PM is supersolvable group $G$ is a finite group, $G = PM$, where $P$ is a Sylow $p$-subgroup of $G$, $p$ is the largest prime dividing the order of $G$, $P$  is normal in $G$, $M$ is maximal subgroup of $G$, $M$ is supersolvable and $|G/M| = p$.
Is $G$ supersolvable?
(The group $G$ is said super-solvable if $G$ has normal series $0=G_0⊴G_1⊴⋯G_{n−1}⊴G_n=G$ such that $G_i/G_{i−1}$ is cyclic for $i\in\{1,2,…,n\}$.)
Thanks

I don't think you need assume that $M \lhd G$, just that the index $|G:M|=p$.
For a finite group, being supersolvable is equivalent to the chief factors all being cyclic of prime order.
Note that $|P:P \cap M| = p$, so $P \cap M \lhd P$ and $P \cap M \lhd M$ (since $P \lhd G$), so $P \cap M \lhd G$. The terms in the lower central series of $P \cap M$ are characteristic and hence normal in $G$. Refine this series to a chief series of $M$. Then, since $M$ is supersolvable, all factors in this series have prime order. The terms in the series that are contained in $P \cap M$ are all normal in $P$, so they are normal in $G$. Now $P/P\cap M$ has order $p$, so it is cyclic. And $G/P \cong M/(M \cap P)$ is supersolvable, so it has a normal series with cyclic factors.So now we have build up a normal series of $G$ with cyclic factors, and hence $G$ is supersolvable.
But I don't seem to have used the fact that $p$ is the largest prime dividing $|G|$, so perhaps I have  a mistake somewhere!!!
