Abstract Algebra

We note that $(e,e)\in G$ is an identity, that $(\phi (h^{-1})(n^{-1}),h^{-1})$ is inverse to $(n,h)$, and we leave it to the reader to check associativity. Clearly $H,N\le G$ by definition of the multiplication, and we check that for $h\in H$ and $n\in N$, we have

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Any $g\in G$ can be written as $\mathit{𝑛h}$ for some $n\in N$ and $h\in H$ by assumption, so $f$ is onto. For $n,n^{\prime }\in N$ and $h,h^{\prime }\in H$, we have

If $\psi (n,h)=\mathit{𝑛h}=e$, then $n=h^{-1}\in N\cap H$, so $n=h=e$. □

Alternate proof of Proposition 4.11.4. By Proposition 7.1.5, we need only that $\phi :H\to \mathrm{Aut}(N)$ given by $\phi (h)(n)=\mathit{h𝑛}{h}^{-1}$ for $h\in H$ and $n\in N$ is the trivial map. That is, we need that $\mathit{h𝑛}{h}^{-1}=n$, or $[h,n]=e$, for all such $h$ and $n$. This follows as $H$ and $N$ are normal, so $[h,n]\in H\cap N=\{e\}$. □

Note that $f$ has an inverse given by $f^{-1}(n^{\prime },h^{\prime })=({𝜃}^{-1}(n^{\prime }),{\psi }^{-1}(h^{\prime }))$ for all $n^{\prime }\in N$ and $h^{\prime }\in H$, so we need only show that $f$ is a homomorphism. Letting $n_1,n_2\in N$ and $h_1,h_2\in H$, we calculate:

To see that the first coordinates of these expressions are equal, we check that

Thus, $f$ is a homomorphism. □

Let $G$ be a nonabelian group. By Theorem 4.11.5, we have that it has a unique normal subgroup $Q$ of order $q$, and let $P$ be a subgroup of order $p$. By Proposition 7.1.5, we have that $G=Q⋊P$. We have $\mathit{𝑃}\cong \mathit{\mathbb{Z}}∕\mathit{𝑝\mathbb{Z}}$ and $\mathit{𝑄}\cong \mathit{\mathbb{Z}}∕\mathit{𝑞\mathbb{Z}}$. Fixing such isomorphisms and recalling the canonical isomorphism $\mathrm{Aut}(\mathbb{Z}∕\mathit{𝑞\mathbb{Z}})\cong {(\mathbb{Z}∕\mathit{𝑞\mathbb{Z}})}^{\times }$, we are reduced to showing that there is a unique isomorphism class of semi-direct product $(\mathbb{Z}∕\mathit{𝑞\mathbb{Z}}){⋊}_{\phi }(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})$, where $\phi :\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}\to {(\mathbb{Z}∕\mathit{𝑞\mathbb{Z}})}^{\times }$ is a nontrivial homomorphism. The group ${(\mathbb{Z}∕\mathit{𝑞\mathbb{Z}})}^{\times }$ is cyclic by Corollary 6.5.5. Let $a\in {(\mathbb{Z}∕\mathit{𝑞\mathbb{Z}})}^{\times }$ be a generator.

Any nontrivial homomorphism $\phi :\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}\to {(\mathbb{Z}∕\mathit{𝑞\mathbb{Z}})}^{\times }$ must send $1$ to an element of order $p$ in ${(\mathbb{Z}∕\mathit{𝑞\mathbb{Z}})}^{\times }$. If we set $b=a^{(q-1)∕p}$, then $\phi (1)=b^i$ for some $i\in \mathbb{Z}$ with $i\not\equiv 0modp$. Let us denote this particular homomorphism by ${\phi }_i$, and define ${\psi }_i:\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}\to \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}$ to be multiplication by $i$. Then ${\phi }_i={\phi }_1\circ {\psi }_i$ since both maps send $1$ to $b^i$. Proposition 7.1.9 then tells us that the semidirect products $(\mathbb{Z}∕\mathit{𝑞\mathbb{Z}})⋊(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})$ defined by ${\phi }_1$ and ${\phi }_i$ are isomorphic. That is, there is a unique isomorphism class of nonabelian semidirect product of order $\mathit{𝑝𝑞}$. □

Let ${(H_i)}_{i=0}^t$ and ${(K_i)}_{i=0}^s$ be subnormal series in $G$. For $0\le i<t$ and $0\le j<s$, let

Set $M_{\mathit{𝑚𝑛}}=N_{\mathit{𝑚𝑛}}=G$ as well. Then $M_{\mathit{𝑠𝑖}+j}⊴M_{\mathit{𝑠𝑖}+j+1}$ for $0\le i\le t-1$ and $0\le j\le s-2$ as $K_j⊴K_{j+1}$, and $M_{\mathit{𝑠𝑖}+s-1}⊴H_{i+1}=M_{s(i+1)}$ for $0\le i\le t-1$ as $H_i⊴H_{i+1}$ and $H_{s-1}⊴G$. Thus ${(M_i)}_{i=0}^{\mathit{𝑚𝑛}}$ is a subnormal series, as is ${(N_i)}_{i=0}^{\mathit{𝑚𝑛}}$. In fact, we see from this that ${(M_i)}_i$ refines ${(H_i)}_i$ and ${(N_i)}_i$ refines ${(K_i)}_i$.

It remains to see that ${(M_i)}_i$ and ${(N_i)}_i$ are equivalent. For $0\le i\le t-1$ and $0\le j\le s-2$, note that

by the butterfly lemma, and

for $0\le i\le t-1$ since $K_s=G$. Thus, the two refinements are equivalent. □

Let $0\le i\le t-1$. The quotient $(H_i\cap N)∕(H_{i-1}\cap N)$ is a subgroup of the simple group $H_i∕H_{i-1}$ and therefore necessarily trivial or improper. Let $s$ be the number of simple quotients. Let $f(0)=0$, and for $1\le j\le s$, let $f(j)$ be the smallest positive integer greater than $f(j-1)$ and such that $(H_{f(j)}\cap N)∕(H_{f(j)-1}\cap N)$ is simple. Then $H_{f(j)-1}\cap N=H_{f(j-1)}\cap N$, and the result follows.

Similarly, by the third isomorphism theorem, we have

which is a quotient of $H_i∕H_{i-1}$ by the image of $H_{i+1}\cap N$ in it. Since $H_{i+1}∕H_i$ is simple, this image is either trivial or $H_{i+1}∕H_i$. That is, ${\overline{H}}_{i+1}∕{\overline{H}}_i$ is either trivial or simple. Let $s$ be the number of simple terms. Set $f^{\prime }(0)=0$, and for $1\le j\le r$, take $f^{\prime }(j)$ to be the smallest integer greater than $f^{\prime }(j-1)$ such that ${\overline{H}}_{f^{\prime }(j)}∕{\overline{H}}_{f^{\prime }(j)-1}$ is simple. Then ${\overline{H}}_{f^{\prime }(j)-1}={\overline{H}}_{f^{\prime }(j-1)}$.

Note that ${\overline{H}}_i\ne {\overline{H}}_{i-1}$ if and only if $H_i\cap N=H_{i-1}\cap N$. Then $r+s=t$ and the images of $f$ and $f^{\prime }$ are complementary by construction. □

To show part a, we work by induction on the order $n$ of the group $G$. It is clear in the case that $G$ is trivial, with $t=0$. Now, if $G$ is nontrivial of order $n$, then either it is simple, and the composition series is $1\le G$, or it is not, and there exists a nontrivial normal subgroup $K◃G$, and then $K$ and $G∕K$ have composition series by induction. The result is then immediate from Lemma 7.2.13.

To see that the composition series is unique in the stated sense of part b, start with two composition series as in the statement of the theorem. We work by induction on the minimal length $s$ of a composition series for $G$. If $s=0$, then $G$ is trivial. If $s=1$, then $G$ is simple, so it cannot have a nontrivial normal subgroup, and all composition series must have length $1$. Consider $N=N_{s-1}$, which has the composition series

as well as a composition series

for some $r\le t$ and increasing $f:X_r\to X_t$ by Lemma 7.2.11a. Since the minimal length of a composition series of $N$ is less than $s$, we have by induction that $r=s-1$ and there exists $\sigma \in S_{s-1}$ such that

for all $i$, again by Lemma 7.2.11a.

Let $k<t$ be maximal such that $H_{k-1}\le N$. Then

Since $H_k∕H_{k-1}$ is simple, this forces $H_{k-1}=H_k\cap N$. In particular, $k$ is not in the image of $f$. Moreover, we have

the latter step as $N$ is a maximal normal subgroup of $G$ and $H_k\le ̸N$. As we have found the final composition factor in the series ${(N_i)}_i$ among those of the series ${(H_i)}_i$, it remains only to show that $s=t$. If $(H_i\cap N)∕(H_{i-1}\cap N)$ nontrivial for any for $i\ne k$, then by Lemma 7.2.11c, the group $G∕N$ has a composition series of length at least $2$, but $G∕N$ is simple, so this is impossible. Thus, $r=t-1$ as well, as needed. □

First, Lemma 4.3.15c tells us that $G^{(i)}$ is characteristic in $G^{(i-1)}$ for each $i\ge 1$, and then the result follows recursively from Lemma 4.3.16. □

That (i) implies (ii) is a consequence of the facts that the group $G^{(i)}$ are characteristic, hence normal, and that $G^{(i-1)}∕G^{(i)}$ is the quotient of $G^{(i-1)}$ by its commutator subgroup, hence abelian. That (ii) implies (iii) is obvious. So, suppose (iii) and let

be a subnormal series of length $t$. (Note the reversed indexing, as in Remark 7.2.5.) We claim that $G^{(i)}\le N_i$ for each $i\ge 0$. For $i=0$, we have $G=G^{(0)}=N_0$. In general, suppose inductively that $G^{(i-1)}\le N_{i-1}$. Then $G^{(i)}\le [N_{i-1},N_{i-1}]$ by definition, and we have $[N_{i-1},N_{i-1}]\le N_i$ as $N_{i-1}∕N_i$ is abelian. Therefore, we have that $G_t=N_t=1$, and $G$ is solvable. □

Let $G$ be a group and $N$ a normal subgroup. If $G$ is solvable, then it has a composition series with abelian factors, so $N$ and $G∕N$ are solvable by Lemma 7.2.11. Part (iii) is a corollary of Corollary 7.2.13, since the derived series of $N$ and $G∕N$ have abelian composition factors. □

If $G$ has cyclic Jordan-Hölder factors, then $G$ is solvable by Proposition 7.3.7. If $G$ is solvable and has a composition series, then the composition factors are abelian by Proposition 7.3.7 and the uniqueness in Theorem 7.2.14. As composition factors, they are also simple, hence cyclic of prime order, from which it follows that $G$ is finite. □

This is almost trivial by induction, as

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Let $n$ be the nilpotency class of $G$. Then $G_n$ is nontrivial but central in $G$ since $[G,G_n]=1$. □

The result is clear for abelian groupts, which are the nilpotent groups of nilpotency class $1$. Let $\overline{G}=G∕Z(G)$ for brevity of notation.

Suppose that $G$ is nilpotent of nilpotency class $n\ge 2$. As $G_{n+1}=[G,G_n]=1$, we have $1\ne G_n\le Z(G)$. Since $Z(G)$ is central in $G$, it follows that

for all $i$. Since $G_{n-1}$ is not central in $G$ by definition, we have that ${\overline{G}}_{n-1}\ne 1$. By induction on $n$, we then have

for $1\le i\le n-1$, and $Z^{n-1}(\overline{G})=\overline{G}$. Taking the inverse images of these groups under the quotient map from $G$, we obtain

for $1\le i\le n-1$ as well, and in particular $Z^{n-1}(G)<Z^n(G)=G$.

Conversely, if $Z^n(G)=G$ for some minimial $n\ge 2$, then $Z^{n-1}(\overline{G})=\overline{G}$, so $\overline{G}$ is nilpotent of nilpotency class $n-1$ by induction on $n$. This means that $G_n\le Z(G)$, and therefore $G_{n+1}=1$, so $G$ is nilpotent. □

This follows by induction on the order of a nontrivial $p$-group $P$, since $Z(P)\ne 1$, and $P∕Z(P)$ is a $p$-group which we suppose by induction to be nilpotent. Then $Z^{i+1}(P)$ is the inverse image of $Z^i(P∕Z(P))$ in $P$, so $P$ is nilpotent as well. □