Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
Thanks, although I can't say I fully understand this proof. –  the_fox Jun 25 '12 at 0:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.