# If maximal subgroups of solvable group have equal cores, then they are conjugates

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 $$M \cap H since $$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.

• Sorry for the inconvenience: what is $F(G)$? Thanks. Oct 28 '19 at 20:32
• It is the Fitting subgroup of $G$. Oct 28 '19 at 20:48
• Thank you for taking the time to answer my questions and for the nice explanation! Oct 28 '19 at 20:59
• 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. Oct 28 '19 at 21:12
• 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. Oct 29 '19 at 7:23