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.