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I'd like to ask for a clarification. I came across the following:

Lemma 2. Let $G$ be a finite solvable group and let $p$ be a prime number dividing $|G|$. Suppose that $M_1$ and $M_2$ are inconjugate maximal subgroups of $G$, both of which have $p$-power index in $G$, and neither of which is normal in $G$. Then $(M_1\cap M_2)\mathbf{O}^p(G)=G$. Furthermore, if $P_0\in \mathrm{Syl}_p(\mathbf{O}^p(G))$, then $(M_1\cap P_0)(M_2\cap P_0) = P_0$.

Proof. Note that every maximal subgroup containing $\mathbf{O}^p(G)$ is normal of index $p$. To prove that $(M_1\cap M_2)\mathbf{O}^p(G)$, it therefore suffices to show...

I don't understand the first line of the proof. Any help would be appreciated.

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What does $\mathbf{O}^{\,p}(G)$ stand for? – anon Jun 23 '12 at 9:31
The smallest normal subgroup of $G$ of $p$-power index. It's called the $p$-residual of $G$. – the_fox Jun 23 '12 at 9:39
up vote 3 down vote accepted

The quotient $G/\mathbf{O}^p(G)$ is a $p$-group. By the basic properties of $p$-groups its maximal subgroups are of index $p$. Also all maximal subgroups of a $p$-group are normal (a subgroup of index equal to the smallest prime factor of the order of the group is normal). So it seems to me that the first line of the proof follows from this and the correspondence principle: the subgroups of $G/\mathbf{O}^p(G)$ are in 1-1 correspondence with subgroups of $G$ containing $\mathbf{O}^p(G)$, normality is preserved.

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Thank you. I shouldn't have missed that. – the_fox Jun 23 '12 at 10:21

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