$p$ prime and $G$ non-abelian $p$-group. V is abelian subgroup with $[G:V]=p$. Now I need to show that the number of subgroups of index $p$ in $G$ is $1$ or $p + 1$.
I first started to look what happens for $n=1$. $[G:V]=p \Leftrightarrow |V|=1$ and thus $V$ is the only subgroup with index $p$.
Now I thought maybe there is like, for even n there are $p+1$ subgroups and for odd just $1$. Is there any statement like this?
Well, Other things that might help, I guess the Sylow-sentences, and that all subgroups with index $p$ are normal-subgroups.
Again I don't know how to approach. So some little hint would be awesome :)