I have been asked to

Show that if $U$ and $V$ are maximal subgroups of a soluble group $G$, then the following conditions are equivalent:

(a) - $U$ and $V$ are conjugates

(b) - $U_{G} = V_{G}$, i.e., the subgroups have the same normal core

I was able to show that $(a) \implies (b)$, but I am having trouble in the other direction.

I am trying to find a way to apply the Schur-Zassenhaus theorem to find that $U$ and $V$ are conjugates, but I don't see how this could be connected with the fact that $U_{G} = V_{G}$.


This is Ore's theorem. The theorem is really about primitive soluble groups.

Denote the common core by $K$ and argue by induction on $|G|$.

If $K>1$, then you work in the group $G/K$. Now $G/K$ is soluble and both $U/K$ and $V/K$ are maximal subgroups of $G/K$. Further, both $U/K$ and $V/K$ have trivial core in $G/K$ (do you see why?), thus by the induction hypothesis $U/K$ is conjugate to $V/K$ and it follows easily that $U$ is conjugate to $V$. Therefore, it will suffice to prove the statement when $U_G = V_G=1$ and we do this next.

If $G$ is a finite group I will write $\chi(G)$ to denote the set of prime divisors of $|G|$.

I will assume that you know the following:

  1. Normalisers of proper subgroups of nilpotent groups grow. Also, products of normal nilpotent subgroups are nilpotent.
  2. Minimal normal subgroups are characteristically simple. In particular, a minimal normal subgroup of a soluble group is an elementary abelian $p$-group.
  3. The definition of a chief factor.

Definition. Let $G$ be a finite group. Call $G$ primitive if it has a maximal subgroup $M$ with $M_G=1$. Call this core-free maximal subgroup a stabiliser.

Theorem A. Let $G$ be a primitive group, $M$ a stabiliser of $G$ and $H$ a non-trivial nilpotent normal subgroup of $G$. Then $M$ is a complement of $H$ in $G$.

Proof. We have that $M \cap H <H$ since $M_G=1$. Therefore, $N_H(M \cap H) > M \cap H$ because of the nilpotence of $H$. Together with $M \cap H \unlhd M$, $M$ maximal in $G$ and $M_G=1$ it follows that $M \cap H \unlhd G$ and $M \cap H=1$. That $MH=G$ follows quickly from the fact that $H$ is not contained in $M$ and that $M$ is maximal in $G$.

Corollary B. Let $G$ be a primitive soluble group. Then $G$ has a unique non-trivial nilpotent normal subgroup $N$. In particular, $N$ is the unique minimal normal subgroup of $G$.

Proof. Let $M$ be a stabiliser of $G$, $N$ a minimal normal subgroup of $G$ and $H$ a nilpotent normal subgroup of $G$. Then $NH$ is a non-trivial nilpotent normal subgroup of $G$. From this, by Theorem A, we have $|NH| = |G:M| = |N|$, so $N=NH$ and $H=1$ or $H=N$.

Theorem C. Let $G$ be a primitive soluble group, $N$ a minimal normal subgroup of $G$ and $M_1$, $M_2$ two stabilisers of $G$. Then $M_1$ and $M_2$ are conjugate under $N$.

Proof. If $G=N$, then $M_1 = M_2 =1$ by Theorem A. Otherwise, let $L/N$ be a chief factor of $G$, $\chi(N)=p$ and $\chi(L/N)=q$. Then by Corollary B, $L$ is not nilpotent, so $p \neq q$. Since $M_1$, $M_2$ are complements of $N$ in $G$ by Theorem A, it follows that $M_1 \cap L \in \mathrm{Syl}_q(L)$ and $M_2 \cap L \in \mathrm{Syl}_q(L)$. By Sylow's theorem, let $M_1 \cap L = (M_2 \cap L)^{g}$ with $g \in L$. Then $M_1 \cap L \unlhd M_1$, $M_2 \cap L \unlhd M_2$, and $M_1 \cap L = (M_2 \cap L)^g \unlhd M_2^g$. If $M_1 \neq M_2^g$, then $\langle M_1, M_2^g \rangle = G$, and $M_1 \cap L$ would be a non-trivial nilpotent normal subgroup of $G$, in contradiction to Corollary B.

  • $\begingroup$ Sorry for the inconvenience: what is $F(G)$? Thanks. $\endgroup$ Oct 28 '19 at 20:32
  • 2
    $\begingroup$ It is the Fitting subgroup of $G$. $\endgroup$
    – the_fox
    Oct 28 '19 at 20:48
  • $\begingroup$ Thank you for taking the time to answer my questions and for the nice explanation! $\endgroup$ Oct 28 '19 at 20:59
  • 2
    $\begingroup$ There are certain things that I explain only briefly or claim without proof. It is difficult to give a proof of that theorem without assuming some knowledge. If you need further explanations, please tell me what you know and also what you can't figure out. $\endgroup$
    – the_fox
    Oct 28 '19 at 21:12
  • 2
    $\begingroup$ I have rewritten the proof to make it more organised. I am also stating explicitly what I am using. Note that this proof does not use the Schur-Zassenhaus theorem. $\endgroup$
    – the_fox
    Oct 29 '19 at 7:23

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