Abstract Algebra — Chapter 2 Group theory

Chapter 2
Group theory

2.1. Groups

In this section, we introduce groups, which can briefly be defined as associative binary structures with identities and inverses. We begin by defining the two latter terms.

Definition 2.1.1. §

Suppose that $(X, ⋆)$ is a binary structure.

a.: A left (resp., right) identity element of $X$ is an element $e \in X$ that satisfies $e ⋆ x = x$ (resp., $x ⋆ e = x$ ).
b.: If $e \in X$ is both a left and a right identity element of $X$ , we say that it is an identity element of $X$ .

Examples 2.1.2. §

a.

Under addition, $0$ is a left and right identity element in $ℤ$ , $ℚ$ , $ℝ$ , $ℂ$ , $ℝ^{n}$ , $M_{n} (ℝ)$ , and $Maps (ℝ, ℝ)$ , with $0$ in the latter three examples being the zero vector, zero matrix, and constant function with value $0$ . Similarly, under multiplication, $1$ is a left and right identity element in all of the latter sets.

b.

Under subtraction on the sets from part $a$ , the element $0$ is a right identity but there is no left identity element.

c.

Under composition, $f (x) = x$ is an identity element in $Maps (ℝ, ℝ)$ .

d.

Under union, $∅$ is an identity element in $𝒫_{X}$ .

e.

Multiplication is a binary operation on the even integers $2 ℤ$ but $2 ℤ$ has no left and no right identity elements.

f.

For the binary structure defined on ${a, b}$ by the table

$⋆$	$a$	$b$



$a$	$a$	$b$

$b$	$a$	$b$

$a$ and $b$ are both left identity elements, but there is no right identity element.

One could ask whether or not there can be more than one (left and right) identity element in a binary structure. The following provides the answer.

Lemma 2.1.3. §

Let $(X, ⋆)$ be a binary structure. Suppose that $e \in X$ is a left identity element and that $f \in X$ is a right identity element. Then $e = f$ , and in particular $e$ is an identity element in $X$ .

Proof.

If $f$ is a right identity element, we have $e ⋆ f = e$ . On the other hand, since $e$ is a left identity element, we have $e ⋆ f = f$ . Therefore, we have $e = f$ . □

The following is an immediate corollary.

Corollary 2.1.4. §

Let $(X, ⋆)$ be a binary structure that contains an identity element $e$ . Then every (left or right) identity element in $X$ is equal to $e$ .

Definition 2.1.5. §

Suppose that $(X, ⋆)$ is a binary structure with an identity element $e \in X$ .

a.: A left (resp., right) inverse of $x \in X$ is an element $y \in X$ such that $y ⋆ x = e$ (resp., $x ⋆ y = e$ ).
b.: An element that is both a left and a right inverse to $x \in X$ is called an inverse of $x \in X$ .

Examples 2.1.6. §

a.

In $ℤ$ , $ℚ$ , $ℝ$ , $ℂ$ , $M_{n} (ℝ)$ , and $Maps (ℝ, ℝ)$ , the negative $- x$ of an element called $x$ is the inverse under addition. Under multiplication, $x^{- 1} = 1 ∕ x$ is the inverse of any $x \neq 0$ in $ℚ$ , $ℝ$ , and $ℂ$ . The elements that have multiplicative inverses in $ℤ$ are $\pm 1$ , in $M_{n} (ℝ)$ they are the matrices with nonzero determinant, and in $Maps (ℝ, ℝ)$ they are the nowhere vanishing functions.

b.

Under subtraction on the sets of part a, an element $x$ is its own left and right inverse.

c.

Under composition, an element $f \in Maps (ℝ, ℝ)$ has an inverse $f^{- 1}$ if and only if it is a bijection.

d.

Under union on $𝒫_{X}$ , only $∅$ has an inverse, which is itself.

e.

For the binary structure defined on ${a, b, c}$ by the table

$⋆$	$a$	$b$	$c$



$a$	$a$	$b$	$c$

$b$	$b$	$a$	$a$

$c$	$c$	$b$	$c$

$a$ is an identity element and is its own inverse, $b$ is an inverse of itself, $c$ is a right inverse of $b$ and therefore $b$ is a left inverse of $c$ , but $c$ has no right inverse.

Lemma 2.1.7. §

Let $(X, ⋆)$ be a binary structure with an identity element $e$ . Suppose that $x \in X$ has a left inverse $y$ and a right inverse $z$ . Then $y = z$ .

Proof.

We need only write down the chain of equalities

y = y ⋆ e = y ⋆ (x ⋆ z) = (y ⋆ x) ⋆ z = e ⋆ z = z .

□

With the concepts of identity elements and inverses in hand, we now give the full definition of a group.

Definition 2.1.8. §

A group is a set $G$ together with a binary operation $⋆ : G \times G \to G$ such that

i.: $⋆$ is associative,
ii.: there exists an element $e \in G$ such that $e ⋆ x = x = x ⋆ e$ , and
iii.: for every $x \in G$ , there exists an element $y \in G$ such that $x ⋆ y = e = y ⋆ x$ .

In other words, a group is a set with an associative binary operation, an identity element, and inverses with respect to that identity element.

Here are some examples of groups.

Examples 2.1.9. §

a.: Under addition, $ℤ$ , $ℚ$ , $ℝ$ , $ℂ$ , $M_{n} (ℝ)$ , and $Maps (ℝ, ℝ)$ are all groups.
b.: For $X = ℚ$ , $ℝ$ , or $ℂ$ , we set $X^{\times} = X - {0}$ . Under multiplication, $ℚ^{\times}$ , $ℝ^{\times}$ , and $ℂ^{\times}$ are groups.
c.: Under multiplication, the set ${GL}_{n} (ℝ)$ of invertible $n$ by $n$ -matrices (i.e., those with nonzero determinant) forms a group, known as the general linear group.
d.: Under multiplication, the set of nowhere vanishing functions in $Maps (ℝ, ℝ)$ forms a group.
e.: The set ${e}$ consisting of a single element is a group under the binary operation $⋆$ defined by $e ⋆ e = e$ . This group is known as the trivial group.

On the other hand, here are some of many binary structures that are not groups.

Examples 2.1.10. §

a.: The integers are not a group under multiplication, nor are $ℚ$ , $ℝ$ , or $ℂ$ before removing $0$ .
b.: The set $Maps (ℝ, ℝ)$ is not a group under composition, as not every function has an inverse.
c.: The set $𝒫_{X}$ of subsets of a set is not a group under union.

The following theorem is used in showing the uniqueness of inverses.

Proposition 2.1.11 (Cancellation theorem). §

Let $G$ be a group, and let $x, y, z \in G$ be such that

x ⋆ y = x ⋆ z (resp., y ⋆ x = z ⋆ x) .

Then $y = z$ .

Proof.

We prove the first statement. Let $x^{'}$ be any (left) inverse to $x$ . Under the given assumption, we have

y = e ⋆ y = (x^{'} ⋆ x) ⋆ z = x^{'} ⋆ (x ⋆ y) = x^{'} ⋆ (x ⋆ z) = (x^{'} ⋆ x) ⋆ z = e ⋆ z = z .

□

The following is now quickly derived.

Lemma 2.1.12. §

Let $G$ be a group. If $y, z \in G$ are both inverses to $x \in G$ on either the left or the right (or both), then $y = z$ .

Proof.

Suppose first that $y$ and $z$ are right inverses to $x$ . Then we have

x ⋆ y = e = x ⋆ z,

and the result now follows from the cancellation theorem. A similar argument holds if both $y$ and $z$ are right inverses. In fact, even if $y$ is a left inverse and $z$ is a right inverse, there is by definition of the group a third element $x^{'}$ in the group that is both a left an a right inverse, and so equals both $y$ and $z$ by what we have just proven. So $y$ and $z$ must be equal. □

Notation 2.1.13. §

Let $G$ be a group and $x \in G$ an element. Suppose the operation on $G$ is not denoted $+$ . Then we (almost invariably) use the following notation.

a.: The unique inverse to $x$ is written $x^{- 1}$ .
b.: Let $n \in ℤ$ . We set $x^{0} = e$ . If $n \geq 1$ , we usually write $x^{n}$ for $x ⋆ x ⋆ \dots ⋆ x$ , the product being of $n$ copies of $x$ , which is unambiguously defined by the associativity of $⋆$ .

If the binary operation on the group is denoted $+$ , then we write the inverse of $x$ as $- x$ and $𝑛𝑥$ instead of $x^{n}$ .

Remark 2.1.14. §

Let $G$ be a group and $e$ an element for which the operation is not denoted as $+$ . The reader should be able to check that for $x \in G$ and $m, n \in ℤ$ , one has

\begin{array}{rcl} x^{m + n} = x^{m} x^{n}, x^{𝑚𝑛} = {(x^{m})}^{n}, x^{n} = {(x^{- 1})}^{- n} = {(x^{- n})}^{- 1}, and e^{n} = e . \end{array}

Definition 2.1.15. §

Let $G$ be a group.

a.: We say that $G$ is abelian if its binary operation is commutative.
b.: We say that $G$ is nonabelian if its binary operation is not commutative.

Examples 2.1.16. §

a.: All of $ℤ$ , $ℚ$ , $ℝ$ , $ℂ$ , $M_{𝑚𝑛} (ℝ)$ , and $Maps (ℝ, ℝ)$ are abelian groups under addition.
b.: The groups $ℚ^{\times}$ , $ℝ^{\times}$ , and $ℂ^{\times}$ are abelian (under multiplication).
c.: The group ${GL}_{n} (ℝ)$ is nonabelian if $n \geq 2$ .

Remark 2.1.17. §

From now on, we will drop the use of $⋆$ for an arbitrary binary operation, and simply use the more conventional symbol $\cdot$ . However, the reader should keep in mind that this does not mean that the operation in question is multiplication. Moreover, we shall often write $x \cdot y$ more simply as $𝑥𝑦$ .

Lemma 2.1.18. §

Let $G$ be a group. For $x, y \in G$ , we have ${(𝑥𝑦)}^{- 1} = y^{- 1} x^{- 1}$ .

Proof.

We have

(y^{- 1} x^{- 1}) (𝑥𝑦) = y^{- 1} (x^{- 1} (𝑥𝑦)) = y^{- 1} ((x^{- 1} x) y) = y^{- 1} (𝑒𝑦) = y^{- 1} y = e .

Therefore $y^{- 1} x^{- 1}$ is left inverse to $𝑥𝑦$ , and so by Lemma 2.1.12 it equals ${(𝑥𝑦)}^{- 1}$ . □

We end this section with a few more examples of groups.

Example 2.1.19. §

The set $ℤ ∕ 𝑛ℤ$ of congruence classes modulo $n$ forms a group under the addition law

\bar{a} + \bar{b} = \bar{a + b} .

The identity is $\bar{0}$ , and the inverse of $\bar{a}$ is $\bar{- a}$ .

Clearly, $ℤ ∕ 𝑛ℤ$ is an abelian group.

Remark 2.1.20. §

Usually, we simply write $a$ for $\bar{a}$ . We have kept up the distinction to this point to make clear the difference between $a$ and its equivalence class. From now on, however, if we understand that we are working in $ℤ ∕ 7 ℤ$ , e.g., from context, we will write equations such as $5 + 2 = 0$ , with the fact that we are working with equivalence classes as above being understood.

Definition 2.1.21. §

The symmetric group $S_{X}$ on a set $X$ is the set

S_{X} = {f : X \to X ∣ f is bijective}

with the binary operation $\circ$ of composition.

Definition 2.1.22. §

Let $X$ be a set. An element of $S_{X}$ is referred to as a permutation of $X$ . We say that $σ \in S_{X}$ permutes the elements of $X$ .

Remark 2.1.23. §

The group $S_{X}$ is alternately referred to as the group of permutations of a set $X$ .

Remark 2.1.24. §

The group $S_{X}$ is nonabelian if $X$ has at least three elements.

Example 2.1.25. §

If $X = ℝ$ , then $f (x) = x + 1$ and $g (x) = x^{3}$ both lie in $S_{ℝ}$ , but do not commute.

Definition 2.1.26. §

When $X = {1, 2, \dots, n}$ , then we set $S_{n} = S_{X}$ , and we refer to $S_{n}$ as the symmetric group on $n$ letters.

Remark 2.1.27. §

The notion of isomorphism of binary structures carries over to groups. An isomorphism of groups is just an isomorphism of the underlying binary structures, i.e., a bijection $f : G \to G^{'}$ between groups $G$ and $G^{'}$ such that

f (x \cdot y) = f (x) \cdot f (y)

for each $x, y \in G$ . If $G$ and $G^{'}$ are isomorphic, we write $𝐺≅ G^{'}$ (noting that the property of being isomorphic forms an equivalence class on any set of groups).

Examples 2.1.28. §

a.

The group ${GL}_{1} (ℝ)$ is isomorphic to $ℝ^{\times}$ via the map $f : ℝ^{\times} \to {GL}_{1} (ℝ)$ defined by $f (a) = (a)$ .

b.

Let $X$ be a set with exactly $n$ elements, say $X = {x_{1}, x_{2}, \dots, x_{n}}$ . Then we define an isomorphism

f : S_{n} \to_{}^{\sim} S_{X}, f (σ) (x_{i}) = x_{σ (i)},

which is to say that $f$ takes a permutation $σ \in S_{n}$ that takes $i$ to some other number $j$ to the permutation in $S_{X}$ that maps $x_{i}$ to $x_{j}$ . In other words, it doesn’t matter whether we’re permuting $n$ cars or $n$ apples: the groups are isomorphic.

To every group, we have an associated opposite group.

Definition 2.1.29. §

The opposite group $G^{op}$ of a group $G$ is the set $G$ together with the operation $x ⋆ y = 𝑦𝑥$ for $x, y \in G$ .

Example 2.1.30. §

The opposite group of an abelian group is the original group.

2.2. Subgroups

Definition 2.2.1. §

A subset $H$ of a group $G$ is a subgroup if it is closed under the binary operation on $G$ and is a group with respect to the restriction of that operation to a binary operation on $H$ . If $H$ is a subgroup of $G$ , we write $H \leq G$ .

More succinctly, a subset of a group is a subgroup if it is a group with respect to the operation on the group.

Remark 2.2.2. §

The relation $\leq$ is a partial ordering on the set of subgroups of a group.

Definition 2.2.3. §

a.: The set ${e}$ containing only the identity element of $G$ is a subgroup of $G$ known as the trivial subgroup (as it is a trivial group that is also a subgroup).
b.: A subgroup $H$ of $G$ that is not the trivial subgroup is called nontrivial.

Definition 2.2.4. §

If $H$ is a subgroup of $G$ with $H \neq G$ , then we say that $H$ is a proper subgroup of $G$ , and we write $H < G$ .

Examples 2.2.5. §

The groups $ℤ$ , $ℚ$ , and $ℝ$ under addition are all subgroups of $ℂ$ .

To check that a group is a subgroup, one usually employs the following criteria.

Theorem 2.2.6. §

A subset $H$ of a group $G$ is a subgroup under the restriction on the binary operation $\cdot$ on $G$ if and only if

(0): $e \in H$ ,
(1): $H$ is closed under $\cdot$ ,
(2): if $h \in H$ , then $h^{- 1} \in H$ .

Proof.

If $H$ is a subgroup of $G$ with respect to $\cdot$ , then it is by definition closed under $\cdot$ . Since $H$ is a group under $\cdot$ , there exists an element $f \in H$ with $f \cdot h = h$ for all $h \in H$ . By the cancellation theorem, we then have $f = e$ , so $e \in H$ . Also, for each $h \in H$ , we have an element $h^{'} \in H$ with $h \cdot h^{'} = e$ . As $e = h \cdot h^{- 1}$ , the cancellation theorem again tells us that $h^{'} = h^{- 1}$ , so $h^{- 1} \in H$ . Therefore, the conditions (0)-(2) hold.

Conversely, if conditions (0)-(2) hold, then $H$ is a binary structure under $\cdot$ by (1) and (0) and (2) leave us only to verify associativity in the definition of a group. However, this follows automatically on $H$ from the associativity of $\cdot$ on the larger set $G$ . □

Examples 2.2.7. §

The subset $2 ℤ$ of $ℤ$ is a subgroup under $+$ . To see this, note that $0$ is even, the sum of two even integers is even, and the negative of an even integer is also even.

Example 2.2.8. §

The subset

{SL}_{n} (ℝ) = {A \in {GL}_{n} (ℝ) ∣ \det (A) = 1}

of ${GL}_{n} (ℝ)$ is a subgroup under $\cdot$ , known as the special linear group. We use Theorem 2.2.6 to check this:

(0)

We have $\det I_{n} = 1$ , so $I_{n} \in {SL}_{n} (ℝ)$ .

(1)

If $A, B \in {SL}_{n} (ℝ)$ , then

\det (A \cdot B) = \det (A) \cdot \det (B) = 1

so $A \cdot B \in {SL}_{n} (ℝ)$ .

(2)

If $A \in {SL}_{n} (R)$ , then

\det (A^{- 1}) = \det {(A)}^{- 1} = 1

so $A^{- 1} \in {SL}_{n} (ℝ)$ .

Example 2.2.9. §

Let

S^{1} = {z \in ℂ ∣ | z | = 1} = {e^{2 𝜋𝑖𝜃} ∣ 𝜃 \in ℝ} .

Here $e^{2 𝜋𝑖𝜃}$ corresponds to the point $(\cos 𝜃, \sin 𝜃)$ on the unit circle in the usual model of the complex plane. In fact, recall that

e^{2 𝜋𝑖𝜃} = \cos 𝜃 + i \sin 𝜃 \in ℂ

and

| \cos 𝜃 + i \sin 𝜃 | = \sqrt{\cos^{2} (𝜃) + \sin^{2} (𝜃)} = 1 .

Then $S^{1}$ is a subgroup of $ℂ^{\times}$ under $\cdot$ . To see this, we check:

(0): We have $| 1 | = 1$ .
(1): If $z, w \in S^{1}$ , then $z = e^{2 𝜋𝑖𝜃}$ and $w = e^{2 𝜋𝑖𝜓}$ for some $𝜃, ψ \in ℝ$ . We have
$𝑧𝑤 = e^{2 𝜋𝑖 (𝜃 + ψ)} \in S^{1} .$
(2): If $z = e^{2 𝜋𝑖𝜃}$ , then
$z^{- 1} = e^{2 𝜋𝑖 (- 𝜃)} \in S^{1} .$

Theorem 2.2.6 has the following shorter formulation.

Corollary 2.2.10. §

A nonempty subset $H$ of a group is a subgroup under the restriction of the binary operation $\cdot$ on $G$ if and only if $h \cdot k^{- 1} \in H$ for all $h, k \in H$ .

Proof.

If $H$ is a subgroup of $G$ and $h, k \in H$ , then $k^{- 1} \in H$ and, consequently, $h k^{- 1} \in H$ by Theorem 2.2.6. Conversely, suppose $h k^{- 1} \in H$ for all $h, k \in H$ . As $H$ is nonempty, let $h \in H$ . Using this critersion, we have successively that $e = h h^{- 1} \in H$ , $k^{- 1} = e k^{- 1} \in H$ , and $h𝑘 = h {(k^{- 1})}^{- 1} \in H$ , so Theorem 2.2.6 implies that $H$ is a subgroup. □

Definition 2.2.11. §

a.: A group $G$ is finite if its underlying set is finite. Otherwise, we say that $G$ is infinite.
b.: The order $| G |$ of a finite group $G$ is the order (number of elements in) of the underlying set. If $G$ is infinite, we say that its order is infinite.

The following provides an interesting example.

Lemma 2.2.12. §

Let $n \geq 1$ . The group $S_{n}$ is finite of order $n!$ .

Proof.

For an arbitrary element $σ \in S_{n}$ , we have $n$ choices for the value $σ (1)$ . Then $σ (2)$ can be any of the remaining $n - 1$ values, and $σ (3)$ is one of the then remaining $n - 2$ values, and so forth, until one value is left for $σ (n)$ . Therefore, the order of $S_{n}$ is $n \cdot (n - 1) \dots 1 = n!$ . □

Example 2.2.13. §

As a further subgroup of $S^{1}$ (so also a subgroup of $ℂ^{\times}$ ), we have

μ_{n} = {z \in ℂ^{\times} ∣ z^{n} = 1} = {e^{2 𝜋𝑖𝑘 ∕ n} ∣ k \in ℤ} .

To see the equality of the latter two sets, note that ${(e^{2 𝜋𝑖𝑘 ∕ n})}^{n} = 1$ . On the other hand $z^{n} = 1$ implies that $| z |^{n} = 1$ , so $| z | = 1$ , which means that $z = e^{2 𝜋𝑖𝜃}$ for some $𝜃 \in ℝ$ . But the only way that ${(e^{2 𝜋𝑖𝜃})}^{n} = 1$ can hold is for $𝑛𝜃$ to be an integer, which means exactly that $𝜃 = k ∕ n$ for some $n \in ℤ$ . Note that $| μ_{n} | = n$ , since $e^{2 𝜋𝑖𝑘 ∕ n} = e^{2 𝜋𝑖𝑗 ∕ n}$ if and only if $j \equiv k mod n$ . That this order equals $| ℤ ∕ 𝑛ℤ | = n$ is no coincidence. In fact, these two groups are isomorphic, as well shall see in the following section.

2.3. Cyclic groups

Definition 2.3.1. §

Let $G$ be a group, and let $g \in G$ . The cyclic subgroup of $G$ generated by $g$ is the group

⟨ g ⟩ = {g^{n} ∣ n \in ℤ} .

Lemma 2.3.2. §

Let $g \in G$ . Then $⟨ g ⟩$ is the smallest subgroup of $G$ containing $g$ .

Proof.

Since the smallest subgroup of $G$ containing $G$ is itself a group, it must contain $g^{n}$ for all $n \in ℤ$ , so it contains $⟨ g ⟩$ . On the other hand, we see that $⟨ g ⟩$ is a subgroup of $G$ since it contains $e = g^{0}$ , is closed under multiplication (as $g^{m} \cdot g^{n} = g^{m + n}$ ), and contains inverses (as ${(g^{n})}^{- 1} = g^{- n}$ ). Being that $⟨ g ⟩$ is a subgroup of $G$ contained in the smallest subgroup containing $g$ , it is itself the smallest subgroup. □

Examples 2.3.3. §

a.

The cyclic subgroup $⟨ 2 ⟩$ of $ℤ$ generated by $2$ is $2 ℤ$ .

b.

The cyclic subgroup of ${GL}_{2} (ℝ)$ generated by

A = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})

⟨ A ⟩ = {I_{2}, A, - I_{2}, - A} .

Definition 2.3.4. §

a.: A group $G$ is called cyclic if there exists $g \in G$ with $G = ⟨ g ⟩$ .
b.: An element of $g$ of a group $G$ is called a generator if $G = ⟨ g ⟩$ . We then say that $g$ generates $G$ and that $G$ is generated by $g$ .

Remark 2.3.5. §

Of course, any cyclic subgroup of a group $G$ is itself a cyclic group.

Examples 2.3.6. §

a.: The group $ℤ$ is cyclic, generated by $1$ .
b.: The group $ℤ ∕ 𝑛ℤ$ is cyclic for any $n \geq 0$ , again generated by $1$ .
c.: The group $μ_{n}$ is cyclic with generator $e^{2 𝜋𝑖 ∕ n}$ .
d.: The trivial group is a cyclic group of order $1$ .

Remark 2.3.7. §

Every cyclic group is abelian, since powers of a generator commute.

Definition 2.3.8. §

Let $G$ be a group. The order of an element $g \in G$ is the smallest positive integer $n$ such that $g^{n} = e$ , if it exists. If such an $n$ exists, then $g$ is said to have finite order, and otherwise $g$ is said to have infinite order.

Proposition 2.3.9. §

Let $g$ be an element in a group. Then the order of $⟨ g ⟩$ and the order of $g$ are equal if either is finite (and both infinite otherwise). Moreover, for any $i, j \in ℤ$ , we have $g^{i} = g^{j}$ if and only if

$i \equiv j mod n$ , if $g$ is finite of order $n$ , and
$i = j$ , if $g$ has infinite order.

Proof.

First, suppose that $g$ has finite order $n$ . If $g^{i} = g^{j}$ , then $g^{i - j} = e$ . Note that $g^{n} = e$ as well. Dividing $i - j$ by $n$ , we have

i - j = 𝑞𝑛 + r

for some quotient $q \in ℤ$ and remainder $0 \leq r \leq n - 1$ . We then have

e = g^{i - j} = g^{𝑞𝑛 + r} = {(g^{n})}^{q} g^{r} = g^{r},

but $r < n$ and $n$ is minimal, so $r = 0$ . That is, $i - j$ is a multiple of $n$ , so $i \equiv j mod n$ . In particular, the distinct elements of $⟨ g ⟩$ are exactly $e, g, \dots, g^{n - 1}$ , so $⟨ g ⟩$ has order $n$ .

If $g$ has infinite order, then for $g^{i} = g^{j}$ to hold, one must have $g^{i - j} = e$ , which forces $i = j$ . Therefore, all of the powers of $g$ are distinct, and $⟨ g ⟩$ is infinite. □

Lemma 2.3.10. §

Suppose that $G$ is a cyclic group. If $G$ is infinite, then $G$ is isomorphic to $ℤ$ . Otherwise, $G$ is isomorphic to $ℤ ∕ 𝑛ℤ$ , where $n = | G |$ .

Proof.

Let $g$ be a generator of $G$ . Suppose first that $G$ is infinite. We define a map

f : ℤ \to G, f (i) = g^{i} for all i \in ℤ .

This is one-to-one since $f (i) = f (j)$ implies $g^{i} = g^{j}$ , which can only happen if $i = j$ by Proposition 2.3.9. It is onto as every element of $⟨ g ⟩$ has the form $g^{i} = f (i)$ for some $i$ . It is then an isomorphism of groups as

f (i + j) = g^{i + j} = g^{i} g^{j} = f (i) f (j) .

If $| G | = n$ , then we define

f : ℤ ∕ 𝑛ℤ \to G, f (i) = g^{i} for all i \in ℤ .

This is well-defined as $f (i + 𝑞𝑛) = g^{i + 𝑞𝑛} = g^{i}$ , so it is independent of the choice of representative of $i$ modulo $n$ . It is one-to-one as $g^{i} = g^{j}$ implies $i = j$ in $ℤ ∕ 𝑛ℤ$ by Proposition 2.3.9. It is then onto and an isomorphism for the same reasons as in the infinite case. □

As a result, the group $μ_{n}$ is isomorphic to $ℤ ∕ 𝑛ℤ$ (under the map taking $e^{2 𝜋𝑖𝑘 ∕ n}$ to $k$ ). The groups $𝑛ℤ$ for $n \geq 1$ are all isomorphic to $ℤ$ itself, but for this one must take the map $ℤ \to 𝑛ℤ$ that is multiplcation by $n$ .

Theorem 2.3.11. §

Every subgroup of a cyclic group is cyclic.

Proof.

Let $H$ be a subgroup of a cyclic group $G$ with generator $g$ . Let $k \geq 1$ be minimal such that $g^{k} \in H$ . We claim that $H = ⟨ g^{k} ⟩$ . Since $H$ is closed under multiplication and inverses, it must contain every power of $g^{k}$ , so it contains the subgroup $⟨ g^{l} ⟩$ . Now suppose that $g^{i} \in H$ for some $i \in ℤ$ . Again, divide $i$ by $k$ and get $q \in ℤ$ and $0 \leq r \leq k - 1$ with $i = 𝑞𝑘 + r$ . Then $g^{i} = {(g^{k})}^{q} g^{r}$ , so

g^{r} = g^{i} {(g^{k})}^{- q} \in H,

in that $H$ is a subgroup. But minimality forces $r = 0$ , so $i$ is a multiple of $k$ , proving the claim. □

Corollary 2.3.12. §

The subgroups of $ℤ$ are exactly the $𝑛ℤ = ⟨ n ⟩$ with $n$ a nonnegative integer.

Let’s consider the subgroups of $ℤ ∕ 𝑛ℤ$ for some $n \geq 1$ , which we now know to be cyclic. Recall that the greatest common divisor $\gcd (i, j)$ of two integers $i$ and $j$ that are not both zero is defined to be the smallest positive integer dividing both $i$ and $j$ . We also set $\gcd (0, 0) = 0$ .

Lemma 2.3.13. §

Given $i, j \in ℤ$ , we have

⟨ \gcd (i, j) ⟩ = {𝑎𝑖 + 𝑏𝑗 ∣ a, b \in ℤ} .

Proof.

In the case that $i = j = 0$ , we have that both sides equal $⟨ 0 ⟩$ , so the lemma holds, and therefore we may assume that at least one is nonzero. Since $\gcd (i, j)$ divides both $i$ and $j$ , we have $i, j \in ⟨ \gcd (i, j) ⟩$ . As a subgroup, the latter group is closed under addition and taking of negatives, so $𝑎𝑖 + 𝑏𝑗$ is in it as well. In other words, $H = {𝑎𝑖 + 𝑏𝑗 ∣ a, b \in ℤ}$ is contained in $⟨ \gcd (i, j) ⟩$ .

Conversely, note that the set $H$ is a (nontrivial) subgroup of $ℤ$ in that it satisfies all of the properties of one, so it equals $⟨ d ⟩$ for some $d \geq 1$ . Since $i, j \in ⟨ d ⟩$ by definition, we have that $d$ divides both $i$ and $j$ , and therefore is less than or equal to $\gcd (i, j)$ . On the other hand, we know that $d \in ⟨ \gcd (i, j) ⟩$ , so $\gcd (i, j) \leq d$ , and therefore $d = \gcd (i, j)$ . In other words, we have $H = ⟨ \gcd (i, j) ⟩$ . □

Proposition 2.3.14. §

Every subgroup of $ℤ ∕ 𝑛ℤ$ has the form $⟨ d ⟩$ for some $d \geq 1$ dividing $n$ . In fact, for any $j \in ℤ$ , we have $⟨ j ⟩ = ⟨ \gcd (j, n) ⟩$ .

Proof.

The second statement implies the first, so we focus on it. Since $\gcd (j, n)$ divides $j$ , we have that $⟨ j ⟩ \leq ⟨ \gcd (j, n) ⟩$ . On the other hand, we have by Lemma 2.3.13 that

\gcd (j, n) \in {𝑎𝑗 + 𝑏𝑛 ∣ a, b \in ℤ}

inside $ℤ$ , which means that $\gcd (j, n) \equiv 𝑎𝑗 mod n$ for some $a \in ℤ$ . In other words, in $ℤ ∕ 𝑛ℤ$ , we have $\gcd (j, n) \in ⟨ j ⟩$ , so $⟨ \gcd (j, n) ⟩ \leq ⟨ j ⟩$ , as desired. □

Remark 2.3.15. §

The subgroup $⟨ n ⟩$ of $ℤ ∕ 𝑛ℤ$ is just the trivial subgroup $⟨ 0 ⟩ = {0}$ .

Recall that two integers are said to be relatively prime if their greatest common divisor is $1$ .

Corollary 2.3.16. §

Let $G$ be a group and $g \in G$ an element of order $n$ .

a.: For $i \in ℤ$ , the order of $g^{i}$ is $n ∕ d$ , where $d = \gcd (i, n)$ , and $⟨ g^{i} ⟩ = ⟨ g^{d} ⟩$ .
b.: The generators of $⟨ g ⟩$ are the $g^{i}$ with $i$ relatively prime to $n$ .

Proof.

Consider the isomorphism $ϕ : G \to ℤ ∕ 𝑛ℤ$ under which $g^{i}$ is taken to $i$ . This carries the subgroup $⟨ g^{i} ⟩$ bijectively to the subgroup $⟨ i ⟩$ , which by Proposition 2.3.14 equals $⟨ d ⟩$ . But the latter group has elements $0, d, 2 d, \dots, (n ∕ d - 1) d$ , so has order $n ∕ d$ . As $ϕ$ is a bijection, part a is then seen to hold. Part b then follows immediately from part a, as the $i$ for which $⟨ g^{i} ⟩ = ⟨ g ⟩$ are the $i$ with $\gcd (i, n) = 1$ . □

Definition 2.3.17. §

The Euler phi-function is the map $φ : ℤ_{> 0} \to ℤ_{> 0}$ such that $φ (n)$ is the number of relatively prime integers to $n$ between $1$ and $n$ .

Remark 2.3.18. §

The Euler phi-function $φ$ has the properties that $φ (𝑚𝑛) = φ (m) φ (n)$ whenever $\gcd (m, n) = 1$ and that $φ (p^{r}) = p^{r - 1} (p - 1)$ for a prime number $p$ and $r \geq 1$ . Its values on $1, 2, 3, 4, 5, \dots$ are $1, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8, 8, 16, \dots$ .

Remark 2.3.19. §

It follows from Corollary 2b that the number of generators of a cyclic group $G$ of order $n$ is exactly $φ (n)$ , where $φ$ is the Euler phi-function.

2.4. Generators

The relation $\leq$ is a partial ordering on any set of subgroups of a group. The following proposition asserts the existence of minimal elements of certain such subsets. It is a consequence of Lemma 1.2.24, but prove it here for convenience.

Proposition 2.4.1. §

Let $G$ be a group, and let $S$ be a nonempty subset of $G$ . Then there exists a smallest subgroup $⟨ S ⟩$ of $G$ containing $S$ .

Proof.

The set $P_{S}$ of subgroups of $G$ containing $S$ is nonempty, for it contains $G$ itself. Set

⟨ S ⟩ = ⋂_{H \in X} H .

As each $H \in X$ contains $G$ , so does $⟨ S ⟩$ . Moreover, an arbitrary intersection of subgroups of $G$ is easily verified to itself be a subgroup of $G$ , so $⟨ S ⟩$ is a subgroup. Finally, if $H$ is any subgroup of $G$ containing $S$ , then $H \in X$ , so $⟨ S ⟩ \leq H$ by definition of the intersection, so $⟨ S ⟩$ is the smallest such subgroup (i.e., the unique minimal element of $X$ ). □

Definition 2.4.2. §

The smallest subgroup $⟨ S ⟩$ containing a set $S$ is the subgroup of $G$ generated by $S$ .

While this definition is rather abstract, we do have the following more concrete description of the elements of $⟨ S ⟩$ .

Proposition 2.4.3. §

Let $S$ be a nonempty subset of $G$ . An element $g \in G$ is contained in $⟨ S ⟩$ if and only if $g$ may be written as a product of powers of elements of $S$ : i.e.,

g = s_{1}^{m_{1}} s_{2}^{m_{2}} \dots s_{k}^{m_{k}}

for some $k \geq 0$ , $s_{i} \in S$ and $m_{i} \in ℤ$ for $1 \leq i \leq k$ .

Proof.

First, let $g$ be an element that is a product of powers of elements of $S$ . Since $⟨ S ⟩$ is a subgroup, it is closed under integer powers and products, so $g \in ⟨ S ⟩$ .

Conversely, note that the set $H$ of elements that are products of powers of elements of $S$ is a subgroup of $G$ , as it contains $e = s^{0}$ for $s \in S$ , is closed under products by definition, and is closed under inverses as

{(s_{1}^{m_{1}} s_{2}^{m_{2}} \dots s_{k}^{m_{k}})}^{- 1} = s_{k}^{- m_{k}} \dots s_{2}^{- m_{2}} s_{1}^{- m_{1}} .

As $H$ is a subgroup of $G$ containing $S$ but contained in $⟨ S ⟩$ and $⟨ S ⟩$ is the minimal such subgroup, we have $H = ⟨ S ⟩$ . Thus, any element of $⟨ S ⟩$ may be written as a product of powers of elements of $S$ , as desired. □

Definition 2.4.4. §

We say that a subset $S$ of $G$ generates $G$ if $G = ⟨ S ⟩$ , and then $S$ is said to be a set of generators of $G$ .

Definition 2.4.5. §

We say that a group $G$ is finitely generated if there exists a finite set of generators of $G$ .

Remark 2.4.6. §

If $G$ can be generated by a finite set ${g_{1}, g_{2}, \dots, g_{n}}$ , we usually write

⟨ g_{1}, g_{2}, \dots, g_{n} ⟩

instead of

⟨ {g_{1}, g_{2}, \dots, g_{n}} ⟩,

and we say that $G$ is generated by $g_{1}, g_{2}, \dots, g_{n}$ .

Example 2.4.7. §

A cyclic group is finitely generated: in fact, it is generated by a single element.

Example 2.4.8. §

Any finite group is finitely generated, as it is generated by itself.

Example 2.4.9. §

Consider the subgroup $G$ of ${GL}_{2} (ℝ)$ that is

G = {(\begin{matrix} {(- 1)}^{i} & b \\ 0 & {(- 1)}^{j} \end{matrix}) | i, j, b \in ℤ} .

It can be generated by the set

{(\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}), (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})} .

To see this, note that

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

and

{(\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})}^{j} (\begin{matrix} 1 & b \\ 0 & 1 \end{matrix}) {(\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix})}^{i} = (\begin{matrix} {(- 1)}^{i} & b \\ 0 & {(- 1)}^{j} \end{matrix}) .

The group $G$ is not cyclic as it is infinite but contains elements of order $2$ , but all infinite cyclic groups are isomorphic to $ℤ$ . In fact, $G$ cannot be generated by any two of its elements: the proof of this more tricky fact is left to the reader.

Example 2.4.10. §

The group $ℚ$ can be generated by the set ${\frac{1}{n} ∣ n \geq 1}$ . However, $ℚ$ is not finitely generated. For, any integer $N \geq 1$ and nonzero integers $a_{i}$ , $b_{i}$ with $b_{i} > 0$ for $1 \leq i \leq N$ . Any element of

⟨ \frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \dots, \frac{a_{N}}{b_{N}} ⟩

must have denominator, when put in reduced form, that is a divisor of $b_{1} b_{2} \dots b_{N}$ . But clearly not every fraction has such a denominator, so $ℚ$ cannot be finitely generated.

2.5. Direct products

Given any two groups, we can form a new group out of them, known as the direct product, whose underlying set is in fact exactly the direct product of the underlying sets of the groups in question.

Definition 2.5.1. §

Let $G$ and $G^{'}$ be groups. The direct product of $G$ and $G^{'}$ is the binary structure $G \times G^{'}$ that is the direct product of the sets $G$ and $G^{'}$ together with the binary operation defined by

(a, a^{'}) \cdot (b, b^{'}) = (a \cdot b, a^{'} \cdot b^{'})

for $a, b \in G$ and $a^{'}, b^{'} \in G^{'}$ .

One might expect the direct product of $G$ and $G^{'}$ to be a group, and in fact it is. The straightforward check is left to the reader.

Lemma 2.5.2. §

The direct product $G \times G^{'}$ of two groups is a group.

Of course, using this construction, we can think up more examples of new groups than we can mention, e.g., $S_{m} \times {GL}_{n} (ℝ)$ for any $m, n \geq 1$ . The following remarks are easily verified from the definition of the direct product.

Remark 2.5.3. §

The group $G \times G^{'}$ is abelian if and only if both $G$ and $G^{'}$ are abelian.

Remark 2.5.4. §

If $f : G \to H$ is a group isomorphism and $G^{'}$ is another group, then the map

f^{'} : G \times G^{'} \to H \times G^{'}

given by $f^{'} (g, g^{'}) = (f (g), g^{'})$ for $g \in G$ and $g^{'} \in G^{'}$ is an isomorphism as well.

Remark 2.5.5. §

Direct product forms an associative and commutative binary operation on any set of isomorphism classes of groups. That is, for any groups $G_{1}$ , $G_{2}$ , and $G_{3}$ , we have

\begin{array}{rcl} (G_{1} \times G_{2}) \times G_{3} ≅ G_{1} \times (G_{2} \times G_{3}) & and & G_{1} \times G_{2} ≅ G_{2} \times G_{1} . \end{array}

In particular, the associativity means it makes sense to speak of the group

G_{1} \times G_{2} \times \dots \times G_{n}

for any groups $G_{1}$ , $G_{2}$ , … $G_{n}$ .

Remark 2.5.6. §

If each of the groups $G_{1}, G_{2}, \dots, G_{n}$ is finite, then

| G_{1} \times G_{2} \times \dots \times G_{n} | = \prod_{i = 1}^{n} | G_{i} | .

Notation 2.5.7. §

We write $G^{n}$ for the direct product $G \times G \times \dots \times G$ of $n$ copies of $G$ .

Remark 2.5.8. §

More generally, for any collection

{G_{i} ∣ i \in I}

of groups $G_{i}$ for $i$ in some indexing set $I$ , we can put a binary operation on the direct product set

\prod_{i \in I} G_{i}

given by coordinate-wise multiplication

{(a_{i})}_{i \in I} \cdot {(b_{i})}_{i \in I} = {(a_{i} \cdot b_{i})}_{i \in I},

and the resulting group is known as the direct product of the $G_{i}$ .

Let $n \geq 1$ , and let $G_{i}$ be a group for each $1 \leq i \leq n$ . Let

G = G_{1} \times G_{2} \times \dots \times G_{n} .

For $g \in G_{i}$ , let $g^{(i)} \in G$ denote the element

g^{(i)} = (e, \dots, e, g, e, \dots, e) \in G

that is nontrivial in only the $i$ th coordinate of $G$ and $g$ in the $i$ th coordinate.

Proposition 2.5.9. §

Suppose that $S_{i}$ is a generating set of $G_{i}$ for each $1 \leq i \leq n$ . Then

S = ⋃_{i = 1}^{n} {g^{(i)} ∣ g \in S_{i}}

is a generating set of $G$ .

Proof.

Suppose $g_{i} \in G_{i}$ for each $1 \leq i \leq n$ . Then

(g_{1}, g_{2}, \dots, g_{n}) = g_{1}^{(1)} g_{2}^{(2)} \dots g_{n}^{(n)} \in ⟨ S ⟩ .

□

For example, if each $G_{i}$ is cyclic with generator $g_{i}$ , then the set ${g_{i}^{(i)} ∣ 1 \leq i \leq n}$ generates $G$ . While it is immediate from Proposition 2.5.9 that finite direct products of finitely generated groups are finitely generated, infinite direct products of nontrivial groups are never finitely generated.

Example 2.5.10. §

The group

G = \prod_{i = 1}^{\infty} (ℤ ∕ 2 ℤ)

is not finitely generated. We give a very brief sketch of the proof: one checks that any finite set of elements $X$ in $G$ must have the property that there exist positive integers $j$ and $k$ such that for each $x = (x_{i}) \in X$ , we have $x_{j} = x_{k}$ . Then every element in $⟨ X ⟩$ has this property, and since not every element of $G$ has this property, we have $⟨ X ⟩ \neq G$ .

The following result gives a general recipe for determining the order of $(g_{1}, g_{2}, \dots, g_{n})$ .

Theorem 2.5.11. §

Suppose that $g_{i} \in G_{i}$ for each $1 \leq i \leq n$ . The order of $g = (g_{1}, g_{2}, \dots, g_{n})$ is the least common multiple of the orders of the $g_{i}$ if each of the elements $g_{i}$ has finite order, and otherwise $g$ has infinite order.

Proof.

We have

g^{m} = (g_{1}^{m}, g_{2}^{m}, \dots, g_{n}^{m}),

and this is the identity if and only if $m$ is a multiple of the orders of each of the $g_{i}$ , so infinite if any one of them is infinite, and otherwise a multiple of the least common multiple. □

Example 2.5.12. §

Let $G = ℤ ∕ 2 ℤ \times ℤ ∕ 3 ℤ \times ℤ ∕ 4 ℤ \times ℤ ∕ 4 ℤ$ . Then every element of $G$ has order dividing $lcm (2, 3, 4, 4) = 12$ .

The latter example illustrates a more general phenomenon.

Definition 2.5.13. §

The exponent of a group $G$ is the smallest integer $n \geq 1$ such that $g^{n} = e$ for all $g \in G$ , if it exists. Otherwise, it is infinite.

Corollary 2.5.14. §

If $G_{i}$ has exponent $n_{i}$ for each $1 \leq i \leq n$ , then the exponent of $G$ is the least common multiple of the $n_{i}$ .

We mention the following result, the proof of which we leave to the reader.

Proposition 2.5.15. §

Suppose that ${G_{i} ∣ i \in I}$ is a collection of groups and, for each $i \in I$ , we are given $H_{i} \leq G_{i}$ . Then we have

\prod_{i \in I} H_{i} \leq \prod_{i \in I} G_{i} .

Note, however, that not all subgroups of a direct product are direct products of subgroups.

Example 2.5.16. §

There are $5$ subgroups of the Klein four group $ℤ ∕ 2 ℤ \times ℤ ∕ 2 ℤ$ :

{0}, ℤ ∕ 2 ℤ \times ℤ ∕ 2 ℤ, ⟨ (1, 0) ⟩, ⟨ (0, 1) ⟩, and ⟨ (1, 1) ⟩ .

The first four sit inside $ℤ ∕ 2 ℤ \times ℤ ∕ 2 ℤ$ as direct products of subgroups in the two individual coordinates, while the final subgroup does not.

Finally, we note the following interesting fact.

Theorem 2.5.17. §

Let $m$ and $n$ be relatively prime positive integers. Then the natural map

𝜃_{𝑚𝑛} : ℤ ∕ 𝑚𝑛ℤ \to ℤ ∕ 𝑚ℤ \times ℤ ∕ 𝑛ℤ

induced by $a \mapsto (a, a)$ is an isomorphism. On the other hand, if $m$ and $n$ are not relatively prime, then $ℤ ∕ 𝑚𝑛ℤ$ and $ℤ ∕ 𝑚ℤ \times ℤ ∕ 𝑛ℤ$ are not isomorphic.

Proof.

Suppose that $m$ and $n$ are relatively prime. Note that

𝜃_{𝑚𝑛} (a + b) = (a + b, a + b) = (a, a) + (b, b) = 𝜃_{𝑚𝑛} (a) + 𝜃_{𝑚𝑛} (b),

so $𝜃_{𝑚𝑛}$ preserves the operation. If $(a, a) = (b, b)$ in $ℤ ∕ 𝑚ℤ \times ℤ ∕ 𝑛ℤ$ , then $m$ and $n$ both divide $a - b$ , so $𝑚𝑛$ does, as they are relatively prime. Therefore, $𝜃_{𝑚𝑛}$ is injective. Since both groups have the same order $𝑚𝑛$ , it is surjective as well.

If $m$ and $n$ are not relatively prime, then their least common multiple is

lcm (m, n) = \frac{𝑚𝑛}{\gcd (m, n)} < 𝑚𝑛 .

Corollary 2.5.14 then implies that the exponent of $ℤ ∕ 𝑚ℤ \times ℤ ∕ 𝑛ℤ$ is less than $𝑚𝑛$ , the exponent of $ℤ ∕ 𝑚ℤ \times ℤ ∕ 𝑛ℤ$ . As the exponent of a group is preserved by an isomorphism, the two groups in question cannot be isomorphic. □

The following equivalent corollary is known as the Chinese remainder theorem (CRT).

Corollary 2.5.18 (Chinese Remainder Theorem). §

Let $k \geq 2$ and $m_{1}, \dots, m_{k}$ be mutually relatively prime positive integers, which is to say that every pair of them is relatively prime. For any $b_{1}, \dots, b_{k} \in ℤ$ , there exists an integer $a$ , unique up to congruence modulo $m_{1} m_{2} \dots m_{k}$ , such that $a \equiv b_{i} mod m_{i}$ for each $1 \leq i \leq k$ .

Proof.

The existence in the case $k = 2$ is equivalent to the surjectivity of $𝜃_{m_{1} m_{2}}$ in Theorem 2.5.17, while the uniqueness is its injectivity. The case of general $k$ follows by an easy induction on $k$ . □

Remark 2.5.19. §

We can give an explicit recipe for the construction of solutions of congruences modulo relatively prime integers (in the case of two congruences, and then by recursion). The construction is contained in the following direct proof that the map $𝜃_{𝑚𝑛}$ in Theorem 2.5.17 is surjective:

Suppose that $b \in ℤ ∕ 𝑚ℤ$ and $c \in ℤ ∕ 𝑛ℤ$ . Let $x, y \in ℤ$ be such that $𝑚𝑥 + 𝑛𝑦 \equiv 1 mod 𝑚𝑛$ , which we can find since $\gcd (m, n) = 1$ . Then $x$ is inverse to $m$ in $ℤ ∕ 𝑛ℤ$ , and $y$ is inverse to $n$ in $ℤ ∕ 𝑚ℤ$ . Therefore, we have that

𝜃_{𝑚𝑛} (𝑐𝑚𝑥 + 𝑏𝑛𝑦) = (b, c) \in ℤ ∕ 𝑚ℤ \times ℤ ∕ 𝑛ℤ .

Example 2.5.20. §

Suppose we want to find $a \in ℤ$ with

\begin{array}{rcl} a \equiv 2 mod 7 & and & a \equiv 1 mod 5 . \end{array}

We note that $3$ is an inverse of $7$ modulo $5$ , and it is also an inverse of $5$ modulo $7$ . So, in the proof of surjectivity in Theorem 2.5.17, we have $m = 7$ , $n = 5$ , $b = 2$ , $c = 1$ , $x = 3$ , and $y = 3$ , so

𝑐𝑚𝑥 + 𝑏𝑛𝑦 = 1 \cdot 7 \cdot 3 + 2 \cdot 5 \cdot 3 = 51 \equiv 16 mod 35 .

Therefore $a = 16$ is the unique integer satisfying the two congruences. Moreover, note that $x$ and $y$ are independent of $b$ and $c$ , so we can use these $x$ and $y$ in solving any two congruences modulo $7$ and $5$ .

Example 2.5.21. §

We can use Theorem 2.5.17 to find isomorphisms between direct products of cyclic groups. For instance, using this and Remarks 2.5.4 and 2.5.5, we have that

ℤ ∕ 12 ℤ \times ℤ ∕ 8 ℤ≅ℤ ∕ 4 ℤ \times ℤ ∕ 3 ℤ \times ℤ ∕ 8 ℤ≅ℤ ∕ 4 ℤ \times ℤ ∕ 24 ℤ .

2.6. Groups of isometries

Definition 2.6.1. §

Let $n \geq 1$ . An isometry of $ℝ^{n}$ is a bijection $f : ℝ^{n} \to ℝ^{n}$ that preserves distances, which is to say, for every $x, y \in ℝ^{n}$ , we have

| f (x) - f (y) | = | x - y |,

where the absolute value denotes the distance from $0$ in $ℝ^{n}$ .

Since the property of preserving distances is preserved by composition, the following lemma is easily seen.

Lemma 2.6.2. §

The set $I_{n}$ of isometries of $ℝ^{n}$ forms a group under composition.

Proposition 2.6.3. §

If $f \in I_{1}$ , then there exists $a \in ℝ$ such that $f (x) = a + x$ or $f (x) = a - x$ .

Remark 2.6.4. §

The function $f (x) = a + x$ is known as a translation, while the function $f (x) = a - x$ is known as a reflection (about the point $x = a$ ).

The group of isometries of $ℝ^{2}$ is a much more complicated group. We state, without proof, the following theorem.

Theorem 2.6.5. §

Every isometry $f$ of $ℝ^{2}$ has one of the following four forms.

i.: $f$ is a translation: there exists $a \in ℝ^{2}$ such that $f (x) = x + a$ .
ii.: $f$ is a reflection: there exists a line $L$ in $ℝ^{2}$ such that $f (x)$ is the reflection of $x$ across $L$ .
iii.: $f$ is a rotation: there exist $a \in ℝ^{2}$ and $𝜃 \in [0, 2 π)$ such that $f$ is given by counterclockwise rotation by $𝜃$ radians about the center $a$ .
iv.: $f$ is a glide reflection: $f$ is the composition of a reflection and followed by a translation by a nonzero distance in a direction parallel to the line of reflection.

Remark 2.6.6. §

Isometries of $ℝ^{2}$ are either orientation-preserving (i.e., the translations and the rotations) or orientation-reversing (i.e., the reflections and glide-reflections). An isometry that preserves orientation will map the letter “S” drawn in the plane to another letter than looks like an “S”, while an orientation-reversing isometry will map it to a backwards “S”. The composition of two orientation-preserving or two orientation-reversing isometries is orientation-preserving, while the composition of an orientation-preserving isometry with an orientation-reversing isometry (in either order) is orientation-reversing.

Definition 2.6.7. §

Suppose that $X \subseteq ℝ^{n}$ . We say that $f \in I_{n}$ is a symmetry of $X$ if for every $x \in ℝ^{n}$ one has $f (x) \in X$ if and only if $x \in X$ .

The condition of $f$ being a symmetry of $X$ insures that the restriction of $f$ to a map from $X$ to $X$ is a bijection.

Definition 2.6.8. §

For $n \geq 3$ , the dihedral group $D_{n}$ is the group of symmetries of a regular $n$ -gon, which we can take to be inscribed about the unit circle around the origin of $ℝ^{2}$ with a vertex at $(1, 0)$ .

Note that a different choice of regular $n$ -gon in $ℝ^{2}$ simply leads to an isomorphic group.

Proposition 2.6.9. §

For $n \geq 3$ , the dihedral group $D_{n}$ is a group of order $2 n$ , consisting of $n$ rotations about the origin by multiples of $\frac{2 π}{n}$ radians and $n$ reflections. In the case that $n$ is odd, these reflections are through lines through a vertex and a midpoint of the opposite side. In the case that $n$ is even, $\frac{n}{2}$ of these reflections are through two opposite vertices and the other $\frac{n}{2}$ of them are through midpoints of two opposite sides.

Proof.

The above-described rotations and reflections are all easily seen to be symmetries of the regular $n$ -gon in question. We must see that these are the only ones. Any nonzero translation or glide reflection moves the origin of $ℝ^{2}$ , and therefore moves the center of the polygon, hence gannot be a symmetry. Any rotation must be about the origin, or it too will move the center, and any rotation about the origin must take a vertex to a vertex, hence be by an angle that is a multiple of $2 π ∕ n$ . Any reflection must for the same reason be a reflection across a line through the origin. If the line determining such a reflection does not cross a midpoint or vertex, then it will move the closest vertex on either side it passes through to a point which is less than the distance of a side of the polygon away, hence not to another vertex. Therefore, it must pass through the origin and either a midpoint or a vertex, and all such lines of reflection are described in the statement of the proposition. □

Proposition 2.6.10. §

Let $r \in D_{n}$ be counterclockwise rotation about the origin by $\frac{2 π}{n}$ radians, and let $s \in D_{n}$ be the reflection across the $x$ -axis. These two elements satisfy $r^{n} = 1$ , $s^{2} = 1$ , and

𝑠𝑟 = r^{- 1} s,

and every element in $D_{n}$ may be written uniquely in the form $r^{j} s^{k}$ with $0 \leq j \leq n - 1$ and $0 \leq k \leq 1$ .

Proof.

The proposition boils down to the assertions that the $r^{j}$ are all of the rotations and the $r^{j} s$ are all of the reflections. Since $r^{j}$ is exactly rotation counterclockwise by $\frac{2 𝜋𝑗}{n}$ radians, the first of these assertions holds. It is easy to see that if we first rotate across the $x$ -axis and then rotate counterclockwise by $\frac{2 𝜋𝑗}{n}$ radians, it is the same as reflecting across the line that is $\frac{𝜋𝑗}{n}$ radians counterclockwise from the $x$ -axis. These lines pass alternately through vertices and midpoints for even and odd $j$ , respectively, and hence are all of the reflections. □

Corollary 2.6.11. §

We have $D_{n} = ⟨ r, s ⟩$ , for $r$ and $s$ as in Proposition 2.6.10.

There are many other interesting objects of which one can consider the symmetries, even in the plane. A pattern of finite, nonzero width and height that is repeated over an over infinitely in one direction has a symmetry group that is known as a “frieze group”, while a pattern of finite, nonzero width and height that is repeated over and over in two non-parallel directions is known as a “wallpaper group”.

2.7. Symmetric groups

Let $n$ be a positive integer. In this section, we study the symmetric group $S_{n}$ . Recall that an element of $S_{n}$ is a bijection $σ : X_{n} \to X_{n}$ , where $X_{n}$ is the set ${1, 2, \dots, n}$ . It is common to denote the element $σ$ of $S_{n}$ by

σ = (\begin{matrix} 1 & 2 & \dots & n \\ σ (1) & σ (2) & \dots & σ (n) \end{matrix}) .

Example 2.7.1. §

The permutation

σ = (\begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 5 & 2 & 1 & 4 \end{matrix})

is the element of $S_{5}$ with values $σ (1) = 3$ , $σ (2) = 5$ , $σ (3) = 2$ , $σ (4) = 1$ , and $σ (5) = 4$ .

This notation for permutations is amenable to composition.

Example 2.7.2. §

Let $σ$ be as in Example 2.7.1, and let

τ = (\begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 1 & 4 & 3 & 5 \end{matrix}) .

To compute $𝜎𝜏$ , we write a three-by-three matrix with the top two rows given by the notation for $τ$ and the next row determined by where $σ$ takes the elements $1$ through $5$ , i.e., we put $σ (i)$ below $i$ for each $i$ in the second row. This reads

(\begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 1 & 4 & 3 & 5 \\ 5 & 3 & 1 & 2 & 4 \end{matrix}) .

The first and third rows of the latter matrix then yield $𝜎𝜏$ :

𝜎𝜏 = (\begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 5 & 3 & 1 & 2 & 4 \end{matrix}) .

Taking inverses is even easier: one merely switches the two rows.

Example 2.7.3. §

Let $σ$ be as in Example 2.7.1. Switching its two rows, we obtain

(\begin{matrix} 3 & 5 & 2 & 1 & 4 \\ 1 & 2 & 3 & 4 & 5 \end{matrix}),

and reordering the top row in the order $1$ through $5$ , while preserving the columns by reordering the bottom row in the same fashion, we obtain

σ^{- 1} = (\begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 4 & 3 & 1 & 5 & 2 \end{matrix}),

Definition 2.7.4. §

Let $σ \in S_{n}$ . The orbit of $x \in X_{n}$ under $σ$ is

O_{σ} (x) = {σ^{j} (x) ∣ j \in ℤ} .

Example 2.7.5. §

Let $σ$ be as in Example 2.7.1 and $τ$ be as in Example 2.7.2. Then

O_{σ} (1) = {1, 2, 3, 4, 5},

while

\begin{array}{rcl} O_{τ} (1) = {1, 2}, & O_{τ} (3) = {3, 4}, & O_{τ} (5) = {5} . \end{array}

Proposition 2.7.6. §

The relation $\sim$ on $X_{n}$ given by $x \sim y$ if and only if $y = σ^{i} (x)$ for some $i \in ℤ$ is an equivalence relation.

Proof.

For $x \in X$ , we have $x = σ^{0} (x)$ , so $x \sim x$ . For $x, y \in X$ with $x \sim y$ , there exists $i \in ℤ$ with $y = σ^{i} (x)$ , and then $x = σ^{- i} (y)$ , so $y \sim x$ . For $x, y, z \in X$ with $x \sim y$ and $y \sim z$ , we have $i, j \in ℤ$ such that $y = σ^{i} (x)$ and $z = σ^{j} (y)$ . We then have

z = σ^{j} (y) = σ^{j} (σ^{i} (x)) = σ^{i + j} (x),

so $x \sim z$ . □

By definition, $O_{σ} (x)$ is the equivalence class of $x$ under the equivalence relation defined in Propositiion 2.7.6.

Definition 2.7.7. §

a.: For $k \geq 2$ , a $k$ -cycle in $S_{n}$ is an element of $S_{n}$ that has one orbit with $k$ elements, and for which all the other orbits have only one element each.
b.: A cycle is a permutation that is a $k$ -cycle for some $k \geq 2$ .
c.: The length of a cycle $σ$ is the integer $k \geq 2$ such that $σ$ is a $k$ -cycle.

Example 2.7.8. §

The element $σ$ of Example 2.7.1 is a $5$ -cycle, but $τ$ as in Example 2.7.2 is not a cycle.

We have another notation for permutations, which depends on their orbit decomposition. We begin with the case of a cycle.

Notation 2.7.9. §

Suppose that $σ$ is a $k$ -cycle, and let $x$ be an element in its largest orbit. We use

(x σ (x) \dots σ^{k - 1} (x))

to denote the element $σ$ .

Remark 2.7.10. §

When a group $G$ is viewed as a subgroup of a symmetric group (i.e., as consisting of permutations of some set), any non-identity element of $G$ is called a nontrivial element of $G$ , and the identity element is called trivial, or the trivial element.

There are $k$ different ways to write a $k$ -cycle in the form of Definition 2.7.9.

Example 2.7.11. §

The $5$ -cycle $σ$ of Example 2.7.1 is equal to

(1 3 2 5 4) = (3 2 5 4 1) = (2 5 4 1 3) = (5 4 1 3 2) = (4 1 3 2 5) .

Example 2.7.12. §

Every nontrivial element of $S_{3}$ is a cycle: these elements are $(12)$ , $(13)$ , $(23)$ , $(123)$ , and $(132)$ .

However, not every nontrivial element of $S_{4}$ is a cycle, as a permutation in $S_{4}$ can have two orbits of order $2$ .

Definition 2.7.13. §

We say that two cycles $σ$ and $τ$ in $S_{n}$ are disjoint if the largest orbit of $σ$ has empty intersection with the largest orbit of $τ$ .

We prove the following lemma.

Lemma 2.7.14. §

Any two disjoint cycles commute.

Proof.

Let $σ, τ \in S_{n}$ be disjoint cycles. Let $O_{σ}$ and $O_{τ}$ denote their largest orbits. If $x \in O_{σ}$ , then $x, σ (x) \notin O_{τ}$ , so

𝜎𝜏 (x) = σ (x) = 𝜏𝜎 (x) .

Similarly, $σ$ and $τ$ commutate on elements of $O_{τ}$ . Finally, if $x$ lies in the complement of $O_{σ} \cup O_{τ}$ , then $𝜎𝜏 (x) = x = 𝜏𝜎 (x)$ . So $τ$ and $σ$ commute. □

Remark 2.7.15. §

We see from the proof of Lemma 2.7.14 that in any product $τ$ of disjoint cycles, an element $x \in X_{n}$ will be fixed by all but at most one of the cycles, and if there is such a cycle, the value $τ (x)$ will equal the value of that cycle on $x$ (and otherwise $τ (x) = x$ ).

We next see that the cycles generate $S_{n}$ .

Proposition 2.7.16. §

Every nontrivial permutation in $S_{n}$ may be written as a product of disjoint cycles in a unique way, up to the order of the cycles.

Proof.

Let $σ \in S_{n}$ , and suppose that $σ$ has $m$ orbits of order greater than $1$ . Choose representatives $x_{i}$ of each of these orbits for $1 \leq i \leq m$ , and set $k_{i} = | O_{σ} (x_{i}) |$ . We then consider the product

τ = (x_{1} σ (x_{1}) \dots σ {(x_{1})}^{k_{1} - 1}) (x_{2} σ (x_{2}) \dots σ {(x_{2})}^{k_{2} - 1}) \dots (x_{m} σ (x_{m}) \dots σ {(x_{m})}^{k_{m} - 1})

of disjoint cycles, and we claim that $τ = σ$ . For $x \in X_{n}$ , we have either $x = σ^{j} (x_{i})$ for some $1 \leq i \leq m$ and $0 \leq j \leq k_{i} - 1$ , or $x$ lies in an orbit of order $1$ . In the former case, we have $τ (x) = σ^{j + 1} (x_{i}) = σ (x)$ . In the latter, we have $τ (x) = x = σ (x)$ as well. Hence, we see that $σ$ may be written as a product of disjoint cycles.

We leave uniqueness primarily to the reader. We merely note that, first, the elements appearing in the individual cycles above are the elements in the orbits and so must be in any such decomposition of $σ$ . Given that, the individual cycles are forced to be as above by the values of $σ$ . □

Example 2.7.17. §

Consider the permutation

λ = (\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 6 & 4 & 3 & 5 & 1 \end{matrix}) .

Then $λ (1) = 2$ , $λ (2) = 6$ , $λ (6) = 1$ , so one of the cycles in the decomposition of $λ$ is $(1 2 6)$ . Also, $λ (3) = 4$ and $λ (4) = 3$ , so another is $(3 4)$ . On the other hand, $λ (5) = 5$ , so $5$ is not moved (or is “fixed”) by $λ$ . We therefore have

λ = (1 2 6) (3 4) .

Proposition 2.7.16 has the following interesting application to orders of elements.

Proposition 2.7.18. §

The order of an element of $S_{n}$ is the least common multiple of the orders of the disjoint cycles of which it is a product.

Proof.

Suppose that $σ \in S_{n}$ decomposes as a product of $m$ disjoint cycles $τ_{1}, τ_{2}, \dots, τ_{m}$ of length $k_{1}, k_{2}, \dots, k_{m}$ . Then Lemma 2.7.14 implies that

σ^{i} = τ_{1}^{i} \dots τ_{m}^{i}

for every $i \in ℤ$ , and by disjointness, the only way for $σ^{i} = e$ to occur is if $τ_{j}^{i} = e$ for all $1 \leq j \leq m$ . But $τ_{j}$ has order $k_{j}$ , so this will happen if and only if $i$ is a multiple of each $k_{j}$ , and therefore of the least common multiple of the $k_{j}$ . Hence the order of $σ$ is this least common multiple. □

Example 2.7.19. §

In $S_{7}$ , the element

σ = (1 4 2 7) (3 6) = (\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 4 & 7 & 6 & 2 & 5 & 3 & 1 \end{matrix})

has order $4$ . Note that

\begin{array}{rcl} σ^{2} = (1 2) (4 7) & and & σ^{3} = (1 7 2 4) (3 6) . \end{array}

Example 2.7.20. §

The exponent of a finite group is the least common multiple of the orders of its elements. Since the order of an element is the least common multiple of the orders of its cycles, which have orders $2, \dots, n$ , the exponent of $S_{n}$ is $lcm (1, 2, \dots, n)$ . E.g., the exponent of $S_{7}$ is $3 \cdot 4 \cdot 5 \cdot 7 = 420$ .

In fact, $G$ has a smaller generating set than the cycles, which is to say the set of transpositions.

Definition 2.7.21. §

A transposition in $S_{n}$ is a $2$ -cycle.

Proposition 2.7.22. §

Every element of $S_{n}$ is a product of transpositions.

Proof.

As every permutation is a product of cycles, we need only show that every cycle is a product of transpositions. In fact,

(x_{1} x_{2} \dots x_{k}) = (x_{1} x_{2}) (x_{2} x_{3}) \dots (x_{k - 1} x_{k}),

as is easily checked. □

Remark 2.7.23. §

In fact, the symmetric group $S_{n}$ is generated by transpositions of the form $(x x + 1)$ with $1 \leq x \leq n - 1$ . That is, the previous proposition tells us that

(x x + 1 \dots y) = (x x + 1) (x + 1 x + 2) \dots (y - 1 y)

for any $1 \leq x < y \leq n$ , and then for such $x$ and $y$ we have

(x y) = (y - 1 y) \cdot (y - 2 y - 1 y) \dots (x x + 1 \dots y),

so all transpositions are contained in the subgroup generated by transpositions of the form $(x x + 1)$ , which again by Proposition 2.7.22 is all that we need.

2.8. Homomorphisms

In order to compare groups, it is useful to consider a generalization of the concept of isomorphism that actually has a simpler definition, as the condition of bijectivity is removed.

Definition 2.8.1. §

Let $G$ and $G^{'}$ be groups. A homomorphism $ϕ$ from $G$ to $G^{'}$ is a function

ϕ : G \to G^{'}

such that

ϕ (𝑎𝑏) = ϕ (a) ϕ (b)

for all $a, b \in G$ .

Examples 2.8.2. §

a.

Let $ψ_{n} : ℤ \to ℤ$ be the multiplication-by- $n$ map, defined by $ψ_{n} (a) = 𝑛𝑎$ for all $a \in ℤ$ . Then $ψ_{n}$ is a homomorphism since

ψ_{n} (a + b) = n (a + b) = 𝑛𝑎 + 𝑛𝑏 = ψ_{n} (a) + ψ_{n} (b) .

b.

The reduction map $ϕ_{n} : ℤ \to ℤ ∕ 𝑛ℤ$ defined by $ϕ_{n} (a) = a + 𝑛ℤ$ is a surjective homomorphism.

c.

The determinant map

\det : {GL}_{n} (ℝ) \to ℝ^{\times}

satisfies

\det (𝐴𝐵) = \det (A) \det (B)

for all $A, B \in {GL}_{n} (ℝ)$ , so is a (surjective) homomorphism.

d.

For $m < n$ , we have an (injective) homomorphism $ι : S_{m} \to S_{n}$ that takes a permutation $σ$ of $X_{m} = {1, 2, \dots, m}$ to the permutation $τ \in S_{n}$ that satisfies $τ (i) = σ (i)$ for $1 \leq i \leq m$ and $τ (j) = j$ for $m < j \leq n$ .

e.

For $A \in M_{𝑟𝑠} (ℝ)$ , we define a left-multiplication-by- $A$ map

ψ_{A} : M_{𝑠𝑡} (ℝ) \to M_{𝑟𝑡} (ℝ)

by $ψ_{A} (B) = 𝐴𝐵$ for $B \in M_{𝑠𝑡} (ℝ)$ . By distributivity of multiplication of matrices, this is a homomorphism. It need not in general be injective or surjective.

f.

The set

C^{1} (ℝ) = {f : ℝ \to ℝ ∣ f is everywhere differentiable}

forms a group under addition. In fact, it is a subgroup of $Maps (ℝ, ℝ)$ . The derivative map

\partial : C^{1} (ℝ) \to Maps (ℝ, ℝ), \partial (f) = f^{'}

is a homomorphism.

Here are several standard homomorphisms between groups.

Definition 2.8.3. §

Let $G$ and $G^{'}$ be groups with identity elements $e$ and $e^{'}$ , respectively.

a.: The trivial homomorphism $ϕ : G \to G^{'}$ is given by $ϕ (g) = e^{'}$ , the identity of $G^{'}$ , for all $g \in G$ .
b.: The identity homomorphism ${id}_{G} : G \to G$ on any group $G$ , given by ${id}_{G} (g) = g$ for all $g \in G$ .
c.: For $H \leq G$ , we the inclusion map $ι_{H} : H \to G$ with $ι_{H} (h) = h$ for all $h \in H$ .

The following easily-proven lemma is useful to know.

Lemma 2.8.4. §

Let $G$ , $G^{'}$ , and $G^{″}$ be groups, and let $ϕ : G \to G^{'}$ and $ψ : G^{'} \to G^{″}$ be homomorphisms. Then

ψ \circ ϕ : G \to G^{″}

is also a homomorphism.

Proof.

For $a, b \in G$ , we have

ψ \circ ϕ (𝑎𝑏) = ψ (ϕ (a) ϕ (b)) = ψ (ϕ (a)) ψ (ϕ (b)) = ψ \circ ϕ (a) \cdot ψ \circ ϕ (b) .

□

In the following, $G$ and $G^{'}$ will be groups, and we will use $e$ and $e^{'}$ to denote their respective identity elements.

Lemma 2.8.5. §

Let $ϕ : G \to G^{'}$ be a homomorphism. Then $ϕ (e) = e^{'}$ , and $ϕ (g^{- 1}) = ϕ {(g)}^{- 1}$ for all $g \in G$ .

Proof.

We have $ϕ (e) = ϕ (e) ϕ (e)$ by the defining property of a homomorphism, and the cancellation theorem then implies that $ϕ (e) = e^{'}$ . Moreover,

ϕ (g) \cdot ϕ (g^{- 1}) = ϕ (e) = e^{'},

again by the homomorphism property, and uniqueness of right inverses in a group then implies that $ϕ (g^{- 1}) = ϕ {(g)}^{- 1}$ . The last statement then follows easily from these and the homomorphism property. □

We have the following easy consequence.

Lemma 2.8.6. §

Let $ϕ : G \to G^{'}$ be a homomorphism. Then

ϕ (g_{1}^{r_{1}} g_{2}^{r_{2}} \dots g_{k}^{r_{k}}) = ϕ {(g_{1})}^{r_{1}} ϕ {(g_{2})}^{r_{2}} \dots ϕ {(g_{k})}^{r_{k}}

for any $k \geq 1$ , $g_{1}, g_{2}, \dots, g_{k} \in G$ , and $r_{1}, r_{2}, \dots, r_{k} \in ℤ$ .

Proof.

For $g \in G$ and $r \in ℤ$ , we have $g^{r} = {(g^{- 1})}^{- r}$ and

ϕ {(g)}^{r} = ϕ {(g^{- 1})}^{- r},

so it suffices to assume that each $r_{i}$ is nonnegative in the theorem. But then, by writing out the powers as products, the result amounts simply to proving the result when each $r_{i} = 1$ . On the other hand,

ϕ (g_{1} g_{2} \dots g_{k}) = ϕ (g_{1}) ϕ (g_{2} \dots g_{k}) = \dots = ϕ (g_{1}) ϕ (g_{2}) \dots ϕ (g_{k})

by iterative use of the defining property of a homomorphism. □

Definition 2.8.7. §

Let $ϕ : G \to G^{'}$ be a homomorphism.

a.: The kernel of $ϕ$ is the subset of $G$ that is
$\ker ϕ = {g \in G ∣ ϕ (g) = e^{'}} .$
b.: The image of $ϕ$ is the subset of $G^{'}$ that is
$im ϕ = {ϕ (g) ∣ g \in G} .$

Proposition 2.8.8. §

Let $ϕ : G \to G^{'}$ be a homomorphism. Then $\ker ϕ$ is a subgroup of $G$ and $im ϕ$ is a subgroup of $G^{'}$ .

Proof.

Since $ϕ (e) = e^{'}$ , we have $e \in \ker ϕ$ . Moreover, if $a, b \in \ker ϕ$ then

ϕ (𝑎𝑏) = ϕ (a) ϕ (b) = e^{'} \cdot e^{'} = e^{'},

so $𝑎𝑏 \in \ker ϕ$ , and if $a \in \ker ϕ$ then

ϕ (a^{- 1}) = ϕ {(a)}^{- 1} = {(e^{'})}^{- 1} = e^{'},

so $a^{- 1} \in \ker ϕ$ . It follows that $\ker ϕ \leq G$ .

Next, note that $e^{'} = ϕ (e)$ , so $e^{'} \in G$ . Also, if $ϕ (a), ϕ (b) \in im ϕ$ for some $a, b \in G$ , then

ϕ (a) ϕ (b) = ϕ (𝑎𝑏) \in im ϕ

and

ϕ {(a)}^{- 1} = ϕ (a^{- 1}) \in im ϕ .

Hence, we have that $im ϕ \leq G^{'}$ . □

Clearly, a homomorphism $ϕ : G \to G^{'}$ is surjective if and only if $im ϕ = G^{'}$ . On the other hand, we have the following less obvious criterion for injectivity of $ϕ$ in terms of its kernel.

Proposition 2.8.9. §

A homomorphism $ϕ : G \to G^{'}$ is injective if and only if $\ker ϕ = {e}$ .

Proof.

If $ϕ$ is injective and $a \in \ker ϕ$ , then $ϕ (a) = e^{'} = ϕ (e)$ , so $a = e$ by injectivity of $ϕ$ . On the other hand, if $\ker ϕ$ is trivial and $ϕ (a) = ϕ (b)$ for some $a, b \in G$ , then

ϕ (a b^{- 1}) = ϕ (a) ϕ {(b)}^{- 1} = e^{'},

so $a b^{- 1} = e$ , and therefore $a = b$ . □

Examples 2.8.10. §

a.: The multiplication-by- $n$ map $ψ_{n}$ is injective, as $ψ_{n} (a) = 𝑛𝑎 = 0$ if and only if $a = 0$ . Its image is $𝑛ℤ$ .
b.: The inclusion map $ι_{H} : H \to G$ of a subgroup $H$ in a group $G$ is obviously injective, and its image is $H$ .
c.: The reduction map $ϕ_{n} : ℤ \to ℤ ∕ 𝑛ℤ$ is surjective, and its kernel is $𝑛ℤ$ .
d.: The determinant map $\det : {GL}_{n} (ℝ) \to ℝ^{\times}$ is surjective with kernel ${SL}_{n} (ℝ)$ .
e.: The derivative map $\partial : C^{1} (ℝ) \to Maps (ℝ, ℝ)$ has kernel equal to the subgroup of constant functions. Its image is difficult to describe explicitly, but it is not surjective.

We can also speak of the image of a subgroup under a homomorphism.

Definition 2.8.11. §

Let $H$ be a subgroup of $G$ . Then image of $H$ under a homomorphism $ϕ : G \to G^{'}$ is

ϕ (H) = {ϕ (h) ∣ h \in H} .

Remark 2.8.12. §

The set $ϕ (H)$ is a subgroup of $G^{'}$ , as it is the image of composition $ϕ \circ ι_{H}$ of the inclusion map $ι_{H} : H \to G$ with $ϕ$ .

Definition 2.8.13. §

The restriction of a homomorphism $ϕ : G \to G^{'}$ to $H \leq G$ is the homomorphism

ϕ |_{H} : H \to G^{'}

that is the composition $ϕ \circ ι_{H}$ , where $ι_{H}$ is the inclusion map. In other words, $ϕ |_{H} (h) = ϕ (h)$ for all $h \in H$ .

We can also speak of the inverse image of a subgroup under a homomorphism.

Definition 2.8.14. §

Let $ϕ : G \to G^{'}$ be a homomorphism. Let $H^{'} \leq G^{'}$ . The inverse image of $H^{'}$ under $ϕ$ is

ϕ^{- 1} (H^{'}) = {h \in H ∣ ϕ (h) \in H^{'}} .

Proposition 2.8.15. §

Let $ϕ : G \to G^{'}$ be a homomorphism, and let $H^{'} \leq G^{'}$ . Then $ϕ^{- 1} (H^{'})$ is a subgroup of $G$ .

Proof.

Note that $ϕ (e) = e^{'} \in H^{'}$ , so $e \in ϕ^{- 1} (H^{'})$ . Also, if $a, b \in ϕ^{- 1} (H^{'})$ , then $ϕ (a b^{- 1}) = ϕ (a) ϕ {(b)}^{- 1} \in H^{'}$ , since $H^{'}$ is a group, so $a b^{- 1} \in ϕ^{- 1} (H^{'})$ . In other words, $ϕ^{- 1} (H^{'})$ is closed under multiplcation and inverses, so is a subgroup of $G$ . □

Example 2.8.16. §

Consider the multiplication-by- $n$ map $ψ_{n} : ℤ \to ℤ$ . We have

ψ_{n}^{- 1} (𝑚ℤ) = {a \in ℤ ∣ 𝑛𝑎 \in 𝑚ℤ} = \frac{m}{\gcd (n, m)} ℤ .

A homomorphism is completely determined by its values on a generating set.

Proposition 2.8.17. §

Let $S$ be a generating set of $G$ , and let $ϕ, ψ : G \to G^{'}$ be homomorphisms. Suppose that $ϕ (s) = ψ (s)$ for all $s \in S$ . Then $ϕ = ψ$ .

Proof.

Since $S$ generates $G$ , every element of $G$ has the form

s_{1}^{r_{1}} s_{2}^{r_{2}} \dots s_{k}^{r_{k}}

for some $k \geq 0$ , $s_{1}, s_{2}, \dots, s_{k} \in G$ , and $r_{1}, r_{2}, \dots, r_{k} \in ℤ$ . We have

ϕ (s_{1}^{r_{1}} s_{2}^{r_{2}} \dots s_{k}^{r_{k}}) = ϕ {(s_{1})}^{r_{1}} ϕ {(s_{2})}^{r_{2}} \dots ϕ {(s_{k})}^{r_{k}} = ψ {(s_{1})}^{r_{1}} ψ {(s_{2})}^{r_{2}} \dots ψ {(s_{k})}^{r_{k}} = ψ (s_{1}^{r_{1}} s_{2}^{r_{2}} \dots s_{k}^{r_{k}}),

as desired. □

This is a very useful property for checking whether or not two homomorphisms are equal. On the other hand, one might be tempted to try to use it to specify a homomorphism by setting its values on a generating set arbitrarily. This in general does not work. For instance, the only homomorphism $ϕ : ℤ ∕ 𝑛ℤ \to ℤ$ is the trivial homomorphism, since one must have $𝑛𝜙 (1) = ϕ (n) = 0$ . That is, one can’t simply take $ϕ (1)$ to be an arbitrary value of $ℤ$ .

We leave it to the reader to check the following easy assertion, which in particular explains the problem just described.

Lemma 2.8.18. §

Suppose that $g \in G$ has finite order and $ϕ : G \to G^{'}$ is a homomorphism. Then the order of $ϕ (g)$ divides the order of $g$ .

2.9. The alternating group

In this section, we study a certain subgroup of $S_{n}$ , known as the alternating group. Let us begin with its definition.

Definition 2.9.1. §

The alternating group $A_{n}$ is the subgroup of $S_{n}$ consisting of permutations that can be written as a product of an even number of transpositions.

Example 2.9.2. §

We have $(1 2 3) \in A_{3}$ , as $(1 2 3) = (1 2) (2 3)$ . Even more obviously, we have $(1 2) (3 4) \in A_{4}$ .

The assertion that $A_{n}$ is a subgroup of $S_{n}$ contained in Definition 2.9.1 is easy to verify, and we leave it to the reader. What is not so immediate is that not every element in $S_{n}$ can be written as a product of an even number of transpositions, which is to say that $A_{n} \neq S_{n}$ . For this reason and others, we give an alternate characterization of $S_{n}$ .

Definition 2.9.3. §

A permutation matrix in ${GL}_{n} (ℝ)$ is a matrix for which the entries are all zero aside from one entry in each row and each column, which is $1$ .

Proposition 2.9.4. §

The set $P_{n}$ of permutation matrices forms a subgroup of ${GL}_{n} (ℝ)$ that is isomorphic to $S_{n}$ . Explicitly, define

κ : S_{n} \to {GL}_{n} (ℝ)

by taking $κ (σ)$ to be the matrix with entry

κ {(σ)}_{𝑖𝑗} = {\begin{matrix} 1 & if σ (j) = i \\ 0 & if σ (j) \neq i . \end{matrix}

in the $i$ th row and $j$ th column. Then $κ$ is injective with image $P_{n}$ .

Proof.

Let $σ \in S_{n}$ . We first remark that $κ (σ)$ is a permutation matrix: since $σ$ is a function, each $j$ is taken to exactly one $i$ , so each column has exactly one $1$ , and since $σ$ is a bijection, each $i$ comes from exactly one $j$ , so each row has exactly one $1$ . Moreover, we have

{(κ (σ) κ (τ))}_{𝑖𝑘} = \sum_{j = 1}^{n} κ {(σ)}_{𝑖𝑗} κ {(τ)}_{𝑗𝑘} = κ {(σ)}_{𝑖𝜏 (k)} κ {(τ)}_{τ (k) k} = {\begin{matrix} 1 & if σ (τ (k)) = i \\ 0 & if σ (τ (k)) \neq i, \end{matrix}

and the latter term is exactly $κ {(𝜎𝜏)}_{𝑖𝑘}$ . Therefore, $κ$ is a homomorphism. It is also clearly one-to-one, since $κ (σ)$ will have a non-diagonal entry that is nonzero if $σ \neq e$ . Finally, for any $A = (a_{𝑖𝑗}) \in P_{n}$ , we have $A = κ (σ)$ , where $σ (j)$ is defined as the unique $i$ such that $a_{𝑖𝑗} = 1$ , so $κ$ is onto. □

We next determine the image of $A_{n}$ under the map $κ$ of Proposition 2.9.4.

Lemma 2.9.5. §

The image of $A_{n}$ under $κ$ is equal to the subgroup of permutation matrices that have determinant $1$ .

Proof.

Then $κ ((a b))$ , where $1 \leq a < b \leq n$ , is exactly the identity matrix after one row operation, which is switching the $i$ th and $j$ th rows. As switching two rows changes the sign of a matrix, we have $\det κ ((a b)) = - 1$ . As $\det \circ κ$ is a homomorphism, we have that the determinant of a product of a product of an even number of transpositions is $1$ , and the determinant of the product of an odd number of transpositions is $- 1$ . □

For the following definition, we note that the determinant of any permutation matrix is either $1$ or $- 1$ .

Definition 2.9.6. §

a.

We define the sign function on $S_{n}$ by

sign = \det \circ κ : S_{n} \to {\pm 1},

with $κ$ as in Proposition 2.9.4. Its value on a permutation is the sign of the permutation.

b.

We say that $σ \in S_{n}$ is even if $sign (σ) = 1$ and odd if $sign (σ) = - 1$ .

Remark 2.9.7. §

By Lemma 2.9.5, we have that $A_{n}$ is exactly the subgroup of even permutations in $S_{n}$ .

Example 2.9.8. §

Cycles of even length are odd, while cycles of odd length are even.

2.10. Cosets

Definition 2.10.1. §

Let $H$ be a subgroup of a group $G$ , and let $a$ be an element of $G$ . The left $H$ -coset of $a$ is the subset of $G$ that is

𝑎𝐻 = {𝑎h ∣ h \in H} .

The right $H$ -coset of $a$ is the set

𝐻𝑎 = {h𝑎 ∣ h \in H} .

Remark 2.10.2. §

If $G$ is abelian, then $𝑎𝐻 = 𝐻𝑎$ for any $H \leq G$ and $a \in G$ , so we may speak simply of cosets (as opposed to left and right cosets). If the operation on $G$ is addition, we write $a + H$ for the $H$ -coset of $a$ .

Example 2.10.3. §

The $2 ℤ$ -coset $1 + 2 ℤ$ of $1$ in $ℤ$ is the set of odd integers.

Example 2.10.4. §

Let $H = ⟨ (12) ⟩ \leq S_{3}$ . Then the left cosets of $H$ are

\begin{array}{rcl} H = (1 2) H = {(1 2), e} \\ (1 2 3) H = (1 3) H = {(1 3), (1 2 3)} \\ (1 3 2) H = (2 3) H = {(2 3), (1 3 2)} . \end{array}

The property of two cosets being equal provides an equivalence relation on a group $G$ , as expressed in the following lemma, the proof of which follows directly from the definitions of left and right cosets.

Lemma 2.10.5. §

Let $H$ be a subgroup of a group $G$ , and let $a, b \in H$ . The relation $a \sim_{l} b$ (resp., $a \sim_{r} b$ ) if and only if $𝑎𝐻 = 𝑏𝐻$ (resp., $𝐻𝑎 = 𝐻𝑏$ ) is an equivalence relation on $G$ , and the equivalence class of $a \in H$ under this relation is $𝑎𝐻$ (resp., $𝐻𝑎$ ).

Corollary 2.10.6. §

If $H \leq G$ , then $G$ is the disjoint union of its distinct left (or right) $H$ -cosets.

We make the following remark.

Lemma 2.10.7. §

Let $H \leq G$ , and let $a, b \in H$ . Then $𝑎𝐻 = 𝑏𝐻$ if and only if $a^{- 1} b \in H$ , and $𝐻𝑎 = 𝐻𝑏$ if and only if $a b^{- 1} \in H$ .

Proof.

Suppose $a^{- 1} b \in H$ , and set $h = a^{- 1} b$ . Then $b = 𝑎h$ , so $𝑏𝑘 = a (h𝑘) \in 𝑎𝐻$ for every $k \in H$ , which implies $𝑏𝐻 \subseteq 𝑎𝐻$ . Moreover, $b h^{- 1} = a$ , so $𝑎𝐻 \subseteq 𝑏𝐻$ as well. Conversely, if $𝑎𝐻 = 𝑏𝐻$ , then there exists $h \in H$ such that $b = 𝑎h$ , so $a^{- 1} b \in H$ . The case of right cosets is similar, noting that $a b^{- 1} \in H$ if and only if $a = h𝑏$ for some $h \in H$ . □

Notation 2.10.8. §

For $H \leq G$ , we let $G ∕ H$ denote the set of left cosets of $H$ in $G$ , and we let $H ∖ G$ be the set of right cosets of $H$ in $G$ .

Example 2.10.9. §

One might notice that the set of left cosets of $𝑛ℤ$ in $ℤ$ is given the notation $ℤ ∕ 𝑛ℤ$ by Corollary 2.10.8, which could in theory lead to some confusion with the group $ℤ ∕ 𝑛ℤ$ . However, this is no coincidence. The cosets of $𝑛ℤ$ in $ℤ$ are the $a + 𝑛ℤ$ with $0 \leq a \leq n - 1$ , which are exactly the elements of the group $ℤ ∕ 𝑛ℤ$ . So, $ℤ ∕ 𝑛ℤ$ as a group is just the set $ℤ ∕ 𝑛ℤ$ (of cosets) with a particular binary operation.

Definition 2.10.10. §

We refer to a set of representatives for the left (resp., right) $H$ -cosets in $G$ as a set of left (resp., right) coset representatives.

Example 2.10.11. §

The elements $e, (1 2 3), (1 3 2)$ form a set of left coset representatives for $⟨ (1 2) ⟩$ in $S_{3}$ .

Example 2.10.12. §

The coset $s ⟨ r ⟩ = ⟨ r ⟩ s$ consists of all reflections in $D_{n}$ . The coset $r^{k} ⟨ s ⟩ = {r^{k}, r^{k} s}$ consists of the counterclockwise rotation $r^{k}$ about the origin by $2 𝜋𝑘 ∕ n$ radians and the reflection $r^{k} s$ across the line through the origin at an angle $𝜋𝑘 ∕ n$ radians counterclockwise from the $x$ -axis. Note that

⟨ s ⟩ r^{k} = {r^{k}, r^{- k} s},

and this is not $r^{k} ⟨ s ⟩$ unless $2 k = n$ .

Proposition 2.10.13. §

There is a canonical bijection

ϕ : G ∕ H \to H ∖ G

given by $ϕ (𝑎𝐻) = H a^{- 1}$ for $a \in G$ .

Proof.

First, we check that the map $ϕ$ is well-defined. For a subset $A$ of $G$ , let us use $A^{- 1}$ to denote

A^{- 1} = {a^{- 1} ∣ a \in A} .

If $𝑎𝐻 = 𝑏𝐻$ , then

H a^{- 1} = {h a^{- 1} ∣ h \in H} = {{(a h^{- 1})}^{- 1} ∣ h \in H} = {{(𝑎𝑘)}^{- 1} ∣ k \in H} = {(𝑎𝐻)}^{- 1} .

(2.10.1)

Since $𝑎𝐻 = 𝑏𝐻$ , we have that ${(𝑎𝐻)}^{- 1} = {(𝑏𝐻)}^{- 1}$ , which implies noting (2.10.1) for both $a$ and $b$ that

H a^{- 1} = {(𝑎𝐻)}^{- 1} = {(𝑏𝐻)}^{- 1} = H b^{- 1} .

Therefore, $ϕ$ is well-defined.

Next, define $ψ : H ∖ G \to G ∕ H$ by $ψ (𝐻𝑎) = a^{- 1} H$ for $a \in G$ . This is also well-defined, as $𝐻𝑎 = 𝐻𝑏$ implies that

a^{- 1} H = {(𝐻𝑎)}^{- 1} = {(𝐻𝑏)}^{- 1} = b^{- 1} H,

and it is clearly inverse to $ϕ$ , so $ϕ$ is a bijection. □

Example 2.10.14. §

Let $H = ⟨ (1 2) ⟩ \leq S_{3}$ . Then the bijection $ϕ$ of Proposition 2.10.13 is given by $ϕ (H) = H$ ,

\begin{array}{rcl} ϕ ((1 2 3) H) = H (1 3 2), & and & ϕ ((1 3 2) H) = H (1 2 3) . \end{array}

Definition 2.10.15. §

Let $H$ be a subgroup of $G$ .

a.: If there are finitely many left cosets of $H$ in $G$ , then we say that $H$ is of finite index in $G$ , and otherwise $H$ is of infinite index.
b.: If $H$ is of finite index in $G$ , then we define the index $[G : H]$ of $H$ in $G$ to be the number of left cosets of $H$ in $G$ .

Examples 2.10.16. §

a.: The index $[S_{3} : H]$ of $H = ⟨ (12) ⟩$ in $S_{3}$ is $3$ .
b.: The index $[ℤ : 𝑛ℤ]$ of $𝑛ℤ$ in $ℤ$ is $n$ .
c.: The group $ℤ$ is not of finite index in $ℚ$ .
d.: We have $[D_{n} : ⟨ s ⟩] = n$ and $[D_{n} : ⟨ r ⟩] = 2$ .
e.: We have $[S_{n} : A_{n}] = 2$ for $n \geq 2$ , and the nonidentity coset is the set of odd permutation which equals, e.g., $(1 2) A_{n}$ .

Example 2.10.17. §

For any group $G$ , we have $[G : G] = 1$ , and if $G$ is finite, we have $[G : ⟨ e ⟩] = | G |$

Remark 2.10.18. §

By Proposition 2.10.13, we could just as well have used right cosets instead of left cosets in the definition of the index.

Theorem 2.10.19 (Lagrange’s theorem). §

Let $H$ be a subgroup of a finite group $G$ . Then we have

| G | = [G : H] | H | .

In particular, the order of $H$ divides the order of $G$ .

Proof.

Since $G$ is finite, so is $H$ , and every coset $𝑎𝐻$ is in bijection with $H$ via the map $𝜃 : H \to 𝑎𝐻$ with $𝜃 (h) = 𝑎h$ . As $G$ is the disjoint union of its left cosets, we have

| G | = \sum_{𝑎𝐻 \in G ∕ H} | 𝑎𝐻 | = \sum_{𝑎𝐻 \in G ∕ H} | H | = [G : H] | H | .

□

We can use Lagrange’s theorem to determine the indices of subgroups when we know both the orders of the group and of the subgroup. Here is an example.

Example 2.10.20. §

There is an injective homomorphism

ι : S_{n - 1} \to S_{n}

as in Example 2.8.2 that takes a permutation of the set $X_{n - 1} \subset X_{n}$ to the permutation that has the same values on the elements of $X_{n - 1}$ and which fixes $n$ . Using $ι$ , we may identify $S_{n - 1}$ with the isomorphic subgroup $ι (S_{n - 1})$ of $S_{n}$ consisting of elements $σ \in S_{n}$ with $σ (n) = n$ . Under this identification, we have

[S_{n} : S_{n - 1}] = \frac{n!}{(n - 1)!} = n .

Example 2.10.21. §

Since $[S_{n} : A_{n}] = 2$ for $n \geq 2$ , we have $| A_{n} | = \frac{1}{2} n!$ for such $n$ .

Corollary 2.10.22. §

Let $G$ be a finite group. Then the order of every element of $G$ divides the order of $G$ .

Example 2.10.23. §

We have already seen that the orders of the subgroups of $ℤ ∕ 𝑛ℤ$ are exactly the positive divisors of $n$ .

Example 2.10.24. §

According to the corollary, every element of $S_{n}$ should have order dividing $n!$ . In fact, we already know from Example 2.7.20 that every element has order dividing the least common multiple of $1, 2, \dots, n$ , which clearly divides $n!$ .

Finally, we mention the following interesting corollary of Lagrange’s theorem.

Corollary 2.10.25. §

Every group of prime order is cyclic.

Proof.

Let $G$ be a group of order a prime $p$ . If $g \in G$ is not the identity, it must generate a nontrivial subgroup of $G$ , which can only have order $p$ by Lagrange’s theorem, and therefore must be $G$ . That is, $G = ⟨ g ⟩$ , finishing the proof. □

We note that the index satisfies the following multiplicative property.

Proposition 2.10.26. §

Let $H$ and $K$ be subgroups of $G$ with $K \leq H$ . Then $K$ has finite index in $G$ if and only if $H$ has finite index in $G$ and $K$ has finite index in $H$ . Moreover, if $K$ has finite index in $G$ , then we have

[G : K] = [G : H] [H : K] .

Proof.

Let $S$ be a set of $H$ -coset representatives in $G$ and $T$ be a set of $K$ -coset representatives in $H$ . Consider the set

U = {𝑠𝑡 ∣ s \in S, t \in T} .

We claim that $U$ is a set of left $K$ -coset representatives in $G$ . For this, note that if $g \in G$ , we may choose $s \in S$ with $g = 𝑠h$ for some $h \in H$ , and we may choose $t \in T$ with $h = 𝑡𝑘$ for some $k \in K$ . In other words, $g \in 𝑠𝑡𝐾$ , so $𝑔𝐾 = 𝑠𝑡𝐾$ . This proves the claim.

Next, note that if $𝑠𝑡𝐾 = s^{'} t^{'} K$ with $s, s^{'} \in S$ and $t, t^{'} \in T$ , then since $𝑡𝐾 \subset H$ and $t^{'} K \subset H$ , we have $𝑠𝐻 = s^{'} H$ , so $s = s^{'}$ . But then $𝑠𝑡𝐾 = s t^{'} K$ , so $𝑡𝐾 = t^{'} K$ , and therefore $t = t^{'}$ . In other words, we have shown that the map $S \times T \to U$ given by $(s, t) \mapsto 𝑠𝑡$ is a bijection. Thus, $S$ and $T$ are finite if and only if $U$ is, and if they are, then $| U | = | S | | T |$ , as desired. □

Remark 2.10.27. §

Proposition 2.10.26 implies Lagrange’s theorem by taking the subgroup $K$ to be the trivial subgroup. That is, for a finite group $G$ and subgroup $H$ , we have

| G | = [G : ⟨ e ⟩] = [G : H] [H : ⟨ e ⟩] = [G : H] | H | .

2.11. Conjugation

Definition 2.11.1. §

Let $G$ be a group.

a.: Let $a$ , $x \in G$ . Then $𝑎𝑥 a^{- 1}$ is known as the conjugate of $x$ by $a$ .
b.: We say that an element $x \in G$ is conjugate to an element $y \in G$ if there exists $a \in G$ with $y = 𝑎𝑥 a^{- 1}$ .

Remark 2.11.2. §

One might recall the related notion of similar matrices in $M_{n} (ℝ)$ .

Lemma 2.11.3. §

The relation $\sim$ on $G$ given by $x \sim y$ if and only if $x$ is conjugate to $y$ is an equivalence relation on $G$ .

Proof.

We have $x = 𝑒𝑥 e^{- 1}$ , so $x \sim x$ . If $x \sim y$ , then there exists $a \in G$ with $y = 𝑎𝑥 a^{- 1}$ , which implies

x = a^{- 1} 𝑦𝑎 = a^{- 1} y {(a^{- 1})}^{- 1},

so $y \sim x$ . Finally, if $x \sim y$ and $y \sim z$ , then there exist $a, b \in G$ with $y = 𝑎𝑥 a^{- 1}$ and $z = 𝑏𝑦 b^{- 1}$ , so

z = 𝑏𝑦 b^{- 1} = b (𝑎𝑥 a^{- 1}) b^{- 1} = (𝑏𝑎) x {(𝑏𝑎)}^{- 1},

and $z \sim x$ . □

Definition 2.11.4. §

The set

C_{x} = {𝑎𝑥 a^{- 1} ∣ a \in G}

of elements of that are conjugate to $x \in G$ is called the conjugacy class of $x$ .

As a consequence of the fact that conjugacy forms an equivalence relation, any two conjugacy classes are either disjoint or equal.

Example 2.11.5. §

Let $a = r^{i}$ and $b = r^{i} s$ in $D_{n}$ . Then we have

\begin{array}{rcl} a r^{j} a^{- 1} = r^{i} \cdot r^{j} \cdot r^{- i} = r^{j}, & a (r^{j} s) a^{- 1} = r^{i} \cdot r^{j} s \cdot r^{- i} = r^{2 i + j} s, \\ b r^{j} b^{- 1} = r^{i} s \cdot r^{j} \cdot s r^{- i} = r^{- j}, & b (r^{j} s) b^{- 1} = r^{i} s \cdot r^{j} s \cdot s r^{- i} = r^{2 i - j} s . \end{array}

Therefore, we have $C_{r^{i}} = {r^{i}, r^{- i}}$ for all $i \in ℤ$ , while

C_{s} = {r^{2 i} s ∣ i \in ℤ},

which is all reflections if $n$ is odd, but only half of them if $n$ is even, in which case the remaining conjugacy class is $C_{𝑟𝑠}$ .

Definition 2.11.6. §

For any $a \in G$ , the conjugation map is the function defined by

γ_{a} : G \to G, γ_{a} (x) = 𝑎𝑥 a^{- 1}

for $x \in G$ .

Remark 2.11.7. §

The process of applying a map $γ_{a}$ to an element of $G$ is referred to as conjugation.

Lemma 2.11.8. §

For $a \in G$ , the conjugation map $γ_{a}$ is an isomorphism.

Proof.

For $x, y \in G$ , we have

γ_{a} (x) γ_{a} (y) = (𝑎𝑥 a^{- 1}) (𝑎𝑦 a^{- 1}) = a (𝑥𝑦) a^{- 1} = γ_{a} (𝑥𝑦) .

Also, $γ_{a^{- 1}}$ is the inverse function to $γ_{a}$ , so $γ_{a}$ is bijective. □

In particular, we have

\begin{array}{rcl} a (x_{1} x_{2} \dots x_{s}) a^{- 1} = a x_{1} a^{- 1} \cdot a x_{2} a^{- 1} \cdot \dots \cdot a x_{s} a^{- 1} & and & {(𝑎𝑥 a^{- 1})}^{- 1} = a x^{- 1} a^{- 1} \end{array}

for any elements of $G$ .

One very interesting example is conjugation in $S_{n}$ . We describe this in the case of a cycle.

Lemma 2.11.9. §

Let $σ \in S_{n}$ , and let $τ = (x_{1} x_{2} \dots x_{k}) \in S_{n}$ be a $k$ -cycle. Then

𝜎𝜏 σ^{- 1} = (σ (x_{1}) σ (x_{2}) \dots σ (x_{k})) .

Proof.

We check this as functions. Let $x \in X_{n}$ . Then

τ (σ^{- 1} (x)) = {\begin{matrix} x_{i + 1} & if σ^{- 1} (x) = x_{i}, 1 \leq i \leq k - 1 \\ x_{1} & if σ^{- 1} (x) = x_{k} \\ σ^{- 1} (x) & otherwise \end{matrix} = {\begin{matrix} x_{i + 1} & if x = σ (x_{i}), 1 \leq i \leq k - 1 \\ x_{1} & if x = σ (x_{k}) \\ σ^{- 1} (x) & otherwise \end{matrix}

It follows that

σ (τ (σ^{- 1} (x))) = {\begin{matrix} σ (x_{i + 1}) & if x = σ (x_{i}), 1 \leq i \leq k - 1 \\ σ (x_{1}) & if x = σ (x_{k}) \\ x & otherwise, \end{matrix}

but the latter just the value of the cycle $(σ (x_{1}) σ (x_{2}) \dots σ (x_{k}))$ on $x$ . □

Remark 2.11.10. §

We can use Lemma 2.11.9 to compute the conjugate of any permutation $τ$ by a permutation $σ$ , as $τ$ can be written as a product of cycles, $τ = τ_{1} τ_{2} \dots τ_{s}$ and

𝜎𝜏 σ^{- 1} = σ τ_{1} σ^{- 1} \cdot σ τ_{2} σ^{- 1} \dots σ τ_{s} σ^{- 1} .

Example 2.11.11. §

In $S_{7}$ , we have

(1 2 3 4) \cdot (2 3 7) (4 5) \cdot {(1 2 3 4)}^{- 1} = (3 4 7) (1 5) .

That is, we have replaced the entries in the permutation $(2 3 7) (4 5)$ that $(1 2 3 4)$ moves to the values it takes them to.

Definition 2.11.12. §

Write $τ \in S_{n}$ as a product of disjoint cycles $τ = τ_{1} τ_{2} \dots τ_{s}$ with lengths $2 \leq k_{1} \leq k_{2} \leq \dots k_{s} \leq n$ . Then $(k_{1}, k_{2}, \dots, k_{s})$ is said to be the cycle type of $τ$ .

We may use Lemma 2.11.9 to prove the following.

Proposition 2.11.13. §

Two elements in $S_{n}$ are conjugate if and only if they have the same cycle type.

Proof.

By Lemma 2.11.9 and Remark 2.11.10, any two conjugate permutations must have the same cycle type. On the other hand, we will show that every permutation with cycle type $(k_{1}, k_{2}, \dots, k_{s})$ is conjugate to a particular permutation $α$ depending only on $(k_{1}, k_{2}, \dots, k_{s})$ . Since conjugacy of elements forms an equivalence relation, we will then have the result.

We first define $α$ . Let $m_{i} = k_{1} + k_{2} + \dots + k_{i - 1}$ for each $1 \leq i \leq s + 1$ . In particular, $m_{1} = 0$ . Define $α_{i} \in S_{n}$ by

α_{i} = (m_{i} + 1 m_{i} + 2 \dots m_{i} + k_{i}) .

Then the $α_{i}$ are disjoint cycles, and we set $α = α_{1} α_{2} \dots α_{s}$ .

Now suppose that $τ$ has cycle type $(k_{1}, k_{2}, \dots, k_{s})$ . We must show that $τ$ is conjugate to $α$ . Write

τ = τ_{1} τ_{2} \dots τ_{s},

where the $τ_{i}$ are disjoint cycles:

τ_{i} = (x_{m_{i} + 1} x_{m_{i} + 2} \dots x_{m_{i} + k_{i}}) .

Now choose any $σ \in S_{n}$ such that $σ (i) = x_{i}$ for each $1 \leq i \leq m_{s}$ . (For each $m_{s} < i \leq n$ , we are free to successively choose the $σ (i)$ as $i$ increases to be any values between $1$ and $n$ not yet chosen.) Then

σ α_{i} σ^{- 1} = (σ (m_{i} + 1) σ (m_{i} + 2) \dots σ (m_{i} + k_{i})) = (x_{m_{i} + 1} x_{m_{i} + 2} \dots x_{m_{i} + k_{i}})

for each $1 \leq i \leq s$ , so $𝜎𝛼 σ^{- 1} = τ$ , as desired. □

Remark 2.11.14. §

In other words, the conjugacy class of a permutation is all permutations with that same cycle type.

2.12. Normal subgroups

We now focus our attention on a very special class of subgroups of a group.

Definition 2.12.1. §

A subgroup $N$ of a group $G$ is said to be normal if $𝑎𝑁 = 𝑁𝑎$ for every $a \in G$ . We also say that $N$ is normal in $G$ , and we write $N ⊴ G$ to indicate this.

Notation 2.12.2. §

If $N$ is a proper normal subgroup of a group $G$ , then we write $N ◃ G$ .

Of course, the trivial subgroup and the improper subgroup of a group $G$ are, by this definition, normal subgroups. If $G$ is abelian, then every subgroup is normal. We also have the following.

Lemma 2.12.3. §

If $H$ is an index $2$ subgroup of a group $G$ , then $H$ is normal in $G$ .

Proof.

Since $G$ has just two left $H$ -cosets, one of which is $H$ , the other must be the complement of $H$ in $G$ . The same holds for the right $H$ -cosets, hence the result. □

Here are a couple more examples, the second of which we can see from Lemma 2.12.3.

Examples 2.12.4. §

a.

We have $⟨ r^{i} ⟩ ◃ D_{n}$ for all $n \geq 3$ and $i \in ℤ$ . To see this, note that $r^{i}$ commutes with every $r^{j}$ , while

r^{j} s \cdot r^{i} = r^{- i} \cdot r^{j} s,

and $r^{- i} \in ⟨ r^{i} ⟩$ .

b.

We have that $A_{n} ◃ S_{n}$ for all $n \geq 2$ .

Remark 2.12.5. §

If $H$ is a subgroup of $G$ that is not normal, then $𝑎𝐻 \neq 𝐻𝑎$ for some $a \in G$ . But note that $a \in 𝑎𝐻 \cap 𝐻𝑎$ , and distinct right cosets are disjoint, so $𝑎𝐻$ cannot equal any right coset of $G$ . Therefore, there exists a left coset that is not equal to a right coset.

We can give another characterization of normal subgroups using conjugation. For this, we use the following definition.

Definition 2.12.6. §

Let $H$ be a subgroup of $G$ . For $a \in G$ , the conjugate subgroup of $H$ by $a$ is the set of conjugates of $a$ :

𝑎𝐻 a^{- 1} = {𝑎h a^{- 1} ∣ h \in H} .

That the conjugate subgroup is, in fact, a subgroup is a corollary of Lemma 2.11.8:

Corollary 2.12.7. §

For $H \leq G$ and $a \in G$ , we have that $𝑎𝐻 a^{- 1} \leq G$ .

Proof.

We have $𝑎𝐻 a^{- 1} = γ_{a} (H)$ , and the latter is the image of a (sub)group under a homomorphism to $G$ , hence a subgroup of $G$ . □

Examples 2.12.8. §

a.: In $D_{n}$ , we have
$\begin{array}{rcl} s ⟨ r ⟩ s^{- 1} = ⟨ r ⟩ & and & r ⟨ s ⟩ r^{- 1} = ⟨ r^{2} s ⟩ . \end{array}$
b.: In $S_{4}$ , we have
$(2 3 4) ⟨ (1 2), (3 4) ⟩ {(2 3 4)}^{- 1} = ⟨ (1 3), (2 4) ⟩ .$

Lemma 2.12.9. §

A subgroup $N$ of $G$ is normal if and only if $𝑎𝑁 a^{- 1} = N$ for all $a \in G$ .

Proof.

Let $a \in G$ . The function $𝜃_{a} : G \to G$ given by right multiplication by $a$ , i.e., $𝜃_{a} (g) = 𝑔𝑎$ is a bijection by the cancellation theorem. Moreover, $𝜃_{a}$ restricts to bijections $𝑎𝑁 a^{- 1} \to 𝑎𝑁$ and $N \to 𝑁𝑎$ , so $𝑎𝑁 a^{- 1} = N$ if and only if $𝑎𝑁 = 𝑁𝑎$ . □

Corollary 2.12.10. §

A subgroup $N$ of $G$ is normal if and only if $𝑎𝑛 a^{- 1} \in N$ for all $a \in G$ and $n \in N$ .

Proof.

The only if direction follows from Lemma 2.12.9. On the other hand, the condition $𝑎𝑛 a^{- 1} \in N$ for all $a \in G$ and $n \in N$ clearly implies that $𝑎𝑁 a^{- 1} \leq N$ , which we have seen implies $𝑎𝑁 \leq 𝑁𝑎$ . But it also means $N a^{- 1} \leq a^{- 1} N$ for all $a \in G$ , and this equation for $a^{- 1}$ reads $𝑁𝑎 \leq 𝑎𝑁$ , which means $𝑎𝑁 = 𝑁𝑎$ , as desired. □

The following proposition gives an extremely useful criterion for a group to be normal.

Proposition 2.12.11. §

Let $ϕ : G \to G^{'}$ be a homomorphism of groups. Then $\ker ϕ$ is a normal subgroup of $G$ .

Proof.

Let $a \in G$ and $n \in \ker ϕ$ . Then we have

ϕ (𝑎𝑛 a^{- 1}) = ϕ (a) ϕ (n) ϕ {(a)}^{- 1} = ϕ (a) ϕ {(a)}^{- 1} = e^{'},

where $e^{'}$ is the identity of $G^{'}$ . In other words, $𝑎𝑛 a^{- 1} \in N$ , so $N$ is normal by Corollary 2.12.10. □

Example 2.12.12. §

The special linear group ${SL}_{n} (ℝ)$ is a normal subgroup of the general linear group ${GL}_{n} (ℝ)$ , as it is the kernel of the determinant map.

Here are two other examples.

Example 2.12.13. §

Consider the group

Aff (ℝ) = {(\begin{matrix} a & b \\ 0 & 1 \end{matrix}) | a \in ℝ^{\times}, b \in ℝ},

which is a subgroup of ${GL}_{2} (ℝ)$ . Then the set

N = {(\begin{matrix} 1 & t \\ 0 & 1 \end{matrix}) | t \in ℝ}

is a normal subgroup of $G$ . To see this is either the following calculation

(\begin{matrix} a & b \\ 0 & 1 \end{matrix}) (\begin{matrix} 1 & t \\ 0 & 1 \end{matrix}) {(\begin{matrix} a & b \\ 0 & 1 \end{matrix})}^{- 1} = (\begin{matrix} a & 𝑎𝑡 + b \\ 0 & 1 \end{matrix}) (\begin{matrix} a^{- 1} & - a^{- 1} b \\ 0 & 1 \end{matrix}) = (\begin{matrix} 1 & 𝑎𝑡 \\ 0 & 1 \end{matrix}) \in N,

or much more simply, that $N$ is the kernel of the restriction of the determinant map to $G$ .

On the other hand, the subgroup

H = {(\begin{matrix} x & 0 \\ 0 & 1 \end{matrix}) ∣ x \in ℝ^{\times}}

is not a normal subgroup of $G$ . In fact,

(\begin{matrix} a & b \\ 0 & 1 \end{matrix}) (\begin{matrix} x & 0 \\ 0 & 1 \end{matrix}) {(\begin{matrix} a & b \\ 0 & 1 \end{matrix})}^{- 1} = (\begin{matrix} 𝑎𝑥 & b \\ 0 & 1 \end{matrix}) (\begin{matrix} a^{- 1} & - a^{- 1} b \\ 0 & 1 \end{matrix}) = (\begin{matrix} x & (1 - x) b \\ 0 & 1 \end{matrix}),

and the latter element is not in $H$ if $x \neq 1$ and $b \neq 0$ .

Example 2.12.14. §

Let $n \geq 3$ . Let $τ \in S_{n}$ be a $k$ -cycle with $k \geq 2$ . Then $⟨ τ ⟩$ is not normal in $S_{n}$ unless $n = k = 3$ . If $⟨ τ ⟩$ were normal in $S_{n}$ , then every conjugate of $τ$ would have to be a nontrivial power of $τ$ , of which there are $k - 1$ . On the other hand, the conjugates of $τ$ are exactly the $k$ -cycles, of which there are

\frac{n!}{k (n - k)!} \geq \frac{(n - 1)!}{(n - k)!} \geq n - 1,

which forces $n = k$ in order that $⟨ τ ⟩$ might possibly be normal. But for $n = k$ , we have

\frac{n!}{k (n - k)!} = (n - 1)!,

and the latter term is greater than $n$ if $n > 3$ , so $n = k = 3$ . On the other hand, we have already seen that

A_{3} = ⟨ (1 2 3) ⟩ = ⟨ (1 3 2) ⟩ ◃ S_{3} .

2.13. Quotient groups

The sets of left and right cosets of a normal subgroup are of course the same set, and in this section we prove that this set can be given the structure of a group.

Definition 2.13.1. §

Let $A$ and $B$ be subsets of a group $G$ . Then we define the product of $A$ and $B$ as

𝐴𝐵 = {𝑎𝑏 ∣ a \in A, b \in B}

Moreover, if $g$ is an element of $G$ , we define

\begin{array}{rcl} 𝑔𝐴 = {𝑔𝑎 ∣ a \in A} & and & 𝐴𝑔 = {𝑎𝑔 ∣ a \in A} . \end{array}

Remark 2.13.2. §

If $H$ is a subgroup of $G$ , then $𝐻𝐻 = H$ .

Theorem 2.13.3. §

Let $N$ be a normal subgroup of a group $G$ . Then the product of cosets as subsets of $G$ provides a binary operation on $G$ that satisfies

𝑎𝑁 \cdot 𝑏𝑁 = 𝑎𝑁𝑏𝑁 = 𝑎𝑏𝑁 .

Moreover, $G ∕ N$ is a group under this operation.

Proof.

Let $a, b \in G$ . Since $N$ is normal in $G$ , we have $𝑁𝑏 = 𝑏𝑁$ . Therefore, as sets we have

𝑎𝑁𝑏𝑁 = a (𝑁𝑏) N = a (𝑏𝑁) N = 𝑎𝑏𝑁𝑁 = 𝑎𝑏𝑁,

as desired. The associativity of the operation is a direct consequence of the associativity of the operation on $G$ , as $(𝑎𝑏) 𝑐𝑁 = a (𝑏𝑐) N$ for any $a, b, c \in G$ . Then $N = 𝑒𝑁$ is easily seen to be the identity element of $G ∕ N$ , and the inverse of $𝑎𝑁$ is $a^{- 1} N$ , since

𝑎𝑁 \cdot a^{- 1} N = a a^{- 1} N = N = a^{- 1} N \cdot 𝑎𝑁 .

Therefore $G ∕ N$ is a group under this operation. □

Definition 2.13.4. §

Let $N$ be a normal subgroup of a group $G$ . The quotient group of $G$ by $N$ is the group that is the set $G ∕ N$ with the binary operation $𝑎𝑁 \cdot 𝑏𝑁 = 𝑎𝑏𝑁$ for $a, b \in G$ .

Examples 2.13.5. §

a.

The set of cosets of $𝑛ℤ$ in $ℤ$ is a group under addition of cosets, and it is exactly the group $ℤ ∕ 𝑛ℤ$ constructed before, since

(a + 𝑛ℤ) + (b + 𝑛ℤ) = (a + b) + 𝑛ℤ

by definition.

b.

The quotient groups $D_{n} ∕ ⟨ r ⟩$ for $n \geq 3$ and $S_{n} ∕ A_{n}$ for $n \geq 2$ are all cyclic groups of order $2$ .

c.

Suppose that $n \geq 4$ is even. Let $H = ⟨ r^{2} ⟩ ◃ D_{n}$ . Then $[D_{n} : H] = 4$ , and the four distinct cosets are $H$ , $𝑟𝐻$ , $𝑠𝐻$ , and $𝑟𝑠𝐻$ . Since the square of each of these cosets is $H$ , we have an isomorphism between $D_{n} ∕ H$ and the Klein four-group

D_{n} ∕ 𝐻≅ℤ ∕ 2 ℤ \times ℤ ∕ 2 ℤ

that takes $𝑟𝐻$ to $(1, 0)$ and $𝑠𝐻 \to (0, 1)$ .

Remark 2.13.6. §

If $N$ is not normal in $G$ , then $G ∕ N$ is not a group under the product of left cosets. In fact, for $a, b \in N$ , the set $𝑎𝑁𝑏𝑁$ will not in general be a left coset. E.g., if one takes $b = a^{- 1}$ and $a$ to be such that $𝑎𝑁 a^{- 1}$ contains an element not in $N$ , then $𝑎𝑁 a^{- 1} N$ will contain but not equal $N$ , so it is not a left coset.

Moreover, if one simply tries to define $𝑎𝑁 \cdot 𝑏𝑁 = 𝑎𝑏𝑁$ , then the resulting operation is not well-defined, as it depends on the choice of coset representatives. E.g., assuming it were well-defined and again taking $a$ and $b$ as above, we can find $n \in N$ such that $𝑎𝑛 a^{- 1} \notin N$ , so

N = 𝑎𝑁 \cdot a^{- 1} N = 𝑎𝑛𝑁 \cdot a^{- 1} N = 𝑎𝑛 a^{- 1} N \neq N,

which contradicts well-definedness.

Note that the function $G \to G ∕ N$ that takes $g \in G$ to its $N$ -coset $𝑔𝑁$ is a homomorphism by definition of the quotient group. We give it a name.

Definition 2.13.7. §

Let $N$ be a normal subgroup of a group $G$ . The quotient map $π_{N} : G \to G ∕ N$ is the homomorphism defined by $π_{N} (g) = 𝑔𝑁$ for $g \in G$ .

Remark 2.13.8. §

The kernel of the quotient map $π_{N} : G \to G ∕ N$ is $N$ .

Corollary 2.13.9. §

A subgroup $N$ of $G$ is normal if and only if there exists a group $G^{'}$ and a group homomorphism $ϕ : G \to G^{'}$ such that $N = \ker ϕ$ .

We end with the following result on the subgroups of quotient groups.

Proposition 2.13.10. §

Let $G$ be a group and $N$ be a normal subgroup of $G$ . Then the subgroups of $G ∕ N$ are exactly the quotient groups $H ∕ N$ , where $H$ is a subgroup of $G$ containing $N$ . Moreover, such a subgroup $H$ of $G$ is normal in $G$ if and only if $H ∕ N$ is normal in $G ∕ N$ .

Proof.

We first note that if $H$ is a subgroup of $G$ containing $N$ , then $N$ is normal in $H$ , so we may form the quotient group $H ∕ N$ . Its binary operation agrees with the restriction of the operation on $G ∕ N$ (multiplication of $N$ -cosets), so it is a subgroup of $G ∕ N$ .

Conversely, if $Q$ is a subgroup of $G ∕ N$ , then set

H = {h \in G ∣ h𝑁 \in Q} .

Then $𝑛𝑁 = N \in Q$ , in that it is the identity element of $G ∕ N$ and $Q$ is a subgroup, so $N \subseteq H$ . That $H$ is a subgroup of $G$ follows directly from the fact that $Q$ is a subgroup of $G ∕ N$ , since if $h𝑁, 𝑘𝑁 \in Q$ , then $h𝑁 \cdot {(𝑘𝑁)}^{- 1} \in Q$ , so $h k^{- 1} N \in Q$ , which means that $h k^{- 1} \in H$ .

Finally, let $H$ be a subgroup of $G$ containing $N$ . Then, for $a \in G$ , we have

𝑎𝑁 \cdot (H ∕ N) = {𝑎h𝑁 ∣ h \in H} = 𝑎𝐻

and

(H ∕ N) \cdot 𝑎𝑁 = {𝑁h𝑎 ∣ h \in H} = 𝐻𝑎,

so $H ∕ N ⊴ G ∕ N$ if and only if $H ⊴ G$ . □

Theorem 2.13.11 (First Isomorphism Theorem). §

Let $ϕ : G \to G^{'}$ be a homomorphism of groups. Then the function

\bar{ϕ} : G ∕ \ker ϕ \to_{}^{\sim} im ϕ, \bar{ϕ} (a \ker ϕ) = ϕ (a)

for $a \in G$ is a well-defined group isomorphism.

Proof.

First, note that if $a, b \in G$ are such that $a \ker ϕ = b \ker ϕ$ , then $a = 𝑏𝑘$ for some $k \in \ker ϕ$ , so

\bar{ϕ} (a \ker ϕ) = ϕ (𝑏𝑘) = ϕ (b) ϕ (k) = ϕ (b) = \bar{ϕ} (b \ker ϕ),

and hence $\bar{ϕ}$ is well-defined. Moreover, if $\bar{ϕ} (a \ker ϕ) = 0$ , then $ϕ (a) = 0$ , so $a \in \ker ϕ$ , and therefore we have that $\bar{ϕ}$ is injective. Since $ϕ$ has image $im ϕ$ , so does $\bar{ϕ}$ , and hence $\bar{ϕ}$ is surjective by definition. □

Remark 2.13.12. §

We have that $ϕ = ι_{im ϕ} \circ \bar{ϕ} \circ π_{\ker ϕ}$ , as is represented in the following diagram:

Factorization through a kernel quotient and image. A full diagram description follows.

Diagram description: Factorization through a kernel quotient and image

The direct map phi is the composite of the quotient projection, the induced map phi bar, and the inclusion of the image, in that order.

Objects, listed by row and column:

Row 1, from left to right: column 1: G; column 4: G prime.
Row 2, from left to right: column 2: G / kernel phi; column 3: image phi.

Arrows and lines:

An arrow from G to G prime, labelled phi.
An arrow from G to G / kernel phi, labelled pi subscript (kernel phi).
An arrow from G / kernel phi to image phi, labelled bar of (phi).
An arrow from image phi to G prime, labelled iota subscript (image phi).

Example 2.13.13. §

The determinant map $\det : {GL}_{n} (ℝ) \to ℝ^{\times}$ induces an isomorphism

\bar{\det} : {GL}_{n} (ℝ) ∕ {SL}_{n} (ℝ) \to_{}^{\sim} ℝ^{\times} .

To give another example, we make the following definition.

Definition 2.13.14. §

Let $G_{1}, G_{2}, \dots, G_{k}$ be groups. The $i$ th projection map is the surjective homomorphism

π_{i} : \prod_{j = 1}^{k} G_{j} \to G_{i}, π_{i} (g_{1}, g_{2}, \dots, g_{k}) = g_{i} .

Remark 2.13.15. §

The projection map $π_{i}$ of Definition 2.13.14 has kernel

\ker π_{i} = {(g_{1}, g_{2}, \dots, g_{k}) \in \prod_{j = 1}^{k} G_{j} | g_{j} = e_{j} for all j \neq i},

where $e_{i}$ is the identity element of $G_{i}$ . By the first isomorphism theorem, $π_{i}$ induces an isomorphism

\begin{array}{rcl} {\bar{π}}_{i} : (\prod_{i = 1}^{k} G_{j}) ∕ \ker π_{i} \to_{}^{\sim} G_{i} . \end{array}

For instance, if $k = 2$ and $i = 1$ , we can think of ${\bar{π}}_{i}$ as an isomorphism

\frac{G_{1} \times G_{2}}{{e_{1}} \times G_{2}} \to_{}^{\sim} G_{1} .

Example 2.13.16. §

We have

\frac{ℤ ∕ 4 ℤ \times ℤ ∕ 4 ℤ}{⟨ 2 ⟩ \times ⟨ 2 ⟩} ≅ℤ ∕ 2 ℤ \times ℤ ∕ 2 ℤ .

To see this, define a map $ϕ : ℤ ∕ 4 ℤ \times ℤ ∕ 4 ℤ \to ℤ ∕ 2 ℤ \times ℤ ∕ 2 ℤ$ by $ϕ (a, b) = (a, b)$ . Then $ϕ$ is surjective with kernel $⟨ 2 ⟩ \times ⟨ 2 ⟩$ , so the first isomorphism theorem applied to $ϕ$ provides the isomorphism.