Group Extensions Complements and Schur Zassenhaus

By [L1], $i(N)=\ker \pi$. Since $\ker \pi$ is normal in $E$ by [L2], the subgroup $i(N)$ is normal in $E$.

The map $\pi$ is surjective because the sequence is exact, so its image is $Q$. Applying [L3] to $\pi$ and using step 1.1 gives $E\mathrm{/}i(N)=E\mathrm{/}\ker \pi \cong Q$.

By [L1], conditions 1, 2, and 3 are equivalent.

Suppose $r:E\to N$ is a retraction of $i$. For $x\in i(N)\cap \ker r$, write $x=i(n)$. Then $1=r(x)=r(i(n))=n$, so $x=i(1)=1$. For any $g\in E$, the element $i(r(g){)}^{-1}g$ lies in $\ker r$ because $r(i(r(g){)}^{-1}g)=r(g{)}^{-1}r(g)=1.$ Therefore $g=i(r(g))(i(r(g){)}^{-1}g)\in i(N)\ker r,$ so $E=i(N)\ker r$ and $\ker r$ is a complement to $i(N)$.

Let $k\in \ker r$ and $x=i(n)\in i(N)$. Because $i(N)$ is normal in $E$ by [L2], the conjugate $kx{k}^{-1}$ still lies in $i(N)$. Applying $r$ gives $r(kx{k}^{-1})=r(k)r(x)r(k{)}^{-1}=n\mathrm{.}$ Since $r$ restricts to the inverse of $i$ on $i(N)$, this forces $kx{k}^{-1}=i(n)=x$. Hence $\ker r$ centralizes $i(N)$, and step 1.2 upgrades the decomposition to a direct product $E\cong i(N)\times \ker r\cong N\times \ker r$.

Step 1.1 is the splitting criterion, while steps 1.2 and 2.1 show that any kernel retraction forces a direct-product splitting.

By [L1], the restriction $\pi {\mathrm{\mid }}_C:C\to Q$ is an isomorphism. Hence for each $q\in Q$ there is a unique $c\in C$ with $\pi (c)=q$, so the displayed formula is well defined.

For each $q\in Q$, conjugation by the corresponding $c$ restricts to an automorphism of $i(N)$ by [L2], because $i(N)⊴E$ and so $ci(N)c^{-1}=i(N)$. Transporting that automorphism across the isomorphism $i:N\to i(N)$ gives the displayed automorphism of $N$. Thus the rule lands in $\mathrm{Aut}(N)$.

If $q_1,q_2\in Q$ correspond to $c_1,c_2\in C$, then $q_1{q}_2$ corresponds to $c_1{c}_2\in C$ because $\pi {\mathrm{\mid }}_C$ is a homomorphism. Conjugation by $c_1{c}_2$ is the composite of conjugation by $c_1$ and conjugation by $c_2$; transporting across $i$ gives $(q_1{q}_2)\cdot n=q_1\cdot (q_2\cdot n)$, and $1_Q\cdot n=n$. So this is an action of $Q$ on $N$ by automorphisms.

By [L1], the split extension is equivalent to the semidirect-product extension defined by the complement-induced action of $Q$ on $N$.

If every element of $C$ commutes with every element of $i(N)$, then each conjugation automorphism from step 1.1 is the identity. So the induced action is trivial, and [L2] makes the decomposition a direct product.

Conversely, if the decomposition is a direct product, then the induced action is trivial by [L2]. Therefore the conjugation of $i(N)$ by every element of $C$ is the identity, so $C$ centralizes $i(N)$.

For $q\in Q$, choose $x\in E$ with $\pi (x)=q$. Since $i(N)=\ker \pi$ is normal, conjugation by $x$ restricts to an automorphism of $i(N)$, hence of $N$, by [L2]. Let ${\omega }_E(q)$ be its class in $\mathrm{Out}(N)$ from [L1].

If $x^{\mathrm{\prime }}=xi(n)$ is another lift of $q$, then for $m\in N$, $x^{\mathrm{\prime }}i(m)x^{\mathrm{\prime }-1}=xi(nm{n}^{-1})x^{-1}\mathrm{.}$ Thus the automorphism from $x^{\mathrm{\prime }}$ differs from the automorphism from $x$ by the inner automorphism of $N$ defined by $n$. So step 1.1 is independent of the chosen lift as an element of $\mathrm{Out}(N)$.

If $q_1,q_2\in Q$ are lifted by $x_1,x_2\in E$, then $x_1{x}_2$ lifts $q_1{q}_2$, and conjugation by $x_1{x}_2$ is the composite of conjugation by $x_1$ and by $x_2$. Passing to classes modulo inner automorphisms with [L1], ${\omega }_E(q_1{q}_2)={\omega }_E(q_1){\omega }_E(q_2)$. So ${\omega }_E$ is a homomorphism.

If $\phi :E\to E^{\mathrm{\prime }}$ is an equivalence of extensions as in [L3], then $\phi$ identifies the chosen copy of $N$ with itself and carries lifts of $q$ in $E$ to lifts of $q$ in $E^{\mathrm{\prime }}$. Therefore the conjugation automorphisms correspond and define the same class in $\mathrm{Out}(N)$. Hence ${\omega }_E$ depends only on the extension class.

Because $N$ is a Hall $\pi$-subgroup, [L1] gives $\gcd (\mathrm{\mid }N\mathrm{\mid },[G:N])=1$.

Since $N⊴G$, the quotient $G\mathrm{/}N$ exists and [L2] gives $\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }=[G:N]$. Therefore step 1.1 says exactly that $\gcd (\mathrm{\mid }N\mathrm{\mid },\mathrm{\mid }G\mathrm{/}N\mathrm{\mid })=1$. The displayed short exact sequence is the standard quotient extension.

We argue by induction on $\mathrm{\mid }G\mathrm{\mid }$. If $N=1$, then $G$ is a complement to $N$. If $N=G$, then $G\mathrm{/}N=1$, so the trivial subgroup is a complement. Assume from now on that $1<N<G$ and that the statement holds for all smaller finite groups.

Suppose $M$ is a normal subgroup of $G$ with $1<M<N$. Then $N\mathrm{/}M$ is a normal Hall subgroup of $G\mathrm{/}M$, so the induction hypothesis gives a subgroup $K\mathrm{/}M\le G\mathrm{/}M$ complementary to $N\mathrm{/}M$. Thus $K<G$, $G=NK$, and $N\cap K=M$. Inside the proper group $K$, the subgroup $M$ is normal Hall because $\mathrm{\mid }K\mathrm{/}M\mathrm{\mid }=\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }$ is coprime to $\mathrm{\mid }M\mathrm{\mid }$. The induction hypothesis applied to $K$ gives a complement $H$ to $M$ in $K$, so $K=MH$. Since $H\le K$, we have $H\cap N=H\cap (N\cap K)=H\cap M=1$, while $NH=N(MH)=NK=G$. Thus $H$ complements $N$ in $G$. For the remainder we may therefore assume that $N$ contains no nontrivial proper subgroup normal in $G$.

Choose a prime $p$ dividing $\mathrm{\mid }N\mathrm{\mid }$ and let $P$ be a Sylow $p$-subgroup of $N$, whose existence is supplied by [L4]; in particular $P\mathrm{\ne }1$. Because $N⊴G$, every $G$-conjugate of $P$ is again a Sylow $p$-subgroup of $N$. Hence [L4] gives, for each $g\in G$, an $n\in N$ with $gP{g}^{-1}=nP{n}^{-1}$. Thus $n^{-1}g\in N_G(P)$ and $G=N{N}_G(P)$. If $P<N$ and $N_G(P)=G$, then $P$ is a nontrivial proper normal subgroup of $G$ contained in $N$, contrary to step 2.1. Thus $P<N$ forces $R:=N_G(P)<G$. Put $A=R\cap N$. The subgroup $A$ is normal in $R$, and $G=NR$ gives $R\mathrm{/}A\cong G\mathrm{/}N$; hence $A$ is a normal Hall subgroup of $R$. Induction gives a complement $H$ to $A$ in $R$, so $R=AH$. Consequently $G=NR=NH$ and $N\cap H=A\cap H=1$, proving the theorem when $P<N$. The only remaining case is $P=N$, so $N$ is a $p$-group.

By [L5], the finite $p$-group $N$ has nontrivial center. The center is characteristic in $N$, so [L6] makes it normal in $G$; step 2.1 therefore forces $Z(N)=N$, and $N$ is abelian. Now ${\mathrm{\Omega }}_1(N)=\{x\in N:x^p=1\}$ is a subgroup, is nontrivial by [L3], and is characteristic because automorphisms preserve $p$th powers. Hence [L6] makes it normal in $G$, and step 2.1 again forces ${\mathrm{\Omega }}_1(N)=N$. Therefore every nonidentity element of $N$ has order $p$, so $N$ is elementary abelian.

Write $Q=G\mathrm{/}N$, choose a set-theoretic section $s:Q\to G$ with $s(1)=1$, and write the abelian group $N$ additively. Because $N$ is abelian, the formula $q\cdot n=s(q)ns(q{)}^{-1}$ does not depend on the chosen lift of $q$, and [L7] makes it an action of $Q$ on $N$ by automorphisms. Define $f:Q\times Q\to N$ by $s(q)s(r)=f(q,r)s(qr)$. This is well defined because $\pi (s(q)s(r))=qr$, so $s(q)s(r)s(qr{)}^{-1}\in N$.

Associativity gives $f(q,r)+f(qr,t)=q\cdot f(r,t)+f(q,rt)$ for all $q,r,t\in Q$. Let $m=\mathrm{\mid }Q\mathrm{\mid }$, choose an integer $b$ with $bm\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p)$, define $\sigma (q)={\sum }_{x\in Q}f(q,x)$ and $c(q)=b\sigma (q)$, and sum the cocycle identity over $x\in Q$. Since $x\mapsto rx$ is a permutation of $Q$, this yields $mf(q,r)+\sigma (qr)=q\cdot \sigma (r)+\sigma (q)$. Because every element of the elementary abelian $p$-group $N$ has order dividing $p$, multiplication by $bm$ is the identity on $N$. Hence $f(q,r)=c(q)+q\cdot c(r)-c(qr)$.

Define $s^{\mathrm{\prime }}:Q\to G$ by $s^{\mathrm{\prime }}(q)=-c(q)+s(q)$, meaning $s^{\mathrm{\prime }}(q)=i(-c(q))s(q)$ inside the extension. Since $i(-c(q))\in N$, we have $\pi (s^{\mathrm{\prime }}(q))=q$. Using the formula from step 6.1, $s^{\mathrm{\prime }}(q)s^{\mathrm{\prime }}(r)=(-c(q)-q\cdot c(r)+f(q,r))s(qr)=-c(qr)s(qr)=s^{\mathrm{\prime }}(qr)$. So $s^{\mathrm{\prime }}$ is a homomorphic section of $G\to Q$. By [L2], the extension splits, equivalently $N$ has a complement in $G$. This completes the induction.

We argue by induction on $\mathrm{\mid }G\mathrm{\mid }$. The claim is trivial when $N=1$ or $N=G$. Assume from now on that $1<N<G$ and that the statement holds for all smaller finite groups satisfying the same solvability hypothesis.

First consider the special case in which $A⊴X$ is an abelian normal Hall subgroup of a finite group $X$, and $U,V$ are complements to $A$. Write $A$ additively. Since $U$ is a complement, every element of $X$ has a unique form $au$ with $a\in A$ and $u\in U$. For each $u\in U$, let $\lambda (u)\in A$ be determined by the unique element $\lambda (u)u\in V$. Then $\lambda (uv)=\lambda (u)+u\cdot \lambda (v)$ because $V$ is a subgroup. Let $m=\mathrm{\mid }U\mathrm{\mid }$, choose $b$ with $bm\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{\mid }A\mathrm{\mid })$, and put $\sigma ={\sum }_{u\in U}\lambda (u)$ and $a=-b\sigma$. Summing $\lambda (ux)=\lambda (u)+u\cdot \lambda (x)$ over $x\in U$ gives $\sigma =m\lambda (u)+u\cdot \sigma$, hence $\lambda (u)=u\cdot a-a$. Therefore every element of $V$ has the form $(u\cdot a-a)u=a^{-1}ua$, so $V=a^{-1}Ua$. Thus complements to an abelian normal Hall subgroup are conjugate by that subgroup.

Suppose $N$ is solvable. By [L3], choose a nontrivial abelian normal subgroup $A\le N$ that is normal in $G$. In $G\mathrm{/}A$, the subgroup $N\mathrm{/}A$ is normal Hall and the images $HA\mathrm{/}A$ and $KA\mathrm{/}A$ are complements. The induction hypothesis applies because $N\mathrm{/}A$ is solvable by [L2]. After conjugating $K$, we may assume $HA=KA=:M$. Inside $M$, the subgroup $A$ is abelian normal Hall and $H,K$ are complements to $A$, so step 1.2 makes them conjugate in $M$, hence in $G$.

Now suppose instead that $G\mathrm{/}N$ is solvable. Since $H\cong G\mathrm{/}N$, the group $H$ is solvable by [L2]. If $\mathrm{\mid }H\mathrm{\mid }$ is prime, then both $H$ and $K$ are Sylow subgroups of that prime order, so [L6] makes them conjugate.

Assume $\mathrm{\mid }H\mathrm{\mid }$ is not prime, and write $Q=G\mathrm{/}N$. Choose a nontrivial proper abelian normal subgroup $\bar{A}⊴Q$: if $Q$ is nonabelian, apply [L3] to the solvable group $Q$ and its normal subgroup $Q$; if $Q$ is abelian of composite order, [L5] gives an element of prime order and [L4] makes the subgroup it generates a proper normal subgroup. Let $M$ be the preimage of $\bar{A}$ under the quotient map $G\to Q$. Then $M⊴G$, $M<G$, and $M\mathrm{/}N\cong \bar{A}$ is solvable. Since $H$ and $K$ are complements to $N$ in $G$, the subgroups $H\cap M$ and $K\cap M$ are complements to $N$ in $M$. The induction hypothesis applied to $M$ therefore lets us conjugate $K$ by an element of $M$ so that $H\cap M=K\cap M=:A$. In particular, $A$ is abelian and normal in both $H$ and $K$.

Both $H$ and $K$ normalize $A$, so $H,K\le N_G(A)$. If $N_G(A)<G$, then every element of $N_G(A)$ has the form $nh$ with $n\in N$ and $h\in H$, and because $h\in N_G(A)$ the condition $nh\in N_G(A)$ forces $n=(nh)h^{-1}\in N\cap N_G(A)$. Hence $N_G(A)=(N\cap N_G(A))H=(N\cap N_G(A))K$. Thus $N\cap N_G(A)$ is a normal Hall subgroup of the proper group $N_G(A)$, and the quotient by it is isomorphic to the solvable group $H$. The induction hypothesis on $N_G(A)$ makes $H$ and $K$ conjugate there.

If instead $N_G(A)=G$, then $A⊴G$. In $G\mathrm{/}A$, the normal Hall subgroup is $NA\mathrm{/}A$, and the images $H\mathrm{/}A$ and $K\mathrm{/}A$ are complements. The quotient $(G\mathrm{/}A)\mathrm{/}(NA\mathrm{/}A)\cong G\mathrm{/}NA\cong Q\mathrm{/}\bar{A}$ is solvable and smaller than $Q$ because $\bar{A}$ is nontrivial. Therefore the induction hypothesis on $G\mathrm{/}A$ lets us conjugate $K$ so that $H\mathrm{/}A=K\mathrm{/}A$. Since $A\le H$ and also $A\le K$ with $A⊴G$, this equality of quotient subgroups means $H=K$. Thus the final conjugacy is proved in this case as well.

Steps 2.1 and 2.2-3.2 cover the two solvability hypotheses. Therefore complements to $N$ are conjugate in $G$ whenever the kernel or quotient is solvable.

By [L1], the kernel $N$ is normal in $G$ and $G\mathrm{/}N\cong Q$, so $[G:N]=\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }=\mathrm{\mid }Q\mathrm{\mid }$. Because $\gcd (\mathrm{\mid }N\mathrm{\mid },\mathrm{\mid }Q\mathrm{\mid })=1$, no prime divisor of $[G:N]$ divides $\mathrm{\mid }N\mathrm{\mid }$. Hence $N$ is a Hall subgroup of $G$.

By [L2], the normal Hall subgroup $N$ has a complement in $G$. Then [L3] turns that complement into a splitting section, so the extension splits.

By [L2], the extension defines a homomorphism $\omega :Q\to \mathrm{Out}(N)$. Since $\mathrm{Out}(N)=1$ by [L1], this homomorphism is trivial. Therefore for every $g\in E$, conjugation by $g$ on $i(N)$ is an inner automorphism.

If $x\in i(N)\cap C_E(i(N))$, then $x$ commutes with every element of $i(N)$. Under the identification of $i(N)$ with $N$, this says that $x$ lies in $Z(N)$, which is trivial by [L1]. So $i(N)\cap C_E(i(N))=1$.

Fix $g\in E$. By step 1.1 there exists $\gamma \in N$ such that $gi(n)g^{-1}=i(\gamma n{\gamma }^{-1})$ for all $n\in N$. Then $i(\gamma {)}^{-1}g$ commutes with every element of $i(N)$, so $i(\gamma {)}^{-1}g\in C_E(i(N))$. Hence every $g\in E$ lies in $i(N)C_E(i(N))$, and therefore $E=i(N)C_E(i(N))$.

Because every element of $C_E(i(N))$ commutes with every element of $i(N)$ by [L3], the two subgroups centralize one another. Together with steps 2.1 and 1.2, this makes $E$ the internal direct product of $i(N)$ and $C_E(i(N))$. In particular $C_E(i(N))$ is a complement to $i(N)$, so [L4] gives a split extension.

The map $s$ is a set-theoretic section because reducing mod $2$ sends $[0]$ to $[0]$ and $[1]$ to $[1]$.

In $C_2$ we have $[1]+[1]=[0]$, but in $C_4$ we get $s([1]+[1])=s([0])=[0]\mathrm{\ne }[2]=[1]+[1]=s([1])+s([1])$. So [L2] shows that $s$ is not a homomorphism. Therefore the claim is false, and [L1] explains why set sections alone do not split extensions.

Define two quotient maps ${\pi }_1,{\pi }_2:E\to Q$ by ${\pi }_1(\bar{x})=\bar{x}(\mathrm{m}\mathrm{o}\mathrm{d}3),{\pi }_2(\bar{x})=2\bar{x}(\mathrm{m}\mathrm{o}\mathrm{d}3)\mathrm{.}$ Both are surjective homomorphisms with kernel $\{\bar{0},\bar{3},\bar{6}\}=\mathrm{im}j$, so they give two extensions of $Q$ by $N$ with the same middle group $E$.

An equivalence in the sense of [L1] would be an automorphism $\phi :E\to E$ such that $\phi \circ j=j$ and ${\pi }_2\circ \phi ={\pi }_1$. Every automorphism of the additive cyclic group $\mathbb{Z}\mathrm{/}9\mathbb{Z}$ has the form $\phi (\bar{x})=u\bar{x}$ for a unit $u\in \{1,2,4,5,7,8\}$. The condition $\phi \circ j=j$ gives $\overline{3u}=\bar{3}$, so $u\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}3)$. But ${\pi }_2\circ \phi ={\pi }_1$ forces $\overline{2u}=\bar{1}$ in $Q$, so $u\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}3)$. This is impossible. Therefore the two extensions are not equivalent.

In $D_4=⟨r,s\mid r^4=s^2=1,sr{s}^{-1}=r^{-1}⟩$, the subgroup $⟨r⟩\cong C_4$ is normal and $⟨s⟩\cong C_2$ is a complement, so the reverse implication of [L2] makes the extension split.

But $sr{s}^{-1}=r^{-1}\mathrm{\ne }r$, so the complement does not centralize the kernel. By [L1], the decomposition is not a direct product. Hence the claim is false.

The subgroup $⟨(12)⟩$ has order $2$ and index $3$, so [L1] makes it a Hall $\{2\}$-subgroup of $S_3$.