Schur-Zassenhaus conjugacy when the kernel or quotient is solvable

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.