GTM148 抄书笔记 Part IV. (不定期更新)

上一篇: GTM148 抄书笔记 Part III.

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Contents

• Contents
• Chapter VII. Extensions and Cohomology
  • The Extension Problem
  • Automorphism Groups
  • Semidirect Products
  • Wreath Products
  • Factor Sets
  • Theorems of Schur-Zassenhaus and Gaschütz

Chapter VII. Extensions and Cohomology

The Extension Problem

Definition 7.1.1 If $K$ and $Q$ are groups, then an extension of $K$ by $Q$ is a group $G$ having a normal subgroup $K_1\cong K$ with $G/K_1\cong Q$.

Example 7.1.2

1. Both $\mathbf Z_6$ and $\mathrm S_3$ are extensions of $\mathbf Z_3$ by $\mathbf Z_2$. However, $\mathbf Z_6$ is an extension of $\mathbf Z_2$ by $\mathbf Z_3$, but $\mathrm S_3$ is not such an extension.
2. For any groups $K$ and $Q$, the direct product $K\times Q$ is an extension of $K$ by $Q$ as well as an extension of $Q$ by $K$.

Proposition 7.1.3

1. If $K$ and $Q$ are finite, then every extension $G$ of $K$ by $Q$ has order $|K||Q|$.
2. If $G$ has a normal series with factor groups $Q_n,\ldots,Q_1$, then $|G|=\prod|Q_i|$.

Proposition 7.1.4 If $(a,b)=1$ and $K$ and $Q$ are abelian groups of orders $a$ and $b$, respectively, then there is only one (to isomorphism) abelian extension of $K$ by $Q$.

Proposition 7.1.5 If $p$ is prime, every nonabelian group of order $p^3$ is an extension of $\mathbb Z_p$ by $\mathbb Z_p\times\mathbb Z_p$.

Automorphism Groups

Definition 7.2.1 The automorphism group of a group $G$, denoted by $\mathrm{Aut}(G)$, is the set of all the automorphisms of $G$ under the operation of composition.

It is easy to check that $\mathrm{Aut}(G)$ is a group; indeed, it is a subgroup of the symmetric group $\mathrm S_G$.

Definition 7.2.2 An automorphism $\varphi$ of $G$ is inner if it is conjugation by some element of $G$; otherwise, it is outer. Denote the set of all inner automorphisms of $G$ by $\mathrm{Inn}(G)$.

Theorem 7.2.3

1. N/C Lemma If $H\le G$, then $\mathrm C_G(H)\lhd\mathrm N_G(H)$ and $\mathrm N_G(H)/\mathrm C_G(H)$ can be imbedded in $\mathrm {Aut}(H)$.
2. $\mathrm {Inn}(G)\lhd\mathrm {Aut}(G)$ and $G/\mathrm Z(G)\cong\mathrm{Inn}(G)$.

Proof

1. If $a\in G$, let $\gamma_a$ denote conjugation by $a$. Define $\varphi\!:\mathrm N_G(H)\rightarrow\mathrm{Aut}(H)$ by $a\mapsto\gamma_a\big|_H$ (note that $\gamma_a\big|_H\in\mathrm{Aut}(H)$ because $a\in\mathrm N_G(H)$); $\varphi$ is easily seen to be a homomorphism. The following statements are equivalent: $a\in\mathrm{ker}\varphi$; $\gamma_a\big|_H$ is the identity on $H$; $aha^{-1}=h$ for all $h\in H$; $a\in\mathrm C_G(H)$. By the first isomorphism theorem, $\mathrm C_G(H)\lhd\mathrm N_G(H)$ and $\mathrm N_G(H)/\mathrm C_G(H)\cong\mathrm{im}\varphi\le\mathrm{Aut}(H)$.
2. If $H=G$, then $\mathrm N_G(G)=G, \mathrm C_G(G)=\mathrm Z(G)$, and $\mathrm{im}\varphi=\mathrm{Inn}(G)$. Therefore, $G/\mathrm Z(G)\cong\mathrm{Inn}(G)$ is a special case of the isomorphism just established. To see that $\mathrm{Inn}(G)\lhd\mathrm{Aut}(G)$, take $\gamma_a\in\mathrm{Inn}(G)$ and $\varphi\in\mathrm{Aut}(G)$. Then $\varphi\gamma_a\varphi^{-1}=\gamma_{\varphi a}\in\mathrm{Inn}(G)$.
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Definition 7.2.4 The group $\mathrm{Aut}(G)/\mathrm{Inn}(G)$ is called the outer automorphism group of $G$.

Example 7.2.6 Nonisomorphic groups can have isomorphic automorphism groups: $\mathrm{Aut}(\mathbf V)\cong\mathrm S_3\cong\mathrm{Aut}(\mathrm S_3)$.

Example 7.2.7 If $G$ is an elementary abelian group of order $p^n$, then $\mathrm{Aut}(G)\cong\mathrm {GL}(n,p)$. (This follows from Proposition 4.1.17: $G$ is a vector space over $\mathbb Z_p$ and every automorphism is a nonsingular linear transformation.)

Theorem 7.2.8 If $G$ is a cyclic group of order $n$, then $\mathrm{Aut}(G)\cong\mathrm U(\mathbb Z_n)$.

Proof

Let $G=\langle a\rangle$. If $\varphi\in\mathrm{Aut}(G)$, then $\varphi(a)=a^k$ for some $k$; moreover, $a^k$ must be a generator of $G$, so that $(k,n)=1$, and $[k]\in\mathrm U(\mathbb Z_n)$. It is routine to show that $\Theta\!:\mathrm{Aut}(G)\rightarrow\mathrm U(\mathbb Z_n)$, defined by $\Theta(\varphi)=[k]$, is an isomorphism.
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Proposition 7.2.9

1. $\mathrm{Aut}(\mathbb Z_2)=\mathbf 1$; $\mathrm{Aut}(\mathbb Z_4)\cong\mathbb Z_2$; if $m\ge 3$, then $\mathrm {Aut}(\mathbb Z_{2^m})\cong\mathbb Z_2\times\mathbb Z_{2^{m-2}}$; $\mathrm{Aut}(\mathbb Z)\cong\mathbb Z_2$.
2. If $p$ is an odd prime, then $\mathrm{Aut}(\mathbb Z_{p^m})\cong\mathbb Z_i$, where $l=(p-1)p^{m-1}$.
3. If $n=p_1^{e_1}\ldots p_t^{e_t}$, where the $p_i$ are distinct primes and the $e_i>0$, then $\mathrm{Aut}(\mathbb Z_n)\cong\prod_i\mathrm{Aut}(\mathbb Z_{q_i})$, where $q_i=p_i^{e_i}$.

Definition 7.2.10 A group $G$ is complete if it is centerless and every automorphism of $G$ is inner.

It follows that $\mathrm {Aut}(G)\cong G$ for every complete group.

Lemma 7.2.11 An automorphism $\varphi$ of $\mathrm S_n$ preserves transpositions ($\varphi(\tau)$ is a transposition whenever $\tau$ is) if and only if $\varphi$ is inner.

Proof

If $\varphi$ is inner, then it preserves the cycle structure of every permutation, by Theorem 3.2.3.

We prove by induction on $t\ge 2$, that there exist conjugations $\gamma_2,\ldots,\gamma_t$ such that $\gamma_t^{-1}\ldots\gamma_2^{-1}\varphi$ fixes $(1\ \ 2),\ldots,(1\ \ t)$. If $\pi\in\mathrm S_n$, we will denote $\varphi(\pi)$ by $\pi^\varphi$ in this proof. By hypothesis, $(1\ \ 2)^\varphi=(i\ \ j)$ for some $i,j$; define $\gamma_2$ to be conjugation by $(1\ \ i)(2\ \ j)$. By Lemma 3.2.2, the quick way of computing conjugates in $\mathrm S_n$, we see that $(1\ \ 2)^\varphi=(1\ \ 2)^{\gamma_2}$, and so $\gamma_2^{-1}\varphi$ fixes $(1\ \ 2)$.

Let $\gamma_2,\ldots,\gamma_t$ be given by the inductive hypothesis, so that $\psi=\gamma_t^{-1}\ldots\gamma_2^{-1}\varphi$ fixes $(1\ \ 2),\ldots,(1\ \ t)$. Since $\psi$ preserves transpositions, $(1\ \ t+1)^\psi=(l\ \ k)$. Now $(1\ \ 2)$ and $(l\ \ k)$ cannot be disjoint, lest $[(1\ \ 2)(1\ \ t+1)]^{\psi}=(1\ \ 2)^\psi(1\ \ t+1)^\psi=(1\ \ 2)(l\ \ k)$ have order $2$, while $(1\ \ 2)(1\ \ t+1)$ has order $3$. Thus, $(1\ \ t+1)^\psi=(1\ \ k)$ or $(1\ \ t+1)^\psi=(2\ \ k)$. If $k\le t$, then $(1\ \ t+1)^\psi\in\langle(1\ \ 2),\ldots,(1\ \ t)\rangle$, and hence it is fixed by $\psi$; this contradicts $\psi$ being injective, for either $(1\ \ t+1)^\psi=(1\ \ k)=(1\ \ k)^\psi=(2\ \ k)=(2\ \ k)^\psi$. Hence, $k\ge t+1$. Define $\gamma_{t+1}$ to be conjugation by $(k\ \ t+1)$. Now $\gamma_{t+1}$ fixes $(1\ \ 2),\ldots,(1\ \ t)$ and $(1\ \ t+1)^{\gamma_{t+1}}=(1\ \ t+1)^\psi$, so that $\gamma_{t+1}^{-1}\cdots\gamma_2^{-1}\varphi$ fixes $(1\ \ 2),\ldots,(1\ \ t+1)$ and the induction is complete. It follows that $\gamma_n^{-1}\cdots\gamma_{2}^{-1}\varphi$ fixes $(1\ \ 2),\ldots,(1\ \ n)$. But these transpositions generate $\mathrm S_n$, by Proposition 2.1.20, and so $\gamma_n^{-1}\ldots\gamma_2^{-1}\varphi$ is the identity. Therefore, $\varphi=\gamma_2\cdots\gamma_n\in\mathrm{Inn}(\mathrm S_n)$.
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Theorem 7.2.12 If $n\neq 2$ or $n\neq 6$, then $\mathrm S_n$ is complete.

Proof

Let $T_k$ denote the conjugacy class in $\mathrm S_n$ consisting of all products of $k$ disjoint transpositions. By Proposition 1.3.4, a permutation in $\mathrm S_n$ is an involultion if and only if it lies in some $T_k$. It follows that if $\theta\in\mathrm{Aut}(\mathrm S_n)$, then $\theta(T_1)=T_k$ for some $k$. We shall show that if $n\neq 6$, then $|T_k|\neq|T_1|$ for $k\neq 1$. Assuming this, then $\theta(T_1)=T_1$, and Lemma 7.2.11 completes the proof since $\mathrm S_{n}$ is centerless for $n\ge 3$ by Proposition 3.1.19.

Now $|T_1|=n(n-1)/2$. To count $T_k$, observe first that there are

$$\frac 12n(n-1)\times \frac 12(n-2)(n-3)\times\cdots\times\frac 12(n-2k+2)(n-2k+1)$$

$k$-tuples of disjoint transpositions. Since disjoint transpositions commute and there are $k!$ orderings obtained from any $k$-tuple, we have

$$|T_k|=n(n-1)(n-2)\cdots(n-2k+1)/k!2^k.$$

The question whether $|T_1|=|T_k|$ leads to the question whether there is some $k>1$ such that

$$\begin{align} (n-2)(n-3)\cdots(n-2k+1)=k!2^{k-1}. \end{align}$$

Since the right side of $(1)$ is positive, we must have $n\ge 2k$. Therefore, for fixed $n$,

$$\text{left side}\ge(2k-2)(2k-3)\cdots(2k-2k+1)=(2k-2)!.$$

An easy induction shows that if $k\ge 4$, then $(2k-2)!>k!2^{k-1}$, and so $(1)$ can hold only if $k=2$ or $k=3$. When $k=2$, the right side is $4$, and it is easy to see that equality never holds; we may assume, therefore, that $k=3$. Since $n\ge 2k$, we must have $n\ge 6$. If $n> 6$, then the left side of $(1)\ge 5\times 4\times 3\times 2=120$, while the right side is $24$. We have shown that if $n\neq 6$, then $|T_1|\neq|T_k|$ for all $k>1$, as desired.
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Corollary 7.2.13 If $\theta$ is an outer automorphism of $\mathrm S_6$, and if $\tau\in\mathrm S_6$ is a transposition, then $\theta(\tau)$ is a product of three disjoint transpositions.

Corollary 7.2.14 If $n\neq 2$ or $n\neq 6$, then $\mathrm{Aut}(\mathrm S_n)\cong \mathrm S_n$.

Lemma 7.2.15 There exists a transitive subgroup $K\le\mathrm S_6$ of order $120$ which contains no transpositions.

Proof

If $\sigma$ is a $5$-cycle, then $P=\langle\sigma\rangle$ is a Sylow $5$-subgroup of $\mathrm S_5$. The Sylow theorem says that if $r$ is the number of conjugates of $P$, then $r\equiv 1\pmod 5$ and $r$ is a divisor of $120$; it follows easily that $r=6$. The representation of $\mathrm S_5$ on $X$, the set of all left cosets of $N=\mathrm N_{\mathrm S_5}(P)$, is a homomorphism $\rho\!:\mathrm S_5\rightarrow\mathrm S_X\cong\mathrm S_6$. Now $X$ is a transitive $\mathrm S_5$-set, by Proposition 4.2.13, and so $|\mathrm{ker}\rho|\le|\mathrm S_5|/r=|\mathrm S_5|/6=20$, by Proposition 3.5.13(iii). Since the only normal subgroups of $\mathrm S_5$ are $\mathrm S_5,\mathrm A_5$, and $\mathbf 1$, it follows that $\mathrm{ker}\rho=\mathbf 1$ and $\rho$ is an injection. Therefore, $\mathrm{im}\rho\cong\mathrm S_5$ is a transtive subgroup of $\mathrm S_X$ of order $120$.

For notational convenience, let us write $K\le\mathrm S_6$ instead of $\mathrm{im}\rho\le\mathrm S_X$. Now $K$ contains an element $\alpha$ of order $5$ which must be a $5$-cycle; say, $\alpha=(1\ \ 2\ \ 3\ \ 4\ \ 5)$. If $K$ contains a transposition $(i\ \ j)$, then transitivity of $K$ provides $\beta\in K$ with $\beta(j)=6$, and so $\beta(i\ \ j)\beta^{-1}=(\beta i\ \ \beta j)=(l\ \ 6)$ for some $l\neq 6$ (of course, $l=\beta i$). Conjugating $(l\ \ 6)\in K$ by the powers of $\alpha$ shows that $K$ contains $(1\ \ 6), (2\ \ 6), (3\ \ 6), (4\ \ 6)$, and $(5\ \ 6)$. But these transpositions generate $\mathrm S_6$, and this contradicts $K(\cong\mathrm S_5)$ being a proper subgroup of $\mathrm S_6$.
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Theorem 7.2.16

1. There exists an outer automorphism of $\mathrm S_6$.
2. $\mathrm{Aut}(\mathrm S_6)/\mathrm{Inn}(\mathrm S_6)\cong\mathbb Z_2$, and so $|\mathrm{Aut}(\mathrm S_6)|=1440$.

Proof

1. Let $K$ be a transitive subgroup of $\mathrm S_6$ of order $120$, and let $Y$ be the family of its left cosets: $Y=\{\alpha_1K,\ldots,\alpha_6K\}$. If $\theta\!:\mathrm S_6\rightarrow\mathrm S_Y$ is the representation of $\mathrm S_6$ on the left cosets of $K$, then $\mathrm{ker}\theta\le K$ is a normal subgroup of $\mathrm S_6$. But $\mathrm A_6$ is the only proper normal subgroup of $\mathrm S_6$, so that $\mathrm{ker}\theta=\mathbf 1$ and $\theta$ is an injection. Since $\mathrm S_6$ is finite, $\theta$ must be a bijection, and so $\theta\in\mathrm{Aut}(\mathrm S_6)$, for $\mathrm S_Y\cong\mathrm S_6$. Were $\theta$ inner, then it would preserve the cycle structure of every permutation in $\mathrm S_6$. In particular, $\theta_{(1\ \ 2)}$, defined by $\theta_{(1\ \ 2)}\!:\alpha_iK\mapsto(1\ \ 2)\alpha_iK$ for all $i$, is a transposition, and hence $\theta$ fixes $\alpha_iK$ for four different $i$. But if $\theta_{(1\ \ 2)}$ fixes even one left coset, say $\alpha_iK=(1\ \ 2)\alpha_iK$, then $\alpha_i^{-1}(1\ \ 2)\alpha_i$ is a transposition in $K$. This contradiction shows that $\theta$ is an outer automorphism.
2. Let $T_1$ be the class of all transpositions in $\mathrm S_6$, and let $T_3$ be the class of all products of $3$ disjoint transpositions. If $\theta$ and $\psi$ are outer automorphisms of $\mathrm S_6$, then both interchange $T_1$ and $T_3$, by Corollary 7.2.13, and so $\theta^{-1}\psi(T_1)=T_1$. Therefore, $\theta^{-1}\psi\in\mathrm{Inn}(\mathrm S_6)$, by Lemma 7.2.11, and $\mathrm{Aut}(S_6)/\mathrm{Inn}(\mathrm S_6)$ has order $2$.
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Definition 7.2.17 A syntheme is a product of $3$ disjoint transpositions. A pentad is a family of $5$ synthemes, no two of which have a common transposition.

Proposition 7.2.18 $\mathrm S_6$ contains exactly $6$ pentads. They are:

$$\begin{aligned} &(12)(34)(56),(13)(25)(46),(14)(26)(35),(15)(24)(36),(16)(23)(45);\\ &(12)(34)(56),(13)(26)(45),(14)(25)(36),(15)(23)(46),(16)(24)(35);\\ &(12)(35)(46),(13)(24)(56),(14)(25)(36),(15)(26)(34),(16)(23)(45);\\ &(12)(35)(46),(13)(26)(45),(14)(23)(56),(15)(24)(36),(16)(25)(34);\\ &(12)(36)(45),(13)(24)(56),(14)(26)(35),(15)(23)(46),(16)(25)(34);\\ &(12)(36)(45),(13)(25)(46),(14)(23)(56),(15)(26)(34),(16)(24)(35). \end{aligned}$$

Theorem 7.2.19 If $\{\sigma_2,\ldots,\sigma_6\}$ is a pentad in some ordering, then there is a unique outer automorphism $\theta$ of $\mathrm S_6$ with $\theta\!:(1\ \ i)\mapsto\sigma_i$ for $i=2,3,4,5,6$. Moreover, every outer automorphism of $\mathrm S_6$ has this form.

Proof

Let $X=\{(1\ \ 2),(1\ \ 3),(1\ \ 4),(1\ \ 5),(1\ \ 6)\}$. If $\theta$ is an outer automorphism of $\mathrm S_6$, then Corollary 7.2.13 shows that each $\theta((1\ \ i))$ is a syntheme. Since $(1\ \ i)$ and $(1\ \ j)$ do not commute for $i\neq j$, it follows that $\theta((1\ \ i))$ and $\theta((1\ \ j))$ do not commute; hence, $\theta(X)$ is a pentad. Let us count the number of possible functions from $X$ to pentads arising from outer automorphisms. Given an outer automorphism $\theta$, there are $6$ choices of pentad for $\theta(X)$; given such a pentad $P$, there are $5!=120$ bijections $X\rightarrow P$. Hence, there are at most $720$ bijections from $X$ to pentads which can possibly arise as restrictions of outer automorphisms. But there are exactly $720$ outer automorphisms, by Theorem 7.2.16, and no two of them can restrict to the same bijection because $X$ generates $\mathrm S_6$. The statements of the theorem follow.
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Corollary 7.2.20 There is an outer automorphism of $\mathrm S_6$ which has order $2$.

Proposition 7.2.21 If $G$ is a finite nonabelian $p$-group, then $p^2\big||\mathrm{Aut}(G)|$.

Proposition 7.2.22 If $G$ is a finite abelian group, then $\mathrm{Aut}(G)$ is abelian if and only if $G$ is cyclic.

Proposition 7.2.23

1. If $G$ is a finite abelian group with $|G|>2$, then $\mathrm{Aut}(G)$ has even order.
2. If $G$ is not abelian, then $\mathrm{Aut}(G)$ is not cyclic.
3. There is no finite group $G$ with $\mathrm{Aut}(G)$ cyclic of odd order $>1$.

Proposition 7.2.24 If $G$ is a finite group and $\mathrm{Aut}(G)$ acts transitively on $G^\#=G-\{\mathbf 1\}$, then $G$ is an elementary abelian group.

Proposition 7.2.25 If $H$ and $K$ are finite groups whose orders are relatively prime, then $\mathrm{Aut}(H\times K)\cong\mathrm{Aut}(H)\times\mathrm{Aut}(K)$.

Theorem 7.2.26 If $G$ is a nonabelian simple group, then $\mathrm {Aut}(G)$ is complete.

Proof

Let $I=\mathrm{Inn}(G)\lhd\mathrm{Aut}(G)=A$. Now $\mathrm Z(G)=\mathbf 1$, because $G$ is simple and nonabelian, and so Theorem 7.2.3(i) gives $I\cong G$. Now $\mathrm Z(A)\le\mathrm C_A(I)=\{\varphi\in A:\varphi\gamma_g=\gamma_g\varphi\text{ for all }g\in G\}$. We claim that $\mathrm C_A(I)=\mathbf 1$; it will then follow that $A$ is centerless. If $\varphi\gamma_g=\gamma_g\varphi$ for all $g\in G$, then $\gamma_g=\varphi\gamma_g\varphi^{-1}=\gamma_{\varphi(g)}$. Therefore, $\varphi(g)g^{-1}\in\mathrm Z(G)=\mathbf 1$ for all $g\in G$, and so $\varphi=\mathbf 1$.

It remains to show that every $\sigma\in\mathrm{Aut}(A)$ is inner. Now $\sigma(I)\lhd A$, bcause $I\lhd A$, and so $I\cap\sigma(I)\lhd\sigma(I)$. But $\sigma(I)\cong I\cong G$ is simple, so that either $I\cap\sigma(I)=\mathbf 1$ or $I\cap\sigma(I)=\sigma(I)$. Since both $I$ and $\sigma(I)$ are normal, $[I,\sigma(I)]\le I\cap\sigma(I)$. In the first case, we have $[I,\sigma(I)]=\mathbf 1$; that is, $\sigma(I)\le\mathrm C_A(I)=\mathbf 1$, and this contradicts $\sigma(I)\cong I$. Hence, $I\cap\sigma(I)=\sigma(I)$, and so $\sigma(I)\le I$. This inclusion holds for every $\sigma\in\mathrm{Aut}(A)$; in particular, $\sigma^{-1}(I)\le I$, and so $\sigma(I)=I$ for every $\sigma\in\mathrm{Aut}(A)$. If $g\in G$, then $\gamma_g\in I$; there is thus $\alpha(g)\in G$ with $\sigma(\gamma_g)=\gamma_{\alpha(g)}$. It may be easily checked that the function $\alpha\!:G\rightarrow G$ is a bijection. We now show that $\alpha$ is an automorphism of $G$; that is, $\alpha\in A$. If $g,h\in G$, then $\sigma(\gamma_g\gamma_h)=\sigma(\gamma_{gh})=\gamma_{\alpha(gh)}$. On the other hand, $\sigma(\gamma_g\gamma_h)=\sigma(\gamma_g)\sigma(\gamma_h)=\gamma_{\alpha(g)}\gamma_{\alpha(h)}=\gamma_{\alpha(g)\alpha(h)}$; hence $\alpha(gh)=\alpha(g)\alpha(h)$.

We claim that $\sigma=\Gamma_\alpha$, conjugation by $\alpha$. To this end, define $\tau=\sigma\circ\Gamma_{\alpha}^{-1}$. Observe, for all $h\in G$, that

$$\begin{aligned} \tau(\gamma_h)&=\sigma\Gamma_\alpha^{-1}(\gamma_h)=\sigma(\alpha^{-1}\gamma_h\alpha)\\&=\sigma(\gamma_{\alpha^{-1}(h)})\\&=\gamma_{\alpha\alpha^{-1}(h)}=\gamma_h. \end{aligned}$$

Thus, $\tau$ fixes everything in $I$. If $\beta\in A$, then for every $g\in G$,

$$\begin{aligned} \beta\gamma_g\beta^{-1}&=\tau(\beta\gamma_g\beta^{-1})\qquad&&(\text{because }\beta\gamma_g\beta^{-1}\in I\text{ and }\tau\text{ fixes }I)\\ &=\tau(\beta)\gamma_g\tau(\beta)^{-1}&&(\text{because }\tau\text{ fixes }I). \end{aligned}$$

Hence $\tau(\beta)\beta^{-1}\in\mathrm C_A(I)=\mathbf 1$, and $\tau(\beta)=\beta$. Therefore, $\tau=\mathbf 1,\sigma=\Gamma_\alpha$, and $A$ is complete.
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It follows, for every nonabelian simple group $G$, that $\mathrm{Aut}(G)\cong\mathrm{Aut}(\mathrm{Aut}(G))$. We know, by Theorem 7.2.3, that every centerless group $G$ can be imbedded in $\mathrm{Aut}(G)$. Moreover, $\mathrm{Aut}(G)$ is also centerless, and so it can also be imbedded in its automorphism group $\mathrm{Aut}(\mathrm{Aut}(G))$. This process may thus be iterated to give the automorphism tower of $G$:

$$G\le\mathrm{Aut}(G)\le\mathrm{Aut}(\mathrm{Aut}(G))\le\cdots.$$

Wielandt proved, for every finite centerless group $G$, that this tower is constant from some point on. Since the last term of an automorphism tower is a complete group, it follows that every finite centerless group can be imbedded in a complete group.

Theorem 7.2.27 If $K\lhd G$ and $K$ is complete, then $K$ is a direct factor of $G$; that is, there is a normal subgroup $Q$ of $G$ with $G=K\times Q$.

Proof

Define $Q=\mathrm C_G(K)=\{g\in G:gk=kg\text{ for all }k\in K\}$. Now $K\cap Q\le\mathrm Z(K)=\mathbf 1$, and so $K\cap Q=\mathbf 1$. To see that $G=KQ$, take $g\in G$. Now $\gamma_g(K)=K$, because $K\lhd G$, and so $\gamma_g\big|_K\in\mathrm{Aut}(K)=\mathrm{Inn}(K)$. There is thus $k\in K$ with $gxg^{-1}=kxk^{-1}$ for all $x\in K$. Hence $k^{-1}g\in\bigcap_{x\in K}\mathrm C_G(x)=\mathrm C_G(K)=Q$, and so $g=k(k^{-1}g)\in KQ$. Finally, $Q\lhd G$, for $gQg^{-1}=k(k^{-1}g)Q(k^{-1}g)k^{-1}=kQk^{-1}$ (because $k^{-1}g\in Q$), and so $g=k(k^{-1}g)\in KQ$. Finally, $Q\lhd G$, for $gQg^{-1}=k(k^{-1}g)Q(g^{-1}g)^{-1}k^{-1}=kQk^{-1}$ (because $k^{-1}g\in Q$), and $kQk^{-1}=Q$ (because every element of $Q$ commutes with $k$). Therefore, $G=K\times Q$.
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Recall that if $a\in K$, then $L_a\!:K\rightarrow K$ denotes left translation by $a$; that is, $L_a(x)=ax$ for all $a\in K$. As in the Cayley theorem, $K$ is isomorphic to $K^l=\{L_a:a\in K\}$, which is a subgroup of $\mathrm S_K$. Similarly, if $R_a\!:K\rightarrow K$ denotes right translation by $a$, that is, $R_a\!:x\mapsto xa^{-1}$, then $K^r=\{R_a:a\in K\}$ is also a subgroup of $\mathrm S_K$ isomorphic to $K$.

Definition 7.2.28 The holomorph of a group $K$, denoted by $\mathrm{Hol}(K)$, is the subgroup of $\mathrm S_K$ generated by $K^l$ and $\mathrm{Aut}(K)$.

Notice, for all $a\in K$, that $R_a=L_{a^{-1}}\gamma_a$, so that $K^r\le\mathrm{Hol}(K)$; indeed, it is easy to see that $\mathrm{Hol}(K)=\langle K^r,\mathrm{Aut}(K)\rangle$.

Lemma 7.2.29 Let $K$ be a group.

1. $K^l\lhd\mathrm{Hol}(K),K^l\mathrm{Aut}(K)=\mathrm{Hol}(K)$, and $K^l\cap\mathrm{Aut}(K)=\mathbf 1$.
2. $\mathrm{Hol}(K)/K^l\cong\mathrm{Aut}(K)$.
3. $\mathrm C_{\mathrm{Hol}(K)}(K^l)=K^r$.
4. If $K$ is complete, then $\mathrm{Hol}(K)=K^l\times K^r$.
5. Every automorphism of $K$ is the restriction of an inner automorphism of $\mathrm{Hol}(K)$.
6. Let $f\in\mathrm S_K$, then $f\in\mathrm{Hol}(K)$ if and only if $f(xy^{-1}z)=f(x)f(y)^{-1}f(z)$ for all $x,y,z\in K$.

Proof

1. It is easy to see that $\varphi L_a\varphi^{-1}=L_{\varphi(a)}$, and that it lies in $K^l$ for every $a\in K$ and $\varphi\in\mathrm{Aut}(K)$; since $\mathrm{Hol}(K)=\langle K^l,\mathrm{Aut}(K)\rangle$, it follows that $K^l\lhd\mathrm{Hol}(K)$ and that $\mathrm{Hol}(K)=K^l\mathrm{Aut}(K)$. If $a\in K$, then $L_a(\mathbf 1)=a$; therefore, $\mathrm L_a\in\mathrm{Aut}(K)$ if and only if $a=\mathbf 1$; that is, $K^l\cap\mathrm{Aut}(K)=\mathbf 1$.
2. $\mathrm{Hol}(K)/K^l=K^l\mathrm{Aut}(K)/K^l\cong\mathrm{Aut}(K)/(K^l\cap\mathrm{Aut}(K))\cong\mathrm{Aut}(K)$.
3. If $a,b,x\in K$, then $L_aR_b(x)=a(xb^{-1})$ and $R_bL_a(x)=(ax)b^{-1}$, so that associativity gives $K^r\le\mathrm C_{\mathrm{Hol}(K)}(K^l)$. For the reverse inclusion, assume that $\eta\in\mathrm{Hol}(K)$ satisfies $\eta L_a=L_a\eta$ for all $a\in K$. Now $\eta=L_b\varphi$ for some $b\in K$ and $\varphi\in\mathrm{Aut}(K)$. If $x\in K$, then $\eta L_a(x)=L_b\varphi L_a(x)=b\varphi(a)\varphi(x)$ and $L_a\eta(x)=L_aL_b\varphi(x)=ab\varphi(x)$. Hence, $b\varphi(a)=ab$ for all $a\in K$; that is, $\varphi=\gamma_{b^{-1}}$. It follows that $\eta(x)=L_b\varphi(x)=b(b^{-1}xb)=xb$, and so $\eta=R_{b^{-1}}\in K^r$, as desired.
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Theorem 7.2.30 If a group $K$ is a direct factor whenever it is (isomorphic to) a normal subgroup of a group, then $K$ is complete.

Proof

We identify $K$ with the subgroup $K^l\le\mathrm{Hol}(K)$. Since $K^l$ is normal, the hypothesis gives a subgroup $B$ with $\mathrm{Hol}(K)=K^l\times B$. Now $B\le\mathrm C_{\mathrm{Hol}(K)}(K^l)=K^r$, because every element of $B$ commutes with each element of $K^l$. It follows that if $\varphi\in\mathrm{Aut}(K)\le\mathrm{Hol}(K)$, then $\varphi=L_aR_b$ for some $a,b\in K$. Hence, $\varphi(x)=axb^{-1}$ for all $x\in K$. But now $axyb^{-1}=\varphi(x)\varphi(y)=axb^{-1}ayb^{-1}$, so that $\mathbf 1=b^{-1}a$; therefore, $\varphi=\gamma_a\in\mathrm{Inn}(K)$.

Since $\mathrm{Hol}(K)=K^l\times B$ and $B\le K^r\le\mathrm{Hol}(K)$, we have $K^r=B\times(K^r\cap K^l)$. If $\varphi\in K^l\cap K^r$, then $\varphi=L_a=R_b$, for $a,b\in K$. For all $c\in K$, $L_a(c)=R_b(c)$ gives $ac=cb^{-1}$; if $c=\mathbf 1$, then $a=b^{-1}$, from which it follows that $a\in\mathrm Z(K)$. Therefore, $K^r\cap K^l=\mathrm Z(K)$ and $K\cong B\times\mathrm Z(K)$. If $\mathbf 1\neq\varphi\in\mathrm{Aut}(\mathrm Z(K))$, then it is easy to see that $\tilde\varphi\!:B\times\mathrm Z(K)\rightarrow B\times\mathrm Z(K)$, defined by $(b,z)\mapsto (b,\varphi z)$, is an automorphism of $K$; $\tilde\varphi$ must be outer, for conjugation by $(\beta,\zeta)\in B\times\mathrm Z(K)\cong K$ sends $(b,z)$ into $(\beta,\zeta)(b,z)(\beta^{-1},\zeta^{-1})=(\beta b\beta^{-1},z)$. But $K$ has no outer automorphisms, so that $\mathrm{Aut}(\mathrm Z(K))=\mathbf 1$ and $|\mathrm Z(K)|\le 2$. If $\mathrm Z(K)\cong\mathbb Z_2$, then it is isomorphic to a normal subgroup $N$ of $\mathbb Z_4$ which is not a direct factor. But $K$ is isomorphic to the normal subgroup $B\times N$ of $B\times\mathbb Z_4$ which is not a direct factor, contradicting the hypothesis. Therefore, $\mathrm Z(K)=\mathbf 1$ and $K$ is complete.
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The holomorph allows one to extend commutator notation. Recall that the commutator $[a,x]=axa^{-1}x^{-1}=x^ax^{-1}$. Now let $G$ be a group and let $A=\mathrm{Aut}(G)$. We may regard $G$ and $A$ as subgroups of $\mathrm{Hol}(G)$ (by identifying $G$ with $G^l$). For $x\in G$ and $\alpha\in A$, define

$$[\alpha,x]=\alpha(x)x^{-1},$$

and define

$$[A,G]=\langle[\alpha,x]:\alpha\in A,x\in G\rangle.$$

Lemma 7.2.31 Let $G$ and $A$ be subgroups of a group $H$, and let $G=G_0\ge G_1\ge \cdots$ be a series of normal subgroups of $G$ such that $[A,G_i]\le G_{i+1}$ for all $i$. Define $A_1=A$ and

$$A_j=\{\alpha\in A:[\alpha,G_i]\le G_{i+j}\text{ for all }i\}.$$

Then $[A_j,A_l]\le A_{j+l}$ for all $j$ and $l$, and $[\mathrm C_{j}(A), G_i]\le G_{i+j+1}$ for all $i$ and $j$.

Proof

The definition of $A_j$ gives $[A_j,G_i]\le G_{i+j}$ for all $i$. It follows that $[A_j,A_l,G_i]=[A_j,[A_l,G_i]]\le[A_j,G_{l+i}]\le G_{j+l+i}$. Similarly, $[A_l,A_j,G_i]\le G_{j+l+i}$. Now $G_{j+l+i}\lhd\langle G,A\rangle$, because both $G$ and $A$ normalize each $G_i$. Since $[A_j,A_l,G_i][A_l,A_j,G_i]\le G_{j+l+i}$, the three subgroups lemma gives $[G_i,[A_j,A_l]]=[[A_j,A_l],G_i]\le G_{j+l+i}$. Therefore, $[A_j,A_l]\le A_{j+l}$, by definition of $A_{j+l}$. It follows, for all $j$, that $A_j\lhd A$, because $[A_j,A]=[A_j,A_l]\le[A_{j+l}]\le A_j$, and so $A=A_1\ge A_2\ge\cdots$ is a central series for $A$. By Proposition 5.5.23(ii), $\mathrm C_{j}(A)\le A_{j+1}$ for all $j$, so that, for all $i$, $[\mathrm C_{j}(A),G_i]\le[A_{j+1},G_i]\le G_{j+i+1}$, as desired.
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Definition 7.2.32 Let $G=G_0\ge G_1\ge\cdots\ge G_r=\mathbf 1$ be a series of normal subgroups of a group $G$. An automorphism $\alpha\in\mathrm{Aut}(G)$ stabilizes this series if $\alpha(G_ix)=G_ix$ for all $i$ and all $x\in G_{i-1}$. The stabilizer $A$ of this series is the subgroup

$$A=\{\alpha\in\mathrm{Aut}(G):\alpha\text{ stabilizes the series}\}\le\mathrm{Aut}(G).$$

Thus, $\alpha$ stabilizes a normal series $G=G_0\ge G_1\ge\cdots\ge G_r=\mathbf 1$ if and only if $\alpha(G_i)\le G_i$ and the induced map $G_i/G_{i+1}\rightarrow G_i/G_{i+1}$, defined by $G_{i+1}x\mapsto G_{i+1}\alpha(x)$, is the identity map for each $i$.

Theorem 7.2.33 The stabilizer $A$ of a series of normal subgroups $G=G_0\ge G_1\ge\cdots\ge G_r=\mathbf 1$ is a nilpotent group of class $\le r-1$.

Proof

Regard both $G$ and $A$ as subgroups of $\mathrm{Hol}(G)$. For all $i$, if $x\in G_i$ and $\alpha\in A$, then $\alpha(x)=g_{i+1}x$ for some $g_{i+1}\in G_{i+1}$, and so $\alpha(x)x^{-1}\in G_{i+1}$. In commutator notation, $[A, G_i]\le G_{i+1}$. By Lemma 7.2.31, $[\mathrm C_{j-1}(A),G_i]\le G_{i+j}$ for all $i$ and $j$. In particular, for $i=0$ and $j=r$, we have $[\mathrm C_{r-1}(A),G]\le G_r=\mathbf 1$; that is, for all $x\in G$ and $\alpha\in\mathrm C_{r-1}(A)$, we have $\alpha(x)x^{-1}=\mathbf 1$. Therefore, $\mathrm C_{r-1}(A)=\mathbf 1$ and $A$ is nilpotent of class $\le r-1$.
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Example 7.2.34 Let $\{v_1,\ldots,v_n\}$ be a basis of a vector space $V$ over a field $k$, and define $V_{i-1}=\langle v_i,v_{i+1},\ldots,v_n\rangle$. Hence,

$$V=V_0>V_1>\cdots>V_n=\mathbf 0$$

is a series of normal subgroups of the (additive abelian) group $V$. If $A\le\mathrm{GL}(V)$ is the group of automorphisms stabilizing this series, then $A$ is a nilpotent group of class $\le n-1$. If each $\alpha\in A\cap\mathrm{GL}(V)$ is regarded as a matrix (relative to the given basis), then it is easy to see that $A\cap\mathrm{GL}(V)=\mathrm{UT}(n,k)$, the group of a unitriangular matrices. Therefore, $\mathrm{UT}(n,k)$ is allso nilpotent of class $\le n-1$.

Proposition 7.2.35 (P.Hall) If $G=G_0\ge G_1\ge\cdots\ge G_r=\mathbf 1$ is any (not necessarily normal) series of a group $G$ (i.e., $G_i$ need not be a normal subgroup of $G_{i-1}$), then the stabilizer of this series is always nilpotent of class $\le\frac 12r(r-1)$.

Semidirect Products

Definition 7.3.1 Let $K$ be a (not necessarily normal) subgroup of a group $G$. Then a subgroup $Q\le G$ is a complement of $K$ in $G$ if $K\cap Q=\mathbf 1$ and $KQ=G$.

A normal subgroup $K$ of a group $G$ need not have a complement and, even if it does, a complement need not be unique. On the other hand, if they exist, complements are unique to isomorphism, for

$$G/K=KQ/K\cong Q/(K\cap Q)=Q/\mathbf 1\cong Q.$$

A group $G$ is the direct product of two normal subgroups $K$ and $Q$ if $K\cap Q=\mathbf 1$ and $KQ=G$.

Definition 7.3.2 A group $G$ is a semidirect product of $K$ by $Q$, denoted by $G=K\rtimes Q$, if $K\lhd G$ and $K$ has a complement $Q_1\cong Q$. One also says that $G$ splits over $K$.

Lemma 7.3.3 If $K$ is a normal subgroup of a group $G$, then the following statements are equivalent:

1. $G$ is a semidirect product of $K$ by $G/K$ (i.e., $K$ has a complement in $G$);
2. there is a subgroup $Q\le G$ so that every element $g\in G$ has a unique expression $g=ax$, where $a\in K$ and $x\in Q$;
3. there exists a homomorphism $s\!:G/K\rightarrow G$ with $vs=\mathbf 1_{G/K}$, where $v\!:G\rightarrow G/K$ is the natural map; and
4. there exists a homomorphism $\pi\!:G\rightarrow G$ with $\mathrm{ker}\pi=K$ and $\pi(x)=x$ for all $x\in\mathrm{im}\pi$ (such a map $\pi$ is called a retraction of $G$ and $\mathrm{im}\pi$ is called a retract of $G$).

Proof

(i)$\Rightarrow$(ii) Let $Q$ be a complement of $K$ in $G$. Let $g\in G$. Since $G=KQ$, there exist $a\in K$ and $x\in Q$ with $g=ax$. If $g=by$ is a second such factorization, then $xy^{-1}=a^{-1}b\in K\cap Q=\mathbf 1$. Hence $b=a$ and $y=x$.

(ii)$\Rightarrow$(iii) Each $g\in G$ has a unique expression $g=ax$, where $a\in K$ and $x\in Q$. If $Kg\in G/K$, then $K=g=Kax=Kx$; define $s\!:G/K\rightarrow G$ by $s(Kg)=x$. The routine verification that $s$ is a well defined homomorphism with $vs=\mathbf 1_{G/K}$ is easy to check.

(iii)$\Rightarrow$(iv) Define $\pi\!:G\rightarrow G$ by $\pi=sv$. If $x=\pi(g)$, then $\pi(x)=\pi(\pi(g))=svsv(g)=sv(g)=\pi(g)=x$ (because $vs=\mathbf 1_{G/K}$). If $a\in K$, then $\pi(a)=sv(a)=\mathbf 1$, for $K=\mathrm{ker}v$. For the reverse inclusion, assume that $\mathbf 1=\pi(g)=sv(g)=s(Kg)$. Now $s$ is an injection, so that $Kg=K$ and so $g\in K$.

(iv)$\Rightarrow$(i) Define $Q=\mathrm{im}\pi$. If $g\in G$, then $\pi(g)=g$; if $g\in K$, then $\pi(g)=\mathbf 1$; a fortiori, if $g\in K\cap Q$, then $g=\mathbf 1$. If $g\in G$, then $g\pi(g^{-1})\in K=\mathrm{ker}\pi$, for $\pi(g\pi(g^{-1}))=\mathbf 1$. Since $\pi(g)\in Q$, we have $g=[g\pi(g^{-1})]\pi(g)\in KQ$. Therefore, $Q$ is a complement of $K$ in $G$ and $G$ is a semidirect product of $K$ by $Q$.
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Example 7.3.4

1. $\mathrm S_n$ is a semidirect product of $\mathrm A_n$ by $\mathbb Z_2$.
2. $\mathrm D_{2n}$ is a semidirect product of $\mathbb Z_n$ by $\mathbb Z_2$.
3. For any group $K$, $\mathrm{Hol}(K)$ is a semidirect product of $K^l$ by $\mathrm{Aut}(K)$.
4. Let $G$ be a solvable group of order $mn$, where $(m,n)=1$. If $G$ contains a normal subgroup of order $m$, then $G$ is a semidirect product of $K$ by a subgroup $Q$ of order $n$.
5. $\mathrm{Aut}(\mathrm S_6)$ is a semidirect product of $\mathrm S_6$ by $\mathbb Z_2$.
6. The group $\mathrm T$ of order $12$ (Definition A.1.18) is a semidirect product of $\mathbb Z_3$ by $\mathbb Z_4$.
7. Both $\mathrm S_3$ and $\mathbb Z_6$ are semidirect products of $\mathbb Z_3$ by $\mathbb Z_2$.

In contrast to direct product, a semidirect product of $K$ by $Q$ is not determined to isomorphism by the two subgroups. However, we see that a semidirect product should depend on "how" $K$ is normal in $G$.

Lemma 7.3.5 If $G$ is a semidirect product of $K$ by $Q$, then there is a homomorphism $\theta\!:Q\rightarrow\mathrm{Aut}(K)$, defined by $\theta_x=\gamma_x\big|_K$; that is, for all $x\in Q$ and $a\in K$,

$$\theta_x(a)=xax^{-1}.$$

Moreover, for all $x,y,\mathbf 1\in Q$ and $a\in K$,

$$\theta_1(a)=a\qquad\text{and}\qquad\theta_x(\theta_y(a))=\theta_{xy}(a).$$

Definition 7.3.6 Let $Q$ and $K$ be groups, and let $\theta\!:Q\rightarrow\mathrm{Aut}(K)$ be a homomorphism. A semidirect product $G$ of $K$ by $Q$ realizes $\theta$ if, for all $x\in Q$ and $a\in K$,

$$\theta_x(a)=xax^{-1}.$$

In this language, Lemma 7.3.5 says that every semidirect product $G$ of $K$ by $Q$ determines some $\theta$ which it realizes. For example, if $\theta$ is the trivial map, that is, $\theta_x=\mathbf 1_K$ for every $x\in G$, then $a=\theta_x(a)=xax^{-1}$ for every $a\in K$, and so $K\le\mathrm C_G(Q)$.

Definition 7.3.7 Given groups $Q$ and $K$ and a homomorphism $\theta\!:Q\rightarrow\mathrm{Aut}(K)$, define $G=K\rtimes_\theta Q$ to be the set of all ordered pairs $(a,x)\in K\times Q$ equipped with the operation

$$(a,x)(b,y)=(a\theta_x(b),xy).$$

Definition 7.3.8 Given groups $Q$ and $K$ and a homomorphism $\theta\!:Q\rightarrow\mathrm{Aut}(K)$, then $G=K\rtimes_\theta Q$ is a semidirect product of $K$ by $Q$ that realizes $\theta$.

Proof

It is easy to prove that $G$ is indeed a group. Define a function $\pi\!:G\rightarrow Q$ by $(a,x)\mapsto x$. Since the only "twist" occurs in the first coordinate, it is routine to check that $\pi$ is a normal subgroup of $G$. We identify $K$ with $\mathrm{ker}\pi$ via the isomorphism $a\mapsto(a,\mathbf 1)$. It is also easy to check that $\{(\mathbf 1,x):x\in Q\}$ is a subgroup of $G$ isomorphic to $Q$ (via $x\mapsto(\mathbf 1,x)$), and we identify $Q$ with this subgroup. Also it is easy to see that $KQ=G$ and $K\cap Q=\mathbf 1$, so that $G$ is a semidirect product of $K$ by $Q$.

Finally, $G$ does realize $\theta$:

$$(\mathbf 1,x)(a,\mathbf 1)(\mathbf 1,x)^{-1}=(\theta_x(a),x)(\mathbf 1,x^{-1})=(\mathrm\theta_x(a),\mathbf 1).$$

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Since $K\rtimes_\theta Q$ realizes $\theta$, that is, $\theta_x(b)=xbx^{-1}$, there can be no confusion if we write $b^x=xbx^{-1}$ instead of $\theta_x(b)$. The operation in $K\rtimes_\theta Q$ will henceforth be written

$$(a,x)(b,y)=(ab^x,xy).$$

Theorem 7.3.9 If $G$ is a semidirect product of $K$ by $Q$, then there exists $\theta\!:Q\rightarrow\mathrm{Aut}(K)$ with $G\cong K\rtimes_\theta Q$.

Proof

Define $\theta_x(a)=xax^{-1}$ (as in Lemma 7.3.5). By Lemma 7.3.3(ii), each $g\in G$ has a unique expression $g=ax$ with $a\in K$ and $x\in Q$. Since multiplication in $G$ satisfies

$$(ax)(by)=a(xbx^{-1})xy=ab^xxy,$$

it is easy to see that the map $K\rtimes_\theta Q\rightarrow G$, defined by $(a,x)\mapsto ax$, is an isomorphism.
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Proposition 7.3.10 The group $\mathbf Q_n$ of generalized quaternions is not a semidirect product.

Proposition 7.3.11 Let $K$ and $Q$ be groups.

1. If $K$ and $Q$ are solvable, then $K\rtimes_\theta Q$ is also solvable.
2. $K\rtimes_\theta Q$ is the direct product $K\times Q$ if and only if $\theta\!:Q\rightarrow\mathrm{Aut}(K)$ is trivial (that is, $\theta_x=\mathbf 1$ for all $x\in Q$).

Proposition 7.3.12 If $|G|=mn$, where $(m,n)=1$, and if $K\le G$ has order $m$, then a subgroup $Q\le G$ is a complement of $K$ if and only if $|Q|=n$.

Wreath Products

Let $D$ and $Q$ be groups, let $\Omega$ be a finite $Q$-set, and let $\{D_\omega:\omega\in\Omega\}$ be a family of isomorphic copies of $D$ indexed by $\Omega$.

Definition 7.4.1 Let $D$ and $Q$ be groups, let $\Omega$ be a $Q$-set, and let $K=\prod_{w\in\Omega}D_{\omega}$, where $D_\omega\cong D$ for all $\omega\in\Omega$. Then the wreath product of $D$ by $Q$, denoted by $D\wr_\Omega Q$, is the semidirect product of $K$ by $Q$, where $Q$ acts on $K$ by $q\cdot(d_\omega)=(d_{q\omega})$ for $q\in Q$ and $(d_\omega)\in\prod_{w\in\Omega}D_\omega$. The normal subgroup $K$ of $D\wr_\Omega Q$ is called the base of the wreath product.

If $D$ is finite, then $|K|=|D|^{|\Omega|}$; if $Q$ is also finite, then $|D\wr_\Omega Q|=|K\rtimes Q|=|D|^{|\Omega|}|Q|$.

If $\Lambda$ is a $D$-set, then $\Lambda\times\Omega$ can be made into a $(D\wr_\Omega Q)$-set. Given $d\in D$ and $\omega\in\Omega$, define a permutation $d_\omega^*$ of $\Lambda\times\Omega$ as follows: for each $(\lambda,\omega')\in\Lambda\times\Omega$, set

$$d_\omega^*(\lambda,\omega')=\begin{cases} (d\lambda,\omega')\qquad&\text{if }\omega'=\omega,\\(\lambda,\omega')&\text{if }\omega'\neq\omega. \end{cases}$$

It is easy to see that $d_{\omega}^*d_\omega'^*=(dd')_\omega^*$, and so $D_\omega^*$, defined by

$$D_\omega^*=\{d_\omega^*:d\in D\},$$

is a subgroup of $\mathrm S_{\Lambda\times\Omega}$; indeed, for each $\omega$, the map $D\rightarrow D_\omega^*$, given by $d\mapsto d_\omega^*$, is an isomorphism.

For each $q\in Q$, define a permutation $q^*$ of $\Lambda\times\Omega$ by

$$q^*(\lambda,\omega')=(\lambda,q\omega'),$$

and define $Q^*=\{q^*:q\in Q\}$. It is easy to see that $Q^*$ is a subgroup of $\mathrm S_{\Lambda\times\Omega}$ and that the map $Q\rightarrow Q^*$, given by $q\mapsto q^*$, is an isomorphism.

Theorem 7.4.2 Given groups $D$ and $Q$, a finite $Q$-set $\Omega$, and a $D$-set $\Lambda$, then the wreath product $D\wr_\Omega Q$ is isomorphic to the subgroup

$$W=\langle Q^*,D_\omega^*:\omega\in\Omega\rangle\le\mathrm S_{\Lambda\times\Omega},$$

and hence $\Lambda\times\Omega$ is a $(D\wr_\Omega Q)$-set.

Proof

We show first that $K^*=\langle\bigcup_{\omega\in\Omega}D_\omega^*\rangle$ is the direct product $\prod_{\omega\in\Omega}D_\omega^*$. It is easy to see that $D_\omega^*$ centralizes $D_\omega^*$ for all $\omega'\neq\omega$, and so $D_{\omega}^*\lhd K^*$ for every $omega$. Each $d_\omega^*\in D_\omega^*$ fixes all $(\lambda,\omega')\in\Lambda\times\Omega$ with $\omega'\neq\omega$, while each element of $\langle\bigcup_{\omega'\neq\omega} D_{\omega'}^*\rangle$ fixes all $(\lambda,\omega)$ for all $\lambda\in\Lambda$. It follows that if $d_\omega^*\in D_\omega^*\cap\langle\bigcup_{\omega'\neq\omega} D_{\omega'}^*\rangle$, then $d_\omega^*=\mathbf 1$.

If $q\in Q$ and $\omega\in\Omega$, then a routine computation gives

$$q^*d_\omega^*q^{*-1}=d_{q\omega}^*$$

for each $\omega\in\Omega$. Hence, $q^*K^*q^{*-1}\le K^*$ for each $q\in Q$, so that $K^*\lhd W$ (because $W=\langle K^*,Q^*\rangle$); it follows that $W=K^*Q^*$. To see that $W$ is a semidirect product of $K^*$ by $Q^*$, it suffices to show that $K^*\cap Q^*=\mathbf 1$. Now $d_\omega^*(\lambda,\omega')=(d\lambda,\omega')\text{ or }(\lambda,\omega')$; in either case, $d_\omega^*$ fixes the second coordinate. If $q^*\in Q^*$, then $q^*(\lambda,\omega')=(\lambda,q\omega')$ and $q^*$ fixes the first coordinate. Therefore, any $g\in K^*\cap Q^*$ fixes every $(\lambda,\omega')$ and hence is the identity.

It is now a simple matter to check that the map $D\wr_\Omega Q\rightarrow W$, given by $(d_\omega)q\mapsto(d_\omega^*)q^*$, is an isomorphism.
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Call the subgroup $W$ of $\mathrm S_{\Lambda\times\Omega}$ the permutation version of $D\wr_\Omega Q$; when we wish to view $D\wr_\Omega Q$ acting on $\Lambda\times \Omega$, then we will think of it as $W$.

Theorem 7.4.3 Let $D$ and $Q$ be groups, let $\Omega$ be a finite $Q$-set, let $\Lambda$ be a $D$-set, and let $W\le \mathrm S_{\Lambda\times\Omega}$ be the permutation version of $D\wr_\Omega Q$.

1. If $\Omega$ is a transitive $Q$-set and $\Lambda$ is a transitive $D$-set, then $\Lambda\times\Omega$ is a transitive $(D\wr_\Omega Q)$-set.
2. If $\omega\in \Omega$, then its stabilizer $Q_\omega$ acts on $\Omega-\{\omega\}$. If $(\lambda,\omega)\in\Lambda\times\Omega$ and $D(\lambda)\le D$ is the stabilizer of $\lambda$, then the stabilizer $W_{(\lambda,\omega)}$ of $(\lambda,\omega)$ is isomorphic to $D(\lambda)\times(D\wr_\Omega Q_\omega)$, and $[W:W_{(\lambda,\omega)}]=[D:D(\lambda)][Q:Q_\omega]$.

Proof

1. Let $(\lambda,\omega),(\lambda',\omega')\in\Lambda\times\Omega$. Since $D$ acts transitively, there is $d\in D$ with $d\lambda=\lambda'$; since $Q$ acts transitively, there is $q\in Q$ with $q\omega=\omega'$. It is easy to check that $q^*d_\omega^*(\lambda,\omega)=(\lambda',\omega')$.
2. Each element of $W$ has the form $(d_\omega^*)q^*$, and $(d_\omega^*)q^*(\lambda,\omega)=(\prod_{\omega'\in\Omega}d_\omega^*)(\lambda,q\omega)=d_{q\omega}^*(\lambda,q\omega)=(d_{q\omega}\lambda,q\omega)$. It follows that $(d_\omega^*)q^*$ fixes $(\lambda,\omega)$ if and only if $q$ fixes $\omega$ and $d_\omega$ fixes $\lambda$. Let $D_\omega^*(\lambda)=\{d_\omega^*:d\in D(\lambda)\}$. Now $D_\omega^*(\lambda)$ is disjoint from $\langle\prod_{\omega'\neq\omega} D_\omega^*,Q_\omega^*\rangle$ and centralizes it: if $q^*\in Q_\omega^*$, then $q^*d_\omega^*q^{*-1}=d_{q\omega}^*=d_\omega^*$; hence

$$\begin{aligned} W_{(\lambda,\omega)}&=\left\langle D_\omega^*(\lambda),\prod_{\omega'\neq\omega}D_\omega^*,Q_\omega^*\right\rangle\\&=D_\omega^*(\lambda)\times\left\langle\prod_{\omega'\neq\omega}D_\omega^*,Q_\omega^*\right\rangle\\ &\cong D(\lambda)\times(D\wr_\Omega Q_\omega). \end{aligned}$$

It follows that $|W_{(\lambda,\omega)}|=|D(\lambda)||D|^{|\Omega|-1}|Q_\omega|$ and

$$[W:W_{(\lambda,\omega)}]=|D|^{|\Omega|}|Q|/|D(\lambda)||D|^{|\Omega|-1}|Q_\omega|=[D:D(\lambda)][Q:Q_\omega].$$

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Theorem 7.4.4 Wreath product is associative: if both $\Omega$ and $\Lambda$ are finite, if $T$ is a group, and if $\Delta$ is a $T$-set, then $T\wr_{\Lambda\times\Omega}(D\wr_\Omega Q)\cong(T\wr_\Lambda D)\wr_\Omega Q$.

Proof

The permutation versions of both $T\wr_{\Lambda\times\Omega}(D\wr_\Omega Q)$ and $(T\wr_\Lambda D)\wr_\Omega Q$ are subgroups of $\mathrm S_{\Delta\times\Lambda\times\Omega}$; we claim that they coincide. The group $T\wr_{\Lambda\times\Omega}(D\wr_\Omega Q)$ is generated by all $t_{(\lambda,\omega)}^*$ and all $f^*$ (for $f\in D\wr_{\Omega}Q$). Note that $t_{(\lambda,\omega)}^*\!:(\delta',\lambda',\omega')\mapsto(t\delta',\lambda',\omega')$ if $(\lambda',\omega')=(\lambda,\omega)$, and fixes it otherwise; also $f^*\!:(\delta',\lambda',\omega')\mapsto(\delta',f(\lambda',\omega'))$. Specializing $f^*$ to $d_\omega^*$ and to $q^*$, we see that $T\wr_{\Lambda\times\Omega}(D\wr_{\Omega}Q)$ is generated by all $t_{\lambda,\omega}^*,d_\omega^*$, and $q^{**}$, where $d_\omega^*\!:(\delta',\lambda',\omega')\mapsto(\delta',d\lambda',\omega')$ if $\omega'=\omega$, and fixes it otherwise, and $q^{**}\!:(\delta',\lambda',\omega')\mapsto(\delta',\lambda',q\omega')$.

A similar analysis of $(T\wr_\Lambda D)\wr_\Omega Q$ shows that it is generated by all $q^{**},d_\omega^*$, and $(t_\lambda)_\omega^*$, where $(t_\lambda)_\omega^*\!:(\delta',\lambda',\omega')\mapsto(t\delta',\lambda',\omega')$ if $\omega'=\omega$ and $\lambda'=\lambda$, and fixes it otherwise. Since $(t_\lambda)_\omega^*=t_{(\lambda,\omega)}^*$, the two wreath products coincide.
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Proposition 7.4.5 If $(a,x)\in D\wr_\Omega Q$ (so that $a\in K=\prod D_\omega$), then

$$(a,x)^n=\left(aa^xa^{x^2}\ldots a^{x^{n-1}},x^n\right).$$

Proposition 7.4.6 If both $D$ and $Q$ are solvable, and $\Omega$ is a $Q$-set, then $D\wr_\Omega Q$ is solvable.

Definition 7.4.7 If $\Omega=Q$ regarded as a $Q$-set acting on itself by left multiplication, then we write $W=D\wr_r Q$, and we call $W$ the regular wreath product. Thus, the base is the direct product of $|Q|$ copies of $D$, indexed by the elements of $Q$, and $q\in Q$ sends a $|Q|$-tuple $(d_x)\in\prod_{x\in Q}D_x$ into $(d_{qx})$, and $|D\wr_r Q|=|D|^{|Q|}|Q|$. It is easy to see that the formation of regular wreath product is not associative when all groups are finite, for $|T\wr_r(D\wr_r Q)|\neq|(T\wr_r D)\wr_r Q|$.

If $\Omega$ is an infinite set and $\{D_\omega:\omega\in\Omega\}$ is a family of groups, then there are two direct product constructions.

1. (sometimes called the complete direct product) consists of all "vectors" $(d_\omega)$ in the cartesian product $\prod_{\omega\in\Omega}D_\omega$ with "coordinatewise" multiplication: $(d_\omega)(d'_\omega)=(d_\omega d'_\omega)$.
2. (called the restricted direct product) is the subgroup of the complete direct product consisting of all those $(d_\omega)$ with only finitely many coordinates $d_\omega\neq\mathbf 1$.

The wreath product using the complete direct product is called the complete wreath product, while using the restricted direct product is the restricted wreath product.

Theorem 7.4.8 (Kaloujnine) If $p$ is a prime, then a Sylow $p$-subgroup of $\mathrm S_{p^n}$ is an iterated regular wreath product $W_n=\mathbb Z_p\wr_r\mathbb Z_p\wr_r\cdots\wr_r\mathbb Z_p$ of $n$ copies of $\mathbb Z_p$, where $W_{n+1}=W_n\wr_r\mathbb Z_p$.

Proof

The proof is by induction on $n$, the case $n=1$ holding naturally. Assume that $n>1$, let $\Lambda$ be a set with $p^n$ elements and let $D$ be a Sylow-$p$ subgroup of $\mathrm S_\Lambda$; thus, $\Lambda$ is a $D$-set. Let $\Omega=\{0,1,\ldots,p-1\}$, and let $Q=\langle q\rangle$ be a cyclic group of order $p$ acting on $\Omega$ by $qi=i+1\pmod p$. The permutation version of the wreath product $P=D\wr_r\mathbb Z_p$ is a subgroup of $\mathrm S_{\Lambda\times\Omega}$; of course, $|\Lambda\times\Omega|=p^{n+1}$. By induction, $D$ is a wreath product of $n$ copies of $\mathbb Z_p$, and so $P$ is a wreath product of $n+1$ copes of $\mathbb Z_p$. To see that $P$ is a Sylow $p$-subgroup, it suffices to see that its order is $p^{\mu(n+1)}$, where $\mu(n+1)=p^n+p^{n-1}+\cdots+p+1$. Now $|D|=p^{\mu(n)}$, so that $|P|=|D\wr_r\mathbb Z_p|=(p^{\mu(n)})^pp=p^{p\mu(n)+1}=p^{\mu(n+1)}$.
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Example 7.4.9 The theorem above can be used to compute the Sylow $p$-subgroup of $\mathrm S_m$ for any $m$. First write $m$ in base $p$:

$$m=a_0+a_1p+a_2p^2+\cdots+a_tp^t, \qquad \text{where } 0\le a_i\le p-1.$$

Partition $X=\{1,2,\ldots,m\}$ into $a_0$ singletons, $a_1$ $p$-subsets, $a_2$ $p^2$-subsets, $\ldots$, and $a_t$ $p^t$-subsets. On each of these $p^i$-subsets $Y$, construct a Sylow $p$-subgroup of $\mathrm S_Y$. Since disjoint permutations commute, the direct product of all these Sylow subgroups is a subgroup of $\mathrm S_X$ of order $p^N$, where $N=a_1+a_2\mu(2)+\cdots+a_t\mu(t)$ (recall that $\mu(i)=p^{i-1}+p^{i-2}+\cdots+p+1$). But it is easy to see that $p^N$ is the highest power of $p$ dividing $m!$. Thus, the direct product has the right order, and so it must be a Sylow $p$-subgroup of $\mathrm S_X\cong\mathrm S_m$.

Proposition 7.4.10 Let $D$ be a (multiplicative) group. A monomial matrix $\mu$ over $D$ is a permutation matrix $P$ whose nonzero entries have been replaced by elements of $D$; we say that $P$ is the support of $\mu$. If $Q$ is a group of $n\times n$ permutation matrices, then denote

$$M(D,Q)=\{\text{all monomial matrices }\mu\text{ over }D\text{ with support in }Q\}.$$

Then $M(D,Q)$ is a group under matrix multiplication, and $M(D,Q)\cong D\wr_\Omega Q$, where $\Omega=\{1,2,\ldots,n\}$ is a faithful $Q$-set here.

Factor Sets

In discussing general extensions $G$ of $K$ by $Q$, it is convenient to use the additive notation for $G$ and its subgroup $K$ (this is one of the rare instances in which one uses additive notation for a nonabelian group). For example, if $k\in K$ and $g\in G$, we shall write the conjugate of $k$ by $g$ as $g+k-g$.

Definition 7.5.1 If $K\le G$, then a (right) transversal of $K$ in $G$ is a subset $T$ of $G$ consisting of one element from each right coset of $K$ in $G$.

If $T$ is a right transversal, then $G$ is the disjoint union $G=\bigcup_{t\in T}(K+t)$. Thus, every element $g\in G$ has a unique factorization $g=k+t$ for $k\in K$ and $t\in T$. There is a similar definition of left transversal; of course, these two notations coincide when $K$ is normal.

If $G$ is a semidirect product and $Q$ is a complement of $K$, then $Q$ is a transversal of $K$ in $G$.

Definition 7.5.2 If $\pi\!:G\rightarrow Q$ is surjective, then a lifting of $x\in Q$ is an element $l(x)\in G$ with $\pi(l(x))=x$.

If one chooses a lifting $l(x)$ for each $x\in Q$, then the set of all such is a transversal of $\mathrm{ker}\pi$. In this case, the function $l\!:Q\rightarrow G$ is also called a right transversal.

Proposition 7.5.3 An extension $G$ of $K$ by $Q$ is a semidirect product if and only if there is a transversal $l\!:Q\rightarrow G$ that is a homomorphism.

Theorem 7.5.4 Let $G$ be an extension of $K$ by $Q$, and let $l\!:Q\rightarrow G$ be a transversal. If $K$ is abelian, then there is a homomorphism $\theta\!:Q\rightarrow\mathrm {Aut}(K)$ with

$$\theta_x(a)=l(x)+a-l(x)$$

for every $a\in K$. Moreover, if $l_1\!:Q\rightarrow G$ is another transversal, then $l(x)+a-l(x)=l_1(x)+a-l_1(x)$ for all $a\in K$ and $x\in Q$.

Proof

Since $K\lhd G$, the restriction $\gamma_g\big|_K$ is an automorphism of $K$ for all $g\in G$, where $\gamma_g$ is conjugation by $g$. The function $\mu\!:G\rightarrow\mathrm{Aut}(K)$, given by $g\mapsto\gamma_g\big|_K$, is easily seen to be a homomorphism; moreover, $K\le\mathrm{ker}\mu$, for $K$ being abelian implies that each conjugation by $a\in K$ is identity. Therefore, $\mu$ induces a homomorphism $\mu_\#\!:G/K\rightarrow\mathrm{Aut}(K)$, namely, $K+g\mapsto\mu(g)$.

The first isomorphism theorem says more than $Q\cong G/K$; it gives an explicit isomorphism $\lambda\!:Q\rightarrow G/K$: if $l\!:Q\rightarrow G$ is a transversal, then $\lambda(x)=K+l(x)$. If $l_1\!:Q\rightarrow G$ is another transversal, then $l(x)-l_1(x)\in K$, so that $K+l(x)=K+l_1(x)$ for every $x\in Q$. It follows that $\lambda$ odes not depend on the choice of transversal. Let $\theta\!:Q\rightarrow\mathrm {Aut}(K)$ be the composite: $\theta=\mu_\#\lambda$. If $x\in Q$, then $\theta_x=\mu_\#\lambda(x)=\mu_\#(K+l(x))=\mu(l(x))\in\mathrm {Aut}(K)$; therefore, if $a\in K$,

$$\theta_x(a)=\mu(l(x))(a)=l(x)+a-l(x)$$

does not depend on the choice of lifting $l(x)$.
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A homomorphism $\theta\!:Q\rightarrow\mathrm{Aut}(K)$ makes $K$ into a $Q$-set, where the action is given by $xa=\theta_x(a)$. (For semidirect products, we denoted $\theta_x(a)$ by $a^x$; since we are now writing $K$ additively, however, the notation $xa$ is more appropriate.) The following formulas are valid for all $x,y,\mathbf 1\in Q$ and $a,b\in K$:

$$\begin{aligned} x(a+b)&=xa+xb,\\ (xy)a&=x(ya),\\ \mathbf 1a&=a. \end{aligned}$$

Definition 7.5.5 Call an ordered triple $(Q,K,\theta)$ data if $K$ is an abelian group, $Q$ is a group, and $\theta\!:Q\rightarrow\mathrm{Aut}(K)$ is a homomorphism. We say that a group $G$ realizes this data if $G$ is an extension of $K$ by $Q$ and, for every tranversal $l\!:Q\rightarrow G$,

$$xa=\theta_x(a)=l(x)+a-l(x)$$

for all $x\in Q$ and $a\in K$.

Using these terms, Theorem 7.5.3 says that when $K$ is abelian, every extension $G$ of $K$ by $Q$ determines a homomorphism $\theta\!:Q\rightarrow\mathrm{Aut}(K)$, and $G$ realizes the data. The intuitive meaning of $\theta$ is that it describes how $K$ is a normal subgroup of $G$.

Let $\pi\!:G\rightarrow Q$ be a surjective homomorphism with kernel $K$, and choose a transversal $l\!:Q\rightarrow G$ with $l(\mathbf 1_Q)=\mathbf 0_G$. Once this transversal has been chosen, every element $g\in G$ has a unique expression of the form

$$g=a+l(x),\qquad a\in K,\ \ x\in Q.$$

There is a formula: for all $x,y\in Q$,

$$\begin{align} l(x)+l(y)=f(x,y)+l(xy)\ \text{ for some }\ f(x,y)\in K, \end{align}$$

because both $l(x)+l(y)$ and $l(xy)$ represent the same coset of $K$.

Definition 7.5.6 If $\pi\!:G\rightarrow Q$ is a surjective homomorphism with kernel $K$, and if $l\!:Q\rightarrow G$ is a transversal with $l(\mathbf 1_Q)=\mathbf 0_G$, then the function $f\!:Q\times Q\rightarrow K$, defined by $(1)$ above, is called a factor set (or cocycle). Of course, the factor set $f$ depends on the transversal $l$.

One may think of a factor set as a "measure" of $G$'s deviation from being a semidirect product, for it describes the obstruction to the transversal $l$ being a homomorphism.

Theorem 7.5.7 Let $\pi\!:G\rightarrow Q$ be a surjective homomorphism with kernel $K$, let $l\!:Q\rightarrow G$ be a transversal with $l(\mathbf 1_Q)=\mathbf 0_G$, and let $f\!:Q\times Q\rightarrow K$ be the corresponding factor set. Then:

1. for all $x,y\in Q$, we have $f(\mathbf 1_Q,y)=\mathbf 0_K=f(x,\mathbf 1_Q)$;
2. the cocycle identity holds for every $x,y,z\in Q$:

$$f(x,y)+f(xy,z)=xf(y,z)+f(x,yz).$$

Proof

The definition of $f$ gives $l(x)+l(y)=f(x,y)+l(xy)$. In particular, $l(\mathbf 1_Q)+l(y)=f(\mathbf 1,y)+l(y)$; since we are assuming that $l(\mathbf 1_Q)=\mathbf 0_G$, we have $f(\mathbf 1_Q,y)=\mathbf 0_G$. A similar calculation shows that $f(x,\mathbf 1_Q)=\mathbf 0_G$. The cocycle identity follows from associativity:

$$\begin{aligned} [l(x)+l(y)]+l(z)&=f(x,y)+l(xy)+l(z)\\ &=f(x,y)+f(xy,z)+l(xyz). \end{aligned}$$

on the other hand,

$$\begin{aligned} l(x)+[l(y)+l(z)]&=l(x)+f(y,z)+l(yz)\\ &=xf(y,z)+l(x)+l(yz)\\ &=xf(y,z)+f(x,yz)+l(xyz). \end{aligned}$$

The cocycle identity follows.
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Theorem 7.5.8 Given data $(Q,K,\theta)$, a function $f\!:Q\times Q\rightarrow K$ is a factor set if and only if it satisfies the cocycle identity

$$xf(y,z)-f(xy,z)+f(x,yz)-f(x,y)=\mathbf 0_K$$

as well as $f(\mathbf 1_Q,y)=\mathbf 0_K=f(x,\mathbf 1_Q)$ for all $x,y,z\in Q$. More precisely, there is an extension $G$ realizing the data and a transversal $l\!:Q\rightarrow G$ such that $f$ is the corresponding factor set.

Proof

To prove sufficiency, let $G$ be the set of all ordered pairs $(a,x)\in K\times Q$ equipped with the operation

$$(a,x)+(b,y)=(a+xb+f(x,y),xy)$$

(note that if $f$ is identically $\mathbf 0_K$, then this is the semidirect product $K\rtimes_\theta Q$).

It is easy to check that $G$ is indeed a group. For every transversal $l\!:Q\rightarrow G$, that $xa=l(x)+a-l(x)$ for all $x\in Q$ and $a\in K$. Now we must have $l(x)=(b,x)$ for some $b\in K$. Therefore,

$$\begin{aligned} l(x)+a-l(x)&=(b,x)+(a,\mathbf 1_Q)-(b,x)\\ &=(b+xa,x)+(-x^{-1}b-x^{-1}f(x,x^{-1}),x^{-1})\\ &=(b+xa+x[-x^{-1}b-x^{-1}f(x,x^{-1})]+f(x,x^{-1}),\mathbf 1_Q). \end{aligned}$$

Since $K$ is abelian, the last term simplifies to $(xa,\mathbf 1_Q)$. As any element of $K$, we identify $xa$ with $(xa,\mathbf 1_Q)$, and so $G$ does realize the data.

Define a transversal $l\!:Q\rightarrow K$ by $l(x)=(\mathbf 0_K,x)$ for all $x\in Q$. The factor set $F$ corresponding to this transversal satisfies $F(x,y)=l(x)+l(y)-l(xy)$. But a straightforward calculation shows that $F(x,y)=(f(x,y),\mathbf 1_Q)$, and so $f$ is a factor set, as desired.
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Notation Denote the extension $G$ constructed in the proof of Theorem 7.5.7 by $G_f$; it realizes $(Q,K,\theta)$ and it has $f$ as a factor set.

Definition 7.5.9 $\mathrm Z_\theta^2(Q,K)$ is the set of all factor sets $f\!:Q\times Q\rightarrow K$.

Theorem 7.5.7 shows that $\mathrm Z_\theta^2(Q,K)$ is an abelian group under pointwise addition: $f+g\!:(x,y)\mapsto f(x,y)+g(x,y)$. If $f$ and $g$ are factor sets then so is $f+g$ (for $f+g$ also satisfies the cocycle identity and vanishes on $(\mathbf 1,y)$ and $(x,\mathbf 1)$). If $G_f$ and $G_g$ are the extensions constructed from them, then $G_{f+g}$ is also an extension; it follows that there is an abelian group structure on the family of all extensions realizing the data $(Q,K,\theta)$ whose identity element is the semidirect product (which is $G_{\mathbf 0}$). This group of all extensions is large, however, because the same extension occurs many times. Take a fixed extension $G$ realizing the data, and choose two different transversals, say, $l$ and $l'$. Each transversal gives a factor set:

$$\begin{aligned} l(x)+l(y)&=f(x,y)+l(xy),\\ l'(x)+l'(y)&=f'(x,y)+l'(xy). \end{aligned}$$

Now the factor sets $f$ and $f'$ are distinct, but both of them have arisen from the same extension.

Lemma 7.5.10 Let $G$ be an extension realizing $(Q,K,\theta)$, and let $l$ and $l'$ be transversals with $l(\mathbf 1_Q)=\mathbf 0_K=l'(\mathbf 1_Q)$ giving rise to factor sets $f$ and $f'$, respectively. Then there is a function $h\!:Q\rightarrow K$ with $h(\mathbf 1_Q)=\mathbf 0_K$ such that

$$f'(x,y)-f(x,y)=xh(y)-h(xy)+h(x)$$

for all $x,y\in Q$.

Proof

For each $x\in Q$, both $l(x)$ and $l'(x)$ are representatives of the same coset of $K$ in $G$; there is thus an element $h(x)\in K$ with

$$l'(x)=h(x)+l(x).$$

Since $l'(\mathbf 1_Q)=\mathbf 0_K=l(\mathbf 1_Q)$, we have $h(\mathbf 1_Q)=\mathbf 0_K$. The main formula is derived as follows.

$$\begin{aligned} l'(x)+l'(y)&=[h(x)+l(x)]+[h(y)+l(y)]\\ &=h(x)+xh(y)+l(x)+l(y)\quad(G\text{ realizes the data})\\ &=h(x)+xh(y)+f(x,y)+l(xy)\\ &=h(x)+xh(y)+f(x,y)-h(xy)+l'(xy). \end{aligned}$$

Therefore, $f'(x,y)=h(x)+xh(y)+f(x,y)-h(xy)$. The desired formula follows because each term lies in the abelian group $K$.
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Definition 7.5.11 Given data $(Q,K,\theta)$, a coboundary is a function $g\!:Q\times Q\rightarrow K$ for which there exists $h\!:Q\rightarrow K$ with $h(\mathbf 1_Q)=\mathbf 0_K$ such that

$$g(x,y)=xh(y)-h(xy)+h(x).$$

The set of all coboundaries is denoted by $\mathrm B_\theta^2(Q,K)$.

It is easy to check that $\mathrm B_\theta^2(Q,K)$ is a subgroup of $\mathrm Z_\theta^2(Q,K)$; that is, every coboundary $g$ satisfies the cocycle identity and $g(x,\mathbf 1_Q)=\mathbf 0_K=g(\mathbf 1_Q,x)$ for all $x\in Q$. Moreover, Lemma 7.5.9 says that factor sets $f$ and $f'$ arising from different transversals of the same extension satisfy $f'-f\in\mathrm B_\theta^2(Q,K)$; that is, they lie in the same coset of $\mathrm B_\theta^2(Q,K)$ in $\mathrm Z_\theta^2(Q,K)$.

Proposition 7.5.12 An extension $G$ realizing data $(Q,K,\theta)$ is a semidirect product if and only if it has a factor set $f\in\mathrm B_\theta^2(Q,K)$.

Definition 7.5.13 Given data $(Q,K,\theta)$, then

$$\mathrm H_\theta^2(Q,K)=\mathrm Z_\theta^2(Q,K)/\mathrm B_\theta^2(Q,K);$$

it is called the second cohomology group of the data.

Definition 7.5.14 Two extensions $G$ and $G'$ realizing data $(Q,K,\theta)$ are equivalent if there are factor sets $f$ of $G$ and $f'$ of $G'$ with $f'-f\in\mathrm B_\theta^2(Q,K)$; that is, the factor sets determine the same element of $\mathrm H_\theta^2(Q,K)$.

Proposition 7.5.15 Any two semidirect products realizing data $(Q,K,\theta)$ are equivalent.

Theorem 7.5.16 Two extensions $G$ and $G'$ realizing data $(Q,K,\theta)$ are equivalent if and only if there exists an isomorphism $\gamma$ making the following diagram commute:

where $i$ and $i'$ are injective, $\pi$ and $\pi'$ are surjective, $\mathrm {im} i=\mathrm{ker}\pi$, and $\mathrm{im}i'=\mathrm{ker}\pi'$.

Remark A homomorphism $\gamma$ making the diagram commute is necessarily an isomorphism.

Proof

Assume that $G$ and $G'$ are equivalent. There are thus factor sets $f,f'\!:Q\times Q\rightarrow K$, arising from lifings $l,l'$, respectively, and a function $h\!:Q\rightarrow K$ with $h(\mathbf 1_Q)=\mathbf 0_K$ such that

$$\begin{align} f'(x,y)-f(x,y)=xh(y)-h(xy)+h(x) \end{align}$$

for all $x,y\in Q$. Each element of $G$ has a unique expression of the form $a+l(x)$, where $a\in K$ and $x\in Q$, and addition is given by

$$[a+l(x)]+[b+l(y)]=a+xb+f(x,y)+l(xy);$$

there is a similar description of addition in $G'$. Define $\gamma\!:G\rightarrow G'$ by

$$\gamma(a+l(x))=a+h(x)+l'(x).$$

Since $l(\mathbf 1)=\mathbf 0$, we have $\gamma(a)=\gamma(a+l(\mathbf 1))=a+h(\mathbf 1)+l'(\mathbf 1)=a$, for all $a\in K$, because $h(\mathbf 1)=\mathbf 0$; that is, $\gamma$ fixes $K$ pointwise. Also, $x=\pi(a+l(x))$, while $\pi'\gamma(a+l(x))=\pi'(a+h(x)+l'(x))=\pi'(l'(x))=x$. We have shown that the diagram commutes. It remains to show that $\gamma$ is a homomorphism. Now

$$\begin{aligned} \gamma([a+l(x)]+[b+l(y)])&=\gamma(a+xb+f(x,y)+l(xy))\\ &= a+xb+f(x,y)+h(xy)+l'(xy), \end{aligned}$$

while

$$\begin{aligned} \gamma(a+l(x))+\gamma(b+l(y))&=[a+h(x)+l'(x)]+[b+h(y)+l'(y)]\\ &= a+h(x)+xb+xh(y)+f'(x,y)+l'(xy). \end{aligned}$$

The element $l'(xy)$ is common to both expressions, and $(3)$ shows that the remaining elements of the abelian group $K$ are equal; thus, $\gamma$ is a homomorphism.

Conversely, assume that there exists an homomorphism $\gamma$ as in the statement. Commutativity of the diagram gives $\gamma(a)=a$ for all $a\in K$. Moreover, if $x\in Q$, then $x=\pi(l(x))=\pi'\gamma(l(x))$; that is, $\gamma l\!:Q\rightarrow G'$ is a lifting. Applying $\gamma$ to the equation $l(x)+l(y)=f(x,y)+l(xy)$ shows that $\gamma f$ is the factor set determined by the lifting $\gamma l$. But $\gamma f(x,y)=f(x,y)$ for all $x,y\in Q$, because $f(x,y)\in K$. Therefore, $\gamma f=f$; that is, $f$ is a factor set of $G'$. But $f'$ is also a factor set of $G'$ (arising from another lifting), and so Lemma 7.5.9 gives $f'-f\in\mathrm B^2$; that is, $G'$ and $G$ are equivalent.
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Definition 7.5.17 If $G$ is a extension of $K$ by $Q$, then $\gamma\in\mathrm{Aut}(G)$ stabilizes the extension if the diagram in Theorem 7.5.15 (with $G'$ replaced by $G$) commutes. The group $A$ of all such $\gamma$ is called the stabilizer of the extension.

Theorem 7.5.18 If $K$ and $Q$ are (not necessarily abelian) groups and $G$ is an extension of $K$ by $Q$, then the stabilizer $A$ of the extension is abelian.

Proof

If $\gamma\in A$, then the hypothesis that $\gamma$ makes the diagram commute is precisely the statement that $\gamma$ stabilizes the series $G> K> \mathbf 1$. It follows from Theorem 7.2.33 that the group $A$ of all such $\gamma$ is nilpotent of class $\le 1$; that is, $A$ is abelian.
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Theorem 7.5.19 (Schreier) There is a bijection from $\mathrm H_\theta^2(Q,K)$ to the set $E$ of all equivalence classes of extensions realizing data $(Q,K,\theta)$ taking the identity $\mathbf 0$ ot the class of the semidirect product.

Proof

Denote the equivalence class of an extension $G$ realizing the data $(Q,K,\theta)$ by $[G]$. Define $\varphi\!:\mathrm H_\theta^2(Q,K)\rightarrow E$ by $\varphi(f+\mathrm B_\theta^2(Q,K))=[G_f]$, where $G_f$ is the extension constructed in Theorem 7.5.7 from a factor set $f$. First, $\varphi$ is well defined, for if $f$ and $G$ are factor sets with $f+\mathrm B^2=g+\mathrm B^2$, then $f-g\in\mathrm B^2$, $G_f$ and $G_g$ are equivalent, and $[G_f]=[G_g]$. Conversely, $\varphi$ is an injection: if $\varphi(f+\mathrm B^2)=\varphi(g+\mathrm B^2)$, then $[G_f]=[G_g], f-g\in \mathrm B^2$, and $f+\mathrm B^2=g+\mathrm B^2$. Now $\varphi$ is a surjection: if $[G]\in E$ and $f$ is a factor set (arising from some choice of transversal), then $[G]=[G_f]$ and $[G]=\varphi(f+\mathrm B^2)$. The last part of the theorem follows from Proposition 7.5.11.
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Thus, there is a unique group structure on $E$ making $\varphi$ an isomorphism, namely, $[G_f]+[G_g]=[G_{f+g}]$.

Corollary 7.5.20 $\mathrm H_\theta^2(Q,K)=\mathbf 0$ if and only if every extension $G$ realizing data $(Q,K,\theta)$ is a semidirect product.

Definition 7.5.21 If both $Q$ and $K$ are abelian and $\theta\!:Q\rightarrow\mathrm{Aut}(K)$ is trivial, then $\mathrm{Ext}(Q,K)$ is the set of all equivalence classes of abelian extensions $G$ of $K$ by $Q$.

Corollary 7.5.22 If both $Q$ and $K$ are abelian and $\theta\!:Q\rightarrow\mathrm{Aut}(K)$ is trivial, then $\mathrm{Ext}(Q,K)\le\mathrm H_\theta^2(Q,K)$.

Proof

If $f\!:Q\times Q\rightarrow K$ is a factor set, then the corresponding extension $G_f$ is abelian if and only if $f(x,y)=f(y,x)$ for all $x,y\in Q$: since $\theta$ is trivial,

$$\begin{aligned} (a,x)+(b,y)&=(a+b+f(x,y),xy)\\ &=(a+b+f(y,x),xy)\\ &=(b+a+f(y,x),yx)\\ &=(b,y)+(a,x). \end{aligned}$$

It is easy to see that the set $\mathrm S_\theta^2(Q,K)$ of all such "symmetric" factor sets forms a subgroup of $\mathrm Z_\theta^2(Q,K)$, and

$$\begin{aligned} \mathrm{Ext}(Q,K)&=(\mathrm Z^2\cap\mathrm S^2)/(\mathrm B^2\cap\mathrm S^2)\\ &\cong[(\mathrm Z^2\cap\mathrm S^2)+\mathrm B^2]/\mathrm B^2\le\mathrm H_\theta^2(Q,K). \end{aligned}$$

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Remark 7.5.23 It is plain that $|\mathrm H_\theta^2(Q,K)|$ is an upper bound for the number of nonisomorphic extensions $G$ of $K$ by $Q$ realizing $\theta$; however, this bound need not be attained.

Theorem 7.5.24 If $D$ and $Q$ are groups with $Q$ finite, then the regular wreath product $D\wr_r Q$ contains an isomorphic copy of every extension of $D$ by $Q$.

Corollary 7.5.25 If $\mathscr C$ is a class of finite groups closed under subgroups and semidirect products (i.e., if $A\in\mathscr C$ and $S\le A$, then $S\in\mathscr C$; if $A,B\in\mathscr C$, then $A\rtimes_\theta B\in\mathscr C$ for all $\theta$), then $\mathscr C$ is closed under extensions.

Theorems of Schur-Zassenhaus and Gaschütz

Theorem 7.6.1 If $K$ is an abelian normal Hall subgroup of a finite group $G$ (i.e., $(|K|,[G:K])=1$), then $K$ has a complement.

Proof

Let $|K|=m$, let $Q=G/K$, and let $|Q|=n$, so that $(n,m)=1$. It suffices to prove, by Corollary 7.5.19, that every factor set $f\!:Q\times Q\rightarrow K$ is a coboundary. Define $\sigma\:Q\rightarrow K$ by

$$\sigma(x)=\sum_{y\in Q}f(x,y);$$

$\sigma$ is well defined since $Q$ is finite and $K$ is abelian. Sum the cocycle identity

$$xf(y,z)-f(xy,z)+f(x,yz)=f(x,y)$$

over all $z\in Q$ to obtain

$$x\sigma(y)-\sigma(xy)+\sigma(x)=nf(x,y)$$

(as $z$ ranges over all of $Q$, so does $yz$). Since $(m,n)=1$, there are integers $s$ and $t$ with $sm+tn=1$. Define $h\!:Q\rightarrow K$ by $h(x)=t\sigma(x)$. Then $h(\mathbf 1_Q)=\mathbf 0_K$ and

$$xh(y)-h(xy)+h(x)=f(x,y)-msf(x,y).$$

But $sf(x,y)\in K$, so that $msf(x,y)=\mathbf 0$. Therefore, $f$ is a coboundary.
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Theorem 7.6.2 If $K$ is an abelian normal Hall subgroup of a finite group $G$, then any two complements of $K$ are conjugate.

Proof

Denote $|K|$ by $m$ and $|G/K|$ by $n$, so that $(m,n)=1$. Let $Q_1$ and $Q_2$ be subgroups of $G$ of order $n$. By Proposition 7.3.12, each of these subgroups is a complement of $K$. By Proposition 7.5.3, there are transversals $l_i\!:G/K\rightarrow G$, for $i=1,2$, with $l_i(G/K)=Q_i$ and with each $l_i$ a homomorphism. It follows that the factor sets $f_i$ determined by $l_i$ are identically zero. If $h(x)$ is defined by $l_1(x)=h(x)+l_2(x)$, then

$$\mathbf 0=f_1(x,y)-f_2(x,y)=xh(x)-h(xy)+h(x).$$

Summing over all $y\in G/K$ gives the equation in $K$:

$$\mathbf 0=xa_0-a_0+nh(x),$$

where $a_0=\sum_{y\in G/K}h(y)$. Let $sm+tn=1$ and define $b_0=ta_0$. Since $K$ has exponent $m$,

$$h(x)=h(x)-smh(x)=-tnh(x)=xta_0-ta_0=xb_0-b_0$$

for all $x\in G/K$. We claim that $-b_0+Q_1+b_0=Q_2$. If $l_1(x)\in Q_1$, then $-b_0+l_1(x)+b_0=-b_0+xb_0+l_1(x)=-h(x)+l_1(x)=l_2(x)-l_1(x)+l_1(x)=l_2(x)$.
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Theorem 7.6.3 (Schur-Zassenhaus Lemma) A normal Hall subgroup $K$ of a finite group $G$ has a complement (and so $G$ is a semidirect product of $K$ by $G/K$).

Proof

Let $|K|=m$ and let $|G|=mn$, where $(m,n)=1$. We prove, by induction on $m\ge 1$, that $G$ contains a subgroup of order $n$. The base step is trivially true. If $K$ contains a proper subgroup $T$ which is also normal in $G$, then $K/T\lhd G/T$ and $(G/T)/(K/T)\cong G/K$ has order $n$; that is, $K/T$ is a normal Hall subgroup of $G/T$. If $|K/T|=m'$, then $m'< m$ and $[G/T:K/T]=n$. The inductive hypothesis gives a subgroup $N/T\le G/T$ of order $n$. Now $|N|=n|T|$ and $(n,|T|)=1$ (for $|T|$ divides $m$), so that $T$ is a normal Hall subgroup of $N$ (with $|T|< m$ and with index $[N:T]=n$). By induction, $N$ and hence $G$ contains a subgroup of order $n$.

We may now assume that $K$ is a minimal normal subgroup of $G$. If $p$ is a prime dividing $m$ and if $P$ is a Sylow $p$-subgroup of $K$, then the Frattini argument gives $G=K\mathrm N_G(P)$. By the second isomorphism theorem,

$$G/K=K\mathrm N_G(P)/K\cong \mathrm N_G(P)/(K\cap\mathrm N_G(P))=\mathrm N_G(P)/\mathrm N_K(P),$$

so that $|\mathrm N_K(P)|n=|\mathrm N_K(P)||G/K|=|\mathrm N_G(P)|$. If $\mathrm N_G(P)$ is a proper subgroup of $G$, then $|\mathrm N_K(P)|<m$, and induction shows that $\mathrm N_G(P)$ contains a subgroup of order $n$. We may assume, therefore, that $\mathrm N_G(P)=G$; that is, $P\lhd G$.

Since $K\ge P$ and $K$ is a minimal normal subgroup of $G$, we have $K=P$. Since $\mathrm Z(P)\text{ char }P$ and $P\lhd G$, we have $\mathrm Z(P)\lhd G$. Minimality applies again, and we get $\mathrm Z(P)=P$. But now $P=K$ is abelian, and the proof follows from Theorem 7.6.1.
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Theorem 7.6.4 Let $K$ be a normal Hall subgroup of a finite group $G$. If either $K$ or $G/K$ is solvable, then any two complements of $K$ in $G$ are conjugate.

Remark The Feit-Thompson theorem says that every group of odd order is solvable. Since $|K|$ and $|G/K|$ are relatively prime, at least one of them has odd order, and so complements of normal Hall subgroups are always conjugate.

Proof

Let $|K|=m$, let $|G/K|=n$, and let $Q_1$ and $Q_2$ be complements of $K$ in $G$; of course, $Q_1\cong G/K\cong Q_2$.

Assume first that $K$ is solvable. Since $K'\text{ char }K$ we have $K'\lhd G$; moreover, $Q_1K'/K'\cong Q_1/(Q_1\cap K')\cong Q_1$, so that $|Q_1K'/K'|=n$. Now $K'< K$, because $K$ is solvable. If $K'=\mathbf 1$, then $K$ is abelian and the result is Theorem 7.6.2; otherwise, $|G/K'|<|G|$, and induction on $|G|$ shows that the subgroups $Q_1K'/K'$ and $Q_2K'/K'$ are conjugate in $G/K'$. Thus, there is $\bar g\in G/K'$ with $\bar g(Q_1K/K')\bar g^{-1}=Q_2K'/K'$; that is, $gQ_1g^{-1}\le Q_2K'$ (where $K'g=\bar g$). But $K'< K$ gives $|Q_1K'|<|G|$, and so the subgroups $gQ_1g^{-1}$ and $Q_2$ of order $n$ are conjugate in $Q_2K'$, hence are conjugate in $G$.

Assume now that $G/K$ is solvable. We do an induction on $|G|$ that any two complements of $K$ are conjugate. Let $M/K$ be a minimal normal subgroup of $G/K$. Since $K\le M$, the Dedekind law gives

$$\begin{align} M=M\cap G=M\cap Q_iK=(M\cap Q_i)K\qquad\text{for }i=1,2; \end{align}$$

note also that $M\cap Q_i\lhd Q_i$. Now solvability of $G/K$ gives $M/K$ a $p$-group for some prime $p$, by Theorem 5.3.14. If $M=G$, then $G/K$ is a $p$-group (indeed, since $M/K$ is a minimal normal subgroup of $G/K$, we must have $|M/K|=p$). Therefore, $Q_1$ and $Q_2(\cong G/K)$ are Sylow $p$-subgroups of $G$, and hence are conjugate, by the Sylow theorem.

We may assume, therefore, that $M<G$. Now $M\cap Q_i$ is a complement of $K$ in $M$, for $i=1,2$ (because $M=(M\cap Q_i)K$, by $(4)$, and $(M\cap Q_i)\cap K\le Q_i\cap K=\mathbf 1$). By induction, there is $x\in M\le G$ with $M\cap Q_1=x(M\cap Q_2)x^{-1}=M\cap xQ_2x^{-1}$. Replacing $Q_2$ by its conjugate $xQ_2x^{-1}$ if necessary, we may assume that

$$M\cap Q_1=M\cap Q_2.$$

If $J=M\cap Q_1=M\cap Q_2$, then $J\lhd Q_i$ for $i=1,2$, and so $Q_i\le \mathrm N_G(J)$. Two applications of the Dedekind law give

$$\mathrm N_G(J)=\mathrm N_G(J)\cap KQ_i=(\mathrm N_G(J)\cap K)Q_i$$

and

$$J[\mathrm N_G(J)\cap K]\cap Q_i=J([\mathrm N_G(J)\cap K]\cap Q_i)=J$$

(because $(\mathrm N_G(J)\cap K)\cap Q_i\le K\cap Q_i=\mathbf 1$). Therefore, $Q_1/J$ and $Q_2/J$ are complements of $J(\mathrm N_G(J)\cap K)/J$ in $\mathrm N_G(J)/J$. By induction, there is $\bar y\in\mathrm N_G(J)/J$ with $Q_1/J=\bar y(Q_2/J)\bar y^{-1}$; it follows that $Q_1=yQ_2y^{-1}$, where $Jy=\bar y$, as desired.
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Theorem 7.6.5 (Gaschütz) Let $K$ be a normal abelian $p$-subgroup of a finite group $G$, and let $P$ be a Sylow $p$-subgroup of $G$. Then $K$ has a complement in $G$ if and only if $K$ has a complement in $P$.

Proof

It is easy to see that if $Q$ is a complement of $K$ in $G$, then $Q\cap P$ is a complement of $K$ in $P$.

For the converse, assume that $Q$ is a complement of $K$ in $P$, so that $Q$ is a transversal of $K$ in $P$. All groups in this proof will be written additively. If $U$ is a transversal of $P$ in $G$ (which need not be a subgroup), then $P=\bigcup_{q\in Q}K+q$ and $G=\bigcup_{u\in U}P+u=\bigcup_{q,u}K+q+u$; thus, $Q+U=\{q+u:q\in Q,u\in U\}$ is a transversal of $K$ in $G$. Thus, $-U-Q=-U+Q=\{-u+q:u\in U,q\in Q\}$ is also a transversal of $K$ in $G$. Let us denote $-U$ by $T$, so that $|T|=[G:P]$ and $T+Q$ is a transversal of $K$ in $G$.

Define $l\!:G/K\rightarrow G$ by $l(K+t+q)=t+q$. The corresponding factor set $f\!:G/K\times G/K\rightarrow K$ is defined by

$$\begin{aligned} l(K+t'+q')+l(K+t+q)&=f(K+t'+q',K+t+q)+l(K+t'+q'+t+q). \end{aligned}$$

In particular, if $t=\mathbf 0$, then

$$f(x,K+q)=\mathbf 0_K\qquad\text{for all }x\in G/K,\quad\text{all }q\in Q$$

Consider the cocycle identity

$$xf(y,z)-f(x+y,z)+f(x,y+z)-f(x,y)=\mathbf 0_K$$

for $x,y\in G/K$ and $z=K+q$ with $q\in Q$. The previous equation shows that the first two terms are zero, and so

$$\begin{align} f(x,y+z)=f(x,y)\qquad\text{for }x,y\in G/K\text{ and }z=K+q. \end{align}$$

Let $T=\{t_1,\ldots,t_n\}$, where $n=[G:P]$. For fixed $y=K+g\in G/K$ and for any $t_i$, $K+g+t_i$ lies in $G/K$; since $T+Q$ is a transversal of $K$ in $G$, there is $t_{\pi i}\in T$ and $q_i\in Q$ with $K+g+t_i=K+t_{\pi i}+q_i$. We claim that $\pi$ is a permutation. If $K+g+t_j=K+t_{\pi i}+q_j$, then

$$g+t_i-(t_{\pi i}+q_i)\in K\quad\text{and}\quad g+t_j-(t_{\pi i}+q_j)\in K$$

give

$$g+t_i-q_i-t_{\pi i}+t_{\pi i}+q_j-t_j-g\in K.$$

Since $K\lhd G$, we have $t_i-q_i+q_j-t_j\in K$; since $Q$ is a subgroup, $-q_i+q_j=q\in Q$, so that $t_i+q-t_j\in K$ and $K+t_j=K+t_i+q$. It follows that $t_j=t_i+q$, so that $j=i$ (and $q=\mathbf 0$). Therefore, $\pi$ is an injection and hence a permutation.

Let $\bar T=\{K+t:t\in T\}$ for $x\in G/K$, define

$$\sigma(x)=\sum_{x\in\bar T}f(x,z);$$

$\sigma$ is well defined because $G/K$ is finite adn $K$ is abelian. Summing the cocycle identity gives

$$x\sigma(y)-\sigma(x)+\sum_{x\in \bar T}f(x,y+z)=[G:P]f(x,y).$$

But $y+z=K+g+K+t_i=K+g+t_i=K+t_{\pi i}+q_i$, so that $f(x,y+z)=f(x,K+t_{\pi i}+q_i)=f(x,K+t_{\pi i})$, by $(5)$; since $\pi$ is a permutation, $\sum_{x\in\bar T}f(x,y+z)=\sigma(x)$. Therefore,

$$x\sigma(y)-\sigma(x+y)+\sigma(x)=[G:P]f(x,y).$$

This is an equation in $K\le P$. Since $([G:P],|K|)=1$, there are integers $a$ and $b$ with $a|K|+b[G:P]=1$. Define $h\!:G/K\rightarrow K$ by $h(x)=b\sigma(x)$. Then $h(\mathbf 1)=\mathbf 0$ and

$$xh(y)-h(x+y)+h(x)=f(x,y);$$

that is, $f$ is a coboundary and so $G$ is a semidirect product.
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