Let $G$ be a finite $p$-group with has more than one maximal subgroup. Prove that $G$ has at least $p+1$ maximal subgroups.

Question

Let $G$ be a finite $p$-group with has more than one maximal subgroup. Prove that $G$ has at least $p+1$ maximal subgroups.

I don't have idea. Help me.

Thanks in advanced.

EDIT: I found a result that group of order $p^2$ has exactly $p+1$ maximal subgroup (see page $27$ of Group Theory I - Michio Suzuki). So for a finite $p$-group we are done that it has at least $p+1$ maximal subgroups. I hope I don't have mistakes. :)

Answer 1

Hint: The intersection of all maximal subgroups $F$ is called a Frattini subgroup. It is known that $G/F$ is Abelian (since $G$ is nilpotent). So it is sufficient to consider maximal subgroups of the Abelian group $G/F$.

Answer 2

An elementary solution, which does not require any knowledge of the Frattini subgroup, is the following.

Let $M$ and $N$ be two distinct maximal subgroups. They are known to be normal in $G$, and the homomorphism $$ G \to G/M \times G/N, \qquad x \mapsto (xM, xN), $$ has kernel $K = M \cap N$, of index $p^2$ in $G$, so $G/K \cong G/M \times G/N$ is elementary abelian of order $p^2$, and its $p+1$ subgroups of index $p$ correspond to $p+1$ distinct maximal subgroups of $G$.