cardinality of a maximal subgroup of a $p$ group

Question

Let $P$ be a $p$ group with $|P|=p^n$. Let $M$ be a maximal subgroup of $P$. Is it true that $|M|=p^{n-1 }$ ?

Answer 1

We proceed by induction

Center of a $p$-group is non-trivial, hence there exists an element of order $p$ in the center of a $p$-group, call this element a.

Hence the quotient $G/<a>$ is a $p$-group of order $p$^$(n-1)$ hence all it's maximal subgroups are of the order $p$^$(n-2)$ (by the induction hypothesis)

Now if the original group G has a maximal subgroup H of order $p$^$a$ where $a<p-1$ then the subgroup $H/<a>$ of $G/<a>$ is also maximal in $G/<a>$ (apply correspondence theorem) and order of $H/<a>$ is less than $p$^$(n-2)$ (a contradiction)

Answer 2

An alternate way is to observe that any maximal subgroup $M$ of a $p$ group is normal. So $G/M$ is a simple $p$ group. But every $p$ group has non trivial centre which implies $G/M$ is abelian and hence has to be isomorphic to $Z/pZ$. This proves the claim.