every group of order $𝑝𝑞$ has a normal subgroup of order $q$ and is in fact cyclic if $q ≢ 1 mod p$ .

Proof.

Note that $n_{q} (G)$ divides $p$ and is $1$ modulo $q$ , which forces $n_{q} (G) = 1$ since $p < q$ . By Corollary 4.10.15, $G$ has a unique, normal Sylow $q$ -subgroup $K$ of order $q$ . On the other hand, $n_{p} (G)$ divides $q$ and is $1$ modulo $p$ . Supposing that $q ≢ 1 mod p$ . then we must have $n_{p} (G) = 1$ as $q$ is a prime. Let $H$ be the unique, normal subgroup of order $p$ . Now, both $H$ and $K$ are cyclic, so let $h, k \in G$ with $H = ⟨ h ⟩$ and $K = ⟨ k ⟩$ . By Proposition 4.11.4, we have that

𝐺≅𝐻 \times 𝐾≅ℤ ∕ 𝑝ℤ \times ℤ ∕ 𝑞ℤ≅ℤ ∕ 𝑝𝑞ℤ,

the last step being the Chinese remainder theorem. □

We can use this to give a more complicated example of a proof that all groups of a given order are cyclic.

Example 4.11.6. §

All groups of order $255$ are cyclic. Since $255 = 3 \cdot 5 \cdot 17$ , the structure theorem for finitely generated abelian groups tells us that every abelian group of order $255$ is cyclic. So, we must show that every group $G$ of order $255$ is abelian. By the third Sylow theorem, $n_{17} (G) = 1$ , so $G$ has a unique subgroup $N$ of order $17$ , which is normal. Then $G ∕ N$ has order $15$ and so is cyclic by Theorem 4.11.5. By Theorem 4.2.8, the subgroup $N$ must contain the commutator subgroup $[G, G]$ of $N$ .

Again by the third Sylow theorem, we have either $n_{3} (G) = 1$ or $n_{3} (G) = 85$ and $n_{5} (G) = 1$ or $n_{5} (G) = 51$ . If $n_{3} (G) = 85$ , then $G$ has at least $170$ elements of order $3$ , and if $n_{5} (G) = 51$ , then $G$ has at least $204$ elements of order $5$ . Clearly, both of these cannot hold at the same time, so either $n_{3} (G) = 1$ or $n_{5} (G) = 1$ . But then $G$ has either a normal subgroup of order $3$ or a normal subgroup of order $5$ . Call this subgroup $Q$ . Then $G ∕ Q$ has order $3 \cdot 17$ or $5 \cdot 17$ , and in either case, Theorem 4.11.5 tells us that it is cyclic. As before, we then have that $Q$ contains $[G, G]$ , but $Q \cap N$ is trivial since $Q$ and $N$ have relatively prime order, so $[G, G] = {e}$ , which is to say that $G$ is abelian.

Let us also expand our study of groups of order $30$ .

Example 4.11.7. §

Every group $G$ of order $30$ has a normal subgroup of order $5$ . To see this, let $P$ be a subgroup of order $3$ and $Q$ a subgroup of order $5$ . By Example 4.11.2, either $P$ or $Q$ is normal, and therefore $𝑃𝑄$ is a subgroup of $G$ of order $15$ . By Theorem 4.11.5, it is cyclic. We thus have that $| N_{G} (Q) |$ is either $15$ or $30$ , so $n_{G} (Q) = 1$ or $2$ , and $2$ is impossible by the third Sylow theorem. Thus, $Q$ is normal.

We can also rule out a whole class of possible orders of simple groups with the following result.

Proposition 4.11.8. §

There are no simple groups of order $p^{2} q$ , where $p$ and $q$ are distinct prime numbers.

Proof.

Let $G$ be a group of order $p^{2} q$ . If $p > q$ , then $n_{p} (G) = 1$ by the third Sylow theorem, so $G$ has a normal Sylow $p$ -subgroup (which is abelian of order $p^{2}$ ). If $q > p$ , then $n_{q} (G) = 1$ or $p^{2}$ . We need only check the latter case. In this case, $p^{2} \equiv 1 mod q$ , so $q$ divides $p^{2} - 1$ , but it does not divide $p - 1$ as $q > p$ , so $q$ divides $p + 1$ , which forces $p = 2$ and $q = 3$ . Then $G$ has order $12$ . Now, if $n_{3} (G) = 4$ , a simple element count shows that one cannot have $n_{2} (G) = 3$ , so $G$ has a normal Sylow $2$ -subgroup. □

Another method for exhibiting the non-simplicity of groups of a given order comes from the use of permutation representations.

Proposition 4.11.9. §

Let $G$ be a finite simple group of order properly divisible by $p$ . Then $G$ is isomorphic to a subgroup of $S_{d}$ where $d = n_{p} (G)$ . In particular, $| G |$ divides $d!$ .

Proof.

Let $G$ act on the set $L$ of left cosets of the normalizer $N_{G} (P)$ of some Sylow $p$ -subgroup $P$ of $G$ by left multiplication. This is a transitive action, so the permutation representation $G \to S_{L}$ is nontrivial, hence injective as $G$ is simple. As $| L | = d$ by the third Sylow theorem, we have $S_{L} ≅ S_{d}$ , so $G$ is isomorphic to a subgroup of $S_{d}$ . □

We provide a couple of examples.

Example 4.11.10. §

There are no simple groups $G$ of order $160$ . That is, if $G$ were such a group, then $n_{2} (G) = 5$ by the third Sylow theorem, and therefore $G$ is isomorphic to a subgroup of $S_{5}$ . But $120$ does not divide $5!$ .

Example 4.11.11. §

There are no simple groups of order $396 = 2^{2} 3^{2} 11$ . If $G$ were such a group, then $n_{11} (G) = 12$ , and the normalizer of a Sylow $11$ -subgroup $P$ in $G$ has order $33 = \frac{396}{12}$ by the third Sylow theorem. It follows by Proposition 4.11.9 that $G$ is isomorphic to and thus may be identified with a subgroup of $S_{12}$ . By definition $N_{G} (P)$ is contained in $N_{S_{12}} (P)$ . But $P$ is generated by an $11$ -cycle, and the number of such Sylow $11$ -subgroups of $S_{12}$ is easily counted to be $12 \cdot 9!$ (as there are $\frac{12!}{11}$ such cycles and $10$ per subgroup), which again by the third Sylow theorem implies that the order $N_{P} (G)$ is $110$ , which is not a multiple of $33$ .

The following weakening of the second isomorphism theorem to allow arbitrary finite subgroups is a useful tool.

Lemma 4.11.12. §

Let $H$ and $K$ be finite subgroups of a group $G$ . Then we have

| 𝐻𝐾 | = \frac{| H | | K |}{| H \cap K |} .

Proof.

By definition, we have $| 𝐻𝐾 | \leq | H | | K |$ . If $h, h^{'} \in H$ and $k, k^{'} \in K$ are such that $h𝑘 = h^{'} k^{'}$ , then setting $a = {(h^{'})}^{- 1} h = k^{'} k^{- 1}$ , we have that $a \in H \cap K$ . Moreover, note that $h^{'} = h a^{- 1}$ and $k^{'} = 𝑎𝑘$ . Conversely, given $h \in H$ , $k \in K$ , and $a \in H \cap K$ , then defining $h^{'} = h a^{- 1}$ and $k^{'} = 𝑎𝑘$ , we see that $h𝑘 = h^{'} k^{'}$ . Therefore, if we define an equivalence relation on the set $H \times K$ by $(h, k) \sim (h^{'}, k^{'})$ if and only if $h𝑘 = h^{'} k^{'}$ , the number of pairs in each equivalence class is $| H \cap K |$ , and as a result there are $| H | | K | ∕ | H \cap K |$ equivalence classes. On the other hand, the number of equivalence classes is by definition $| 𝐻𝐾 |$ , proving the desired equality. □

Let us apply Lemma 4.11.12 to an example.

Example 4.11.13. §

There are no simple groups of order $48$ . Suppose $G$ is a group of order $48$ . By the third Sylow theorem, we have $n_{2} (G) = 1$ or $3$ . If $n_{2} (G) = 1$ , then the unique Sylow $2$ -subgroup is normal. If $n_{2} (G) = 3$ , let $H$ and $K$ be distinct subgroups of $G$ of order $16$ . Then the fact that $| 𝐻𝐾 | \leq 48$ and Lemma 4.11.12 force $| H \cap K | = 8$ . Then $H \cap K$ has index $2$ in $H$ and $K$ , hence is normal in both, so its normalizer $N_{G} (H \cap K)$ contains $𝐻𝐾$ , which has order $32$ , so equals $G$ . But then $H \cap K$ is normal in $G$ , so again $G$ is not simple.

The latter example used a special case of the following, which tells us that a group $G$ of order $48$ with $n_{2} (G) = 3 ≢ 1 mod 4$ has two Sylow $2$ -subgroups with intersection of order $8$ .

Proposition 4.11.14. §

Let $G$ be a finite group, and let $n$ be the exponent of the highest power of $p$ dividing $| G |$ . Let $P$ be a Sylow $p$ -subgroup of $G$ . Let $r \leq n$ be a positive integer such that $| P \cap Q | \leq p^{n - r}$ for every Sylow $p$ -subgroup $Q$ of $G$ with $Q \neq P$ . Then we have $n_{p} (G) \equiv 1 mod p^{r}$ .

Proof.

Let $P$ be a Sylow $p$ -subgroup, and consider the action of $P$ on ${Syl}_{p} (G)$ by conjugation. Let $Q \in {Syl}_{p} (G)$ with $Q \neq P$ . Lemma 4.10.17 tells us that the elements in $P$ that fix $Q$ under conjugation (i.e., the $a \in P$ such that $𝑎𝑄 a^{- 1} = Q$ ) are exactly those in $P \cap Q$ . For $i$ such that $p^{i} = [P : P \cap Q]$ , this implies that there are exactly $p^{i}$ conjugates of $Q$ by elements of $P$ , so the order of the $P$ -orbit of $Q$ is $p^{i}$ . Under the assumption of the proposition, we have that $| P \cap Q |$ divides $p^{n - r}$ , so $i \geq r$ . Therefore, $P$ -orbit of ${Syl}_{p} (G)$ other than the singleton orbit ${P}$ has order divisible by $p^{r}$ , which implies that $n_{p} (G) = | {Syl}_{p} (G) | \equiv 1 mod p^{r}$ . □

4.12. Simplicity of alternating groups

Before we proceed to simplicity, we first show that we can use group actions to give an alternate definition of the sign of a permutation (and therefore of alternating groups) that does not use the determinant map, which we did not define above.

Proposition 4.12.1. §

Let $S_{n}$ act on the set $X$ of polynomials $p = p (x_{1}, x_{2}, \dots, x_{n})$ in $n$ variables $x_{1}, x_{2}, \dots, x_{n}$ by

σ \cdot p = p (x_{σ (1)}, x_{σ (2)}, \dots, x_{σ (n)})

for $σ \in S_{n}$ and $p \in X$ . Take

Δ = \prod_{1 \leq i < j \leq n} (x_{j} - x_{i}) \in X

The function $𝜖 : S_{n} \to {\pm 1}$ given by

σ \cdot Δ = 𝜖 (σ) Δ

for all $σ \in S_{n}$ is equal to the homomorphism $sign$ .

Proof.

For $σ, τ \in S_{n}$ and $p \in X$ , we have

𝜎𝜏 \cdot p = p (x_{σ (τ (1))}, x_{σ (τ (2))}, \dots, x_{σ (τ (n))}) = σ \cdot p (x_{τ (1)}, x_{τ (2)}, \dots, x_{τ (n)}) = σ \cdot (τ \cdot p),

and clearly $e \cdot p = p$ for the identity $e$ , so $S_{n}$ acts on $X$ . Note that since $σ \in S_{n}$ takes each term $x_{j} - x_{i}$ with $i < j$ to plus or minus another term of the same form in a one-to-one fashion (in that $σ$ is one-to-one), we do indeed have $σ \cdot Δ = \pm Δ$ . That $𝜖$ is a homomorphism follows from the fact that $S_{n}$ acts on $X$ , since

𝜖 (𝜎𝜏) Δ = 𝜎𝜏 (Δ) = σ (𝜖 (τ) Δ) = 𝜖 (τ) σ (Δ) = 𝜖 (σ) 𝜖 (τ) Δ .

It remains only to check that $𝜖 (τ) = - 1$ for any transposition $τ = (k ℓ)$ with $1 \leq k < ℓ \leq n$ . For this, note that $τ (x_{j} - x_{i}) = x_{j} - x_{i}$ unless ${i, j} \cap {k, l} = \emptyset$ . If ${i, j} = {k, ℓ}$ , then $τ (x_{ℓ} - x_{k}) = - (x_{ℓ} - x_{k})$ . The remaining terms have the form $\pm (x_{ℓ} - x_{m})$ or $\pm (x_{k} - x_{m})$ for some $m \neq k, ℓ$ . We consider these in pairs. If $m < k$ , then we have

τ ((x_{ℓ} - x_{m}) (x_{k} - x_{m})) = (x_{k} - x_{m}) (x_{ℓ} - x_{m}) = (x_{ℓ} - x_{m}) (x_{k} - x_{m}) .

If $m > ℓ$ , we have

τ ((x_{m} - x_{k}) (x_{m} - x_{ℓ})) = (x_{m} - x_{ℓ}) (x_{m} - x_{k}) = (x_{m} - x_{k}) (x_{m} - x_{ℓ}),

and if $k < m < ℓ$ , we have

τ ((x_{m} - x_{k}) (x_{ℓ} - x_{m})) = (x_{m} - x_{ℓ}) (x_{k} - x_{m}) = (x_{m} - x_{k}) (x_{ℓ} - x_{m}) .

Therefore, the product of the contributions to $𝜖 (τ)$ from the various terms is $- 1$ , as required. □

We once again exhibit that $A_{5}$ is simple, and moreover, that it is the only simple subgroup of order $60$ , up to isomorphism.

Lemma 4.12.2. §

If $G$ is a group of order $60$ with $n_{5} (G) > 1$ , then $G$ is simple.

Proof.

The assumption forces $n_{5} (G) = 6$ by the third Sylow theorem. So, the normalizer of any Sylow $p$ -subgroup has order $10$ . Let $N$ be a proper normal subgroup of $G$ . If $5 ∣ | N |$ , then $N$ contains a Sylow $5$ -subgroup of $G$ and hence all Sylow $5$ -subgroups of $G$ by the second Sylow theorem. But then $N$ has at least $1 + 6 \cdot 4 = 25$ elements, so is of order $30$ . But $N$ has a unique subgroup of order $5$ by Example 4.11.7, which is normal in $G$ by Lemma 4.3.15a and Lemma 4.3.16. This contradicts $n_{5} (G) = 6$ . It follows that $5 ∤ | N |$ . Now, if $N$ has order $6$ or $12$ , then again it has a normal Sylow subgroup which is then by the same reasoning itself normal in $G$ . So, we may assume that $| N | \in {2, 3, 4}$ . Then $| G ∕ N | \in {15, 20, 30}$ , and in all of these cases, $G ∕ N$ has a normal subgroup of order $5$ by the third Sylow theorem and Example 4.11.7. But then $G$ itself has a normal subgroup with order divisible by $5$ , which we have already shown is not the case. Thus $N$ must be the trivial subgroup. □

As a corollary, we recover Proposition 4.6.26 that $A_{5}$ is simple. Let us prove that this is the only subgroup of order $60$ .

Proposition 4.12.3. §

The group $A_{5}$ is isomorphic to every simple group of order $60$ .

Proof.

Let $G$ be a simple group of order $60$ , which we know exists by Proposition 4.6.26. We show that $G$ is isomorphic to $A_{5}$ . From the third Sylow theorem, the possibilities for $n_{2} (G)$ are $3$ , $5$ , and $15$ , which is also the index of the normalizer $N$ of a Sylow $2$ -subgroup $P$ . Since $G$ is not isomorphic to a subgroup of $S_{3}$ , we can eliminate $n_{2} (G) = 3$ .

If $n_{2} (G) = 5$ , then $G$ is isomorphic to a subgroup of $S_{5}$ , so $G$ may be identified with a normal subgroup of $S_{5}$ of index $2$ . It follows that $G \cap A_{5}$ is a normal subgroup of $A_{5}$ which is either $A_{5}$ or of index $2$ in $A_{5}$ . The latter being impossible by the simplicity of $A_{5}$ , we must have that $G = A_{5}$ .

Suppose now that $n_{2} (G) = 15$ . Since $15 ≢ 1 mod 4$ , Proposition 4.11.14 tells us that $| P \cap Q | = 2$ for some $Q \in {Syl}_{2} (G)$ with $Q \neq P$ . Set $M = N_{G} (P \cap Q)$ , which is not $G$ since $G$ is simple. Since $| M |$ is a multiple of $4$ that is greater than $8$ by Lemma 4.11.12, we must have $| M | = 12$ or $20$ , from which it follows that $M$ has index at most $5$ , and therefore $G$ is isomorphic to a subgroup of $S_{5}$ . The same argument as before would tell us that $𝐺≅ A_{5}$ , but note that we assumed $n_{2} (G) = 15$ , so we reach a contradiction. □

We now prove that the alternating groups on at least $5$ elements are simple.

Theorem 4.12.4. §

The groups $A_{n}$ for $n \geq 5$ are simple.

Proof.

We prove this by induction on $n \geq 5$ , the case $n = 5$ having been proven in Proposition 4.6.26. Let $G = A_{n}$ for some $n \geq 6$ . For any $i \in X_{n}$ , the stabilizer $G_{i}$ is isomorphic to $A_{n - 1}$ , which is simple by induction. Suppose that $N$ is a nontrivial normal subgroup of $G$ . If there exists $i \in X_{n}$ and $τ \in N - {e}$ with $τ (i) = i$ , then $N \cap G_{i}$ is a nontrivial normal subgroup of $G_{i}$ , and it follows that $G_{i} \leq N$ by the simplicity of $G_{i}$ . For any $j \in X_{n}$ , we can find $σ \in A_{n}$ with $σ (i) = j$ , and then $G_{j} = σ G_{i} σ^{- 1} \leq N$ by normality of $N$ . As every element of $A_{n}$ can be written as a product of an even number of transpositions, every element of $A_{n}$ may be written as a product of products of two transpositions, and any product of two transpositions lies in $G_{j}$ for some $j \in X_{n}$ since $n > 4$ . Thus, we must have that $N = G = A_{n}$ .

Now, we show that $N$ must contain a permutation that fixes some element of $X_{n}$ . Let $τ \in N$ . If $τ (i) = τ^{'} (i)$ for any $τ^{'} \in N$ and $i \in X_{n}$ , then $τ^{'} τ^{- 1}$ fixes $i$ . If the cycle decomposition of $τ$ contains a $k$ -cycle with $k \geq 3$ , say $(a_{1} a_{2} \dots a_{k})$ , then we may choose $σ \in A_{n}$ that fixes $a_{1}$ and $a_{2}$ but not $a_{3}$ . If $τ$ is a product of disjoint transpositions that does not fix any element, then write $τ = (a_{1} a_{2}) (a_{3} a_{4}) \dots (a_{m - 1} a_{m})$ and take $σ = (a_{1} a_{2}) (a_{3} a_{5})$ (using the fact that $n \geq 6$ ). It follows in both cases that $τ^{'} = 𝜎𝜏 σ^{- 1} \neq τ$ , but $τ^{'} (a_{1}) = τ (a_{1})$ , as desired. □

4.13. Free groups and presentations

We begin with a general definition of a free group by its “universal property”.

Definition 4.13.1. §

A group $F$ is free on a subset $X$ if, whenever $f : X \to G$ is a function, where $G$ is a group, there exists a unique homomorphism

ϕ_{f} : F \to G

such that $ϕ_{f} (x) = f (x)$ for all $x \in G$ . The existence of this unique homomorphism is referred to as the universal property of $F$ .

Proposition 4.13.2. §

Let $F$ be free on a set $X$ and $F^{'}$ be free on a set $X^{'}$ , and suppose $f : X \to X^{'}$ is a bijection. Then the homomorphism $ϕ_{f} : F \to F^{'}$ given by the universal property is an isomorphism.

Proof.

Let $g$ be the inverse to $f$ , and let $ϕ_{g}^{'} : F^{'} \to F$ be the homorphism given by the universal property for $F^{'}$ . Then $ϕ_{g}^{'} \circ ϕ_{f} (x) = x$ for all $x \in X$ and $ϕ_{f} \circ ϕ_{g}^{'} (x^{'}) = x^{'}$ for all $x^{'} \in X$ . Since the identity homomorphisms of $F$ and $F^{'}$ also take elements of $X$ and $X^{'}$ to themselves, respectively, the the universal property for $F$ and for $F^{'}$ imply that $ϕ_{g}^{'} \circ ϕ_{f} = {id}_{F}$ and $ϕ_{f} \circ ϕ_{g}^{'} = {id}_{F^{'}}$ , respectively. Therefore, we have that $ϕ_{g}^{'} = ϕ_{f}^{- 1}$ , so $ϕ_{f}$ is an isomorphism. □

Example 4.13.3. §

The integers $ℤ$ are a free group on the subset ${1}$ , since for any group $G$ and element $x \in G$ , we can define $ϕ : ℤ \to G$ with $ϕ (1) = x$ by $ϕ (n) = x^{n}$ for all $n \in ℤ$ , and this is the unique homomorphism taking $1$ to $x$ .

Example 4.13.4. §

The group $ℤ^{n}$ , although a free abelian group, is not a free group. For example, take $n = 2$ . Then the map $f : {(1, 0), (0, 1)} \to D_{3}$ with $f (1, 0) = r$ and $f (0, 1) = s$ cannot be extended to a homomorphism $ϕ : ℤ^{2} \to D_{3}$ , for such a function would have to satisfy

𝑟𝑠 = ϕ (0, 1) ϕ (1, 0) = ϕ (1, 1) = ϕ (1, 0) ϕ (0, 1) = 𝑠𝑟,

which does not hold in $D_{3}$ .

To show the existence of free groups on larger sets, we construct them explicitly.

Definition 4.13.5. §

A word on in a set $X$ is a symbol

x_{1}^{n_{1}} x_{2}^{n_{2}} \dots x_{k}^{n_{k}}

with $x_{1}, x_{2}, \dots, x_{k} \in X$ and $n_{1}, n_{2}, \dots, n_{k} \in ℤ$ , where $k \geq 0$ . If $k = 0$ , we sometimes denote this word by $e$ , and it is called the empty word.

Remark 4.13.6. §

We write the word $x^{1}$ for $x \in X$ more simply as $x$ .

Definition 4.13.7. §

The product of two words $w = x_{1}^{n_{1}} x_{2}^{n_{2}} \dots x_{k}^{n_{k}}$ and $v = y_{1}^{m_{1}} y_{2}^{m_{2}} \dots y_{l}^{m_{l}}$ in $X$ is the concatenation

w \cdot v = 𝑤𝑣 = x_{1}^{n_{1}} x_{2}^{n_{2}} \dots x_{k}^{n_{k}} y_{1}^{m_{1}} y_{2}^{m_{2}} \dots y_{l}^{m_{l}} .

Clearly, concatenation is an associative binary operation on the set of words, and $e$ is an identity element for this operation.

Recall from Example 1.2.25 that there is a smallest equivalence relation containing any relation on a set. So, let us define an equivalence relation on the set of words by a set of generators.

Definition 4.13.8. §

The standard equivalence relation $\sim$ on the set $W_{X}$ of words on $X$ is the smallest equivalence relation such that

𝑤𝑣 \sim w x^{0} v

(4.13.1)

and

w x^{m + n} v \sim w x^{m} x^{n} v

(4.13.2)

for all $w, v \in W_{X}$ , $x \in X$ , and $m, n \in ℤ$ .

Two words are then equivalent if and only if one can be obtained from the other by a finite sequence of operations on the word consisting each of adding or removing an $x^{0}$ for some $x \in X$ or changing $x^{m + n}$ in a word to $x^{m} x^{n}$ or changing $x^{m} x^{n}$ in a word to $x^{m + n}$ for some $x \in X$ and $m, n \in ℤ$ .

Definition 4.13.9. §

We say that a word $x_{1}^{n_{1}} x_{2}^{n_{2}} \dots x_{k}^{n_{k}}$ in $X$ is reduced if $x_{i} \neq x_{i + 1}$ for all $1 \leq i \leq k - 1$ and $n_{i} \neq 0$ for all $1 \leq i \leq k$ .

Proposition 4.13.10. §

Every word is equivalent to a unique reduced word.

Proof.

The relation (4.13.2) tells us recursively for any $n \geq 1$ that

\begin{array}{rcl} w x^{n} v \sim w (x \cdot x \dots x) v & and & w x^{- n} v \sim w (x^{- 1} \cdot x^{- 1} \dots x^{- 1}) v, & (4.13.3) \end{array}

for all $w, v \in W_{X}$ and $x \in X$ , with $n$ symbols “ $x$ ” appearing on the right-hand sides. If we start with a word $w$ in $X$ , we may use (4.13.3) to expand it and (4.13.1) to remove any $0$ -powers of elements of $X$ , to obtain an equivalent word of the form

x_{1}^{\pm 1} x_{2}^{\pm 1} \dots x_{k}^{\pm 1} .

We may use (4.13.1) and (4.13.3) to remove terms of the form $x_{i} x_{i + 1}^{- 1}$ or $x_{i}^{- 1} x_{i + 1}$ with $x_{i} = x_{i + 1}$ , relabeling after each step, until no such terms exist. We may then gather terms by again applying (4.13.2) to obtain a reduced word equivalent to $w$ .

The process we have described does not change a reduced word. Moreover, the operations of adding in or removing an $x^{0}$ from a word or changing $x^{m + n}$ to $x^{m} x^{n}$ for some $m, n \in ℤ$ or vice-versa do not change the result of the process. Therefore, each word is equivalent to a unique reduced word. □

Notation 4.13.11. §

The set of equivalence classes of words on a set $X$ is denoted $F_{X}$ .

Proposition 4.13.12. §

The set $F_{X}$ is a group under concatenation of words, and it is generated by the set $X$ .

Proof.

We give only a sketch. First, we must check that if $w \sim w^{'}$ and $v \sim v^{'}$ are two pairs of equivalent words in $X$ , then $𝑤𝑣 \sim w^{'} v^{'}$ . This follows quickly from the definition of the equivalence relation $\sim$ . So, the binary operation is well-defined, associative, and has identity $e$ . Moreover, the inverse of the equivalence class of a word $x_{1}^{n_{1}} x_{2}^{n_{2}} \dots x_{k}^{n_{k}}$ is the equivalence class of the word

x_{k}^{- n_{k}} \dots x_{2}^{- n_{2}} x_{1}^{- n_{1}} .

That $X$ generates $F_{X}$ is simply Proposition 2.4.3. □

Remark 4.13.13. §

It is typical to denote an element of $F_{X}$ by any word representing it, which means that we will use the symbol “ $=$ ” instead of “ $\sim$ ” when interpreting these words as elements of $F_{X}$ .

Examples 4.13.14. §

a.

The free group $F_{{x}}$ consists exactly of all $x^{n}$ for $n \in ℤ$ , and only $x^{0} = e$ , so $F_{{x}} ≅ℤ$ .

b.

The free group $F_{{x, y}}$ with $x \neq y$ consists of all words

x^{n_{1}} y^{m_{1}} x^{n_{1}} y^{m_{2}} \dots x^{n_{k}} y^{m_{k}},

where we can take $n_{i} \neq 0$ for $i \geq 2$ and $m_{j} \neq 0$ for $j < k$ . We have, e.g.,

x^{2} y^{- 1} x^{- 3} y^{- 1} \cdot y x^{3} y^{2} x^{5} = x^{2} y x^{5} .

Lemma 4.13.15. §

The group $F_{X}$ is a free group on the set $X$ .

Proof.

By Definition 4.13.1, we must show that for any group $G$ and function $f : X \to G$ , the function

ϕ_{f} (x_{1}^{n_{1}} x_{2}^{n_{2}} \dots x_{k}^{n_{k}}) = f {(x_{1})}^{n_{1}} f {(x_{2})}^{n_{2}} \dots f {(x_{k})}^{n_{k}}

is the unique well-defined homomorphism such that $ϕ_{f} (x) = f (x)$ for all $x \in X$ . That $ϕ_{f}$ is a homomorphism is a direct consequence of its definition and the definition of multiplication of words by concatenation, once it is seen to be well-defined. That it is well-defined is a consequence of the fact that the only relations that are imposed on words are those that exist in any group. That is, for words $v$ and $w$ , $x \in X$ , and $m, n \in ℤ$ , we have

\begin{array}{rcl} ϕ_{f} (v x^{0} w) = ϕ_{f} (v) ϕ_{f} (x^{0}) ϕ_{f} (w) = ϕ_{f} (v) f {(x)}^{0} ϕ_{f} (w) = ϕ_{f} (v) ϕ_{f} (w), \\ ϕ_{f} (v x^{m + n} w) = ϕ_{f} (v) f {(x)}^{m + n} ϕ_{f} (w) = ϕ_{f} (v) f {(x)}^{m} f {(x)}^{n} ϕ_{f} (w) = ϕ_{f} (v x^{m} x^{n} w), \end{array}

so $ϕ_{f}$ is constant on equivalent words. □

Definition 4.13.16. §

The group $F_{X}$ of Proposition 4.13.12 is the free group on a set $X$ .

Proposition 4.13.2 then immediately implies the following.

Corollary 4.13.17. §

If $X$ and $Y$ are sets with the same cardinality, then $F_{X}$ and $F_{Y}$ are isomorphic.

The following then provides an object that is well-defined up to isomorphism.

Notation 4.13.18. §

The free group on a set with $n$ elements is denoted $F_{n}$ .

The following gives the relationship between free groups and free abelian groups.

Proposition 4.13.19. §

The free abelian group on a set $X$ is isomorphic to the abelianization of the free group on $X$ .

Proof.

Define $π : F_{X} \to \oplus_{x \in X} ℤ$ by $π (x) = e_{x}$ , where $e_{x}$ is the standard basis element of $\oplus_{x \in X} ℤ$ corresponding to $x$ . This is a surjective homomorphism. As the image of $π$ is abelian, the map $π$ factors through a sujrective homomorphism $\bar{π} : F_{X}^{ab} \to \oplus_{x \in X} ℤ$ , and therefore the maximal abelian quotient of $F_{X}$ surjects onto $\oplus_{x \in X} ℤ$ . In $F_{X}^{ab}$ , we may rearrange the terms of the image of $x_{1}^{n_{1}} x_{2}^{n_{2}} \dots x_{k}^{n_{k}}$ for $x_{i} \in X$ and $n_{i} \in ℤ$ with $1 \leq i \leq k$ so that it is the image of a like element with $x_{1}, \dots, x_{k}$ are all distinct. Now, such an element is sent to $\sum_{i = 1}^{k} n_{i} e_{x_{i}}$ under $\bar{π}$ , and so it is $0$ if and only if all $n_{i} = 0$ . If follows that $\bar{π}$ is an isomorphism. □

Theorem 4.13.20. §

The free group $F_{n}$ on $n$ elements cannot be generated by fewer than $n$ elements.

Proof.

If $F_{n}$ could be generated by $n - 1$ elements, then $F_{n}^{ab}$ could be generated by $n - 1$ elements by Proposition 4.13.19. But $F_{n}^{ab}$ is isomorphic to the free abelian group on $n$ elements, so Theorem 4.4.16 tells us that $F_{n}^{ab}$ cannot be generated by $n - 1$ elements. □

We omit the proof of the following theorem.

Theorem 4.13.21. §

Every subgroup of a free group is also a free group.

One of the most important uses of free groups is to give presentations of groups. First, we make the following definition, recalling Lemma 1.2.24 to see that it is well-defined.

Definition 4.13.22. §

The normal closure of a subset $S$ of a group $G$ is the smallest normal subgroup of $G$ containing $S$ , equal to the intersection of all normal subgroups of $G$ containing $S$ .

Definition 4.13.23. §

A presentation of a group $G$ on a set $S$ and a subset $T$ of $F_{S}$ is a surjective homomorphism $F_{S} \to G$ with kernel equal to the normal closure of the set $T$ . We say that $G$ is presented by the generating set $S$ and the relation set $T$ , and we write $𝐺≅ ⟨ S ∣ T ⟩$ .

Remark 4.13.24. §

If $G$ is presented by $S$ and $T$ , then the first isomorphism theorem tells us that $𝐺≅ F_{S} ∕ R$ by an isomorphism sending $s \in S$ to the coset of $s$ in $F_{S} ∕ R$ , where $R$ is the normal closure of $T$ .

In fact, we have the following stronger result.

Proposition 4.13.25. §

Suppose that $G$ and $G^{'}$ are groups with $G$ presented by $S$ and $T$ . Suppose we are given a subset ${x_{s} ∣ s \in S}$ of $G^{'}$ . Then there exists a homomorphism $ϕ : G \to G^{'}$ with $ϕ (s) = x_{s}$ for all $s \in S$ if and only if the unique homomorphism $Φ : F_{S} \to G^{'}$ with $Φ (s) = x_{s}$ for all $s \in S$ satisfies $T \subseteq \ker Φ$ .

Proof.

The existence and uniqueness of $Φ$ is by the universal property of $F_{S}$ . If $T \subseteq \ker Φ$ , then the normal closure $R$ of $T$ is contained in $\ker Φ$ since $\ker Φ$ is a normal subgroup of $F_{r, s}$ containing $T$ . In this case, the first isomorphism theorem implies that $Φ$ induces a map $\bar{Φ} : F_{S} ∕ R \to T$ with $\bar{Φ} (𝑠𝑅) = x_{s}$ for all $s \in S$ . Since $𝐺≅ F_{S} ∕ R$ by an isomorphism sending $s$ to $𝑠𝑅$ , the composition $ϕ : G \to G^{'}$ is the desired map. Similarly, if $ϕ$ exists, then we may compose it with the surjection $F_{S} \to_{}^{\sim} G$ taking $s \in S$ to $s$ to obtain a map $Φ : F_{S} \to G^{'}$ with $R$ , and hence $T$ , in its kernel. □

Definition 4.13.26. §

If $G$ is presented by finite sets $S = {s_{1}, s_{2}, \dots, s_{k}}$ and $T = {r_{1}, r_{2}, \dots, r_{d}}$ , then $G$ is said to be finitely presented.

Notation 4.13.27. §

We write

G = ⟨ s_{1}, s_{2}, \dots, s_{k} ∣ r_{1}, r_{2}, \dots, r_{d} ⟩ .

to denote that $G$ has a presentation by sets $S = {s_{1}, s_{2}, \dots, s_{k}}$ and $T = {r_{1}, r_{2}, \dots, r_{d}}$ .

Examples 4.13.28. §

We give several examples of presentations:

a.: $F_{S} ≅ ⟨ S ∣ \emptyset ⟩$ ,
b.: $ℤ^{2} ≅ ⟨ a, b ∣ 𝑎𝑏 a^{- 1} b^{- 1} ⟩$ ,
c.: $ℤ ∕ 𝑛ℤ≅ ⟨ a ∣ a^{n} ⟩$ ,
d.: $D_{n} ≅ ⟨ r, s ∣ r^{n}, s^{2}, 𝑟𝑠𝑟𝑠 ⟩$ .

Remark 4.13.29. §

One sometimes writes

⟨ s_{1}, s_{2}, \dots, s_{k} ∣ r_{1} = r_{1}^{'}, r_{2} = r_{2}^{'}, \dots, r_{d} = r_{d}^{'} ⟩

for a finite presentation

⟨ s_{1}, s_{2}, \dots, s_{k} ∣ r_{1}^{- 1} r_{1}^{'}, r_{2}^{- 1} r_{2}^{'}, \dots, r_{d}^{- 1} r_{d}^{'} ⟩ .

Example 4.13.30. §

We have

ℤ^{n} ≅ ⟨ x_{1}, \dots, x_{n} ∣ x_{i} x_{j} = x_{j} x_{i} for 1 \leq i < j \leq n ⟩ .

Note that we can start with the presentation, rather than a group, in order to define new groups.

Example 4.13.31. §

The quaternion group $Q_{8}$ is the group of order $8$ with the presentation

Q_{8} = ⟨ i, j ∣ i^{4} = e, i^{2} = j^{2}, 𝑖𝑗 = j i^{- 1} ⟩ .

The elements of $Q_{8}$ are usually labelled ${\pm 1, \pm i, \pm j, \pm k}$ , with $k = 𝑖𝑗$ , $- 1 = i^{2}$ , $- i = i^{3}$ , $- j = j^{3}$ , and $- k = 𝑗𝑖 = k^{3}$ . We remark that $Q_{8} ≇ D_{4}$ .

Sometimes, we just end up with complicated presentations of familiar groups.

Example 4.13.32. §

Consider the group

G = ⟨ x, y ∣ x^{2} y, x^{4} y ⟩ .

Then $e = {(x^{2} y)}^{- 1} x^{4} y = y^{- 1} x^{2} y$ , which forces $x^{2} = e$ , and then $y = x^{2} y = e$ . Since in fact $x^{2} = e$ and $y = e$ imply $x^{2} y = x^{4} y = e$ , the group $G$ also has a presentation

G = ⟨ x, y ∣ x^{2}, y ⟩,

and so is just $⟨ x ⟩ ≅ℤ ∕ 2 ℤ$ .

Remark 4.13.33. §

To determine whether or not two presentations yield isomorphic groups is a very difficult question. So difficult, in fact, that it has been shown that there does not exist a single recursive computer algorithm into which one can input any two finite group presentations that will always output whether or not they are yield isomorphic groups. Moreover, there exist presentations of groups $G$ from which one cannot write a computer algorithm to determine whether or not a word in the generators of $G$ is equal to the identity of $G$ .

Definition 4.13.34. §

Let $G_{1}$ and $G_{2}$ be groups. A free product $G_{1} * G_{2}$ of $G_{1}$ and $G_{2}$ is a group for which there exist homomorphisms $ι_{i} : G_{i} \to G_{1} * G_{2}$ for $i \in {1, 2}$ such that for any group $K$ and group homomorphisms $ϕ_{i} : G_{i} \to K$ with $i \in {1, 2}$ , there exists a unique homomorphism $Φ : G_{1} * G_{2} \to K$ with $Φ \circ ι_{i} = ϕ_{i}$ for $i \in {1, 2}$ .

Proposition 4.13.35. §

Let $G_{1}$ and $G_{2}$ be groups. Then the free product of $G_{1}$ and $G_{2}$ exists and is unique up to isomorphism. Moreover, if $G_{i}$ has a presentation $G_{i} ≅ ⟨ S_{i} ∣ T_{i} ⟩$ for each $i \in {1, 2}$ , then the free product is isomorphic to

G_{1} * G_{2} ≅ ⟨ S_{1} ∐ S_{2} ∣ T_{1} ∐ T_{2} ⟩,

where $∐$ denotes the disjoint union.

Proof.

We verify that the group $N = ⟨ S_{1} ∐ S_{2} ∣ T_{1} ∐ T_{2} ⟩$ is a free product of $G_{1}$ and $G_{2}$ . We leave the uniqueness of the free product up to isomorphism as an exercise for the reader. Let $π_{i} : F_{S_{i}} \to G_{i}$ be the surjections defining the presentation of $G_{i}$ for $i \in {1, 2}$ . Define homomorphisms $χ_{i} : F_{S_{i}} \to N$ by letting $χ_{i} (s)$ equal the image of $s$ in $N$ for all $s \in S_{i}$ . By definition of $N$ , we have that $T_{i}$ is contained in the kernel of $χ_{i}$ , so the first isomorphism theorem provides maps $ι_{i} : G_{i} \to N$ such that $ι_{i} (π_{i} (s))$ is the image of $s$ in $N$ for any $s \in S_{i}$ .

Now, for $i \in {1, 2}$ , let $ϕ_{i} : G_{i} \to K$ be a homomorphism to some group $K$ . Then we have a unique map $Ψ : F_{S_{i} ∐ S_{j}} \to K$ determined by $Ψ (s) = ϕ_{i} (π_{i} (s))$ for all $s \in S_{i}$ for $i \in {1, 2}$ . If $t \in T_{i}$ for some $i$ , then $Ψ (t) = ϕ_{i} (e) = e$ , so the the first isomorphism theorem yields a homomorphism $Φ : N \to K$ such that $Φ (ι_{i} (g)) = ϕ_{i} (g)$ for all $g = π_{i} (s)$ for some $s \in S$ for $i \in {1, 2}$ . However, the elements of $π_{i} (S_{i})$ generate $G_{i}$ , so we have that $Φ \circ ι_{i} = ϕ_{i}$ for each $i$ . Moreover, $Φ$ is unique, as its values on the images of the elements of $S_{1} ∐ S_{2}$ are determined by the latter equalities. □

Remark 4.13.36. §

An element of the free product of groups $G$ and $H$ is an equivalence class of words $g_{1} h_{1} g_{2} h_{2} \dots g_{k} h_{k}$ with $g_{i} \in G_{i}$ and $h_{i} \in H_{i}$ for $1 \leq i \leq k$ (under an equivalence relation under which the identity elements of the two groups are each identified with the identity element of the free product and which otherwise only imposes the relations of the original groups within words), with multiplication induced by concatenation.