# Inconjugate maximal subgroups of a soluble group

I am looking for a proof of the following proposition:

If $M_1,M_2$ are inconjugate maximal subgroups of the finite and soluble group $G$, then $M_1\cap M_2$ is maximal in at least one of $M_1$ or $M_2$.

I know there is a proof of this fact in Doerk and Hawkes' "Finite Soluble Groups", but I can't access it, at least for the time being. I was wondering if anyone knows of some other source where a proof is provided.

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## 1 Answer

I couldn't find another reference, but here are Theorem 16.6 and its proof from Doerk's book:

Theorem:

Let $L$ and $M$ be inconjugate maximal subgroups of a finite soluble group $G$. If $M^G\not\leq L^G$, then $L\cap M$ is a maximal subgroup of $L$.

Proof:

Let $R=$ Core$_G(M)$ and $S/R=$ Soc$(G/R)$. The hypothesis implies that $R\not\leq$ Core$_G(L)$ and therefore that $LR=G$ $(*)$.

Since $G/R$ is primitive, $S/R$ is a chief factor of $G$, and since $R$ centralizes $S/R$, it follows from $(*)$ that $S/R$ is $L$-irreducible.

From $(*)$ we also have that $S=S\cap LR=(S\cap L)R$, whence $S/R=(S\cap L)R/R\ \ \stackrel{\cong}{\tiny{L}}\ \ (S\cap L)/(R\cap L)$, and therefore $(S\cap L)/(R\cap L)$ is a chief factor of $L$.

Now $M$ complements $S/R$ in $G$ and $LM=G$.

Hence, $|L:L\cap M|\geq |(S\cap L)(L\cap M):L\cap M|=|S\cap L :R\cap L|=|S:R|=|G:M|=|LM:M|=|L:L\cap M|$.

Consequently $(S\cap L)(L\cap M)=L$, and $L\cap M$ complements the chief factor $(S\cap L)/(R\cap L)$ in $L$. Therefore, $L\cap M$ is a max. subgroup of $L$.$\ \ \$ QED

Since $\leq$ is a partial order, your mentioned theorem follows.

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Thanks, although I can't say I fully understand this proof. – Stefanos Jun 25 '12 at 0:24
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