In a finite p-group,H is a maximal sub group iff H is normal in G and |G:H|=p Let G be a finite p-group,H is a maximal subgroup of G if and only if H is normal in G and |G:H|=p
I tried acting H on right cosets of H in G .... I don't know what to do now... 
 A: One direction is trivial: if $H$ is normal in $G$ and $[G:H]=p$ then $H$ is a maximal subgroup of $G$ by Lagrange's theorem.
Let's prove the other direction by induction on $n$, where $|G|=p^n$. The result is true if $n=1$, because $G$ is simple.
Now recall that the center $Z(G)$ of a $p$-group $G$ is non trivial.
Suppose $n>1$ and that the statement holds for all $p$-groups of order $p^k$, with $k<n$. Let $H$ be a maximal subgroup of $G$. Then we have just two possibilities: (1) $HZ(G)=G$; (2) $HZ(G)=H$.
In case (2), $Z(G)\subseteq H$, so $H/Z(G)$ is a maximal subgroup of $G/Z(G)$ and, by the induction hypothesis, $H/Z(G)$ is normal in $G/Z(G)$ and $[G/Z(G):H/Z(G)]=[G:H]=p$.
In case (1), $H$ is normal in $G$; indeed, if $x\in H$ and $g\in G$, from $HZ(G)=G$ it follows that $g=yz$, for $y\in H$ and $z\in Z(G)$; then
$$
gxg^{-1}=yzxz^{-1}y^{-1}=yxy^{-1}\in H
$$
By maximality, $G/H$ is a simple $p$-group, so $[G:H]=p$.
A: The direction $\;\Longleftarrow\;$ is almost trivial, since if $\;|G|=p^n\;$ , then any maximal (proper!) subgroup $\;M\;$ must be of order $\;p^{n-1}\;$.
The other direction: if $\;M\le G\;$ maximal, then as a finite $\;p  - $ subgroups has a subgroup of any order dividing $\;G\;$ , it must be that $\;M=p^{n-1}\;$ , and normality follows from being  the index $\;[G:M]\;$ equal to the minimal prime dividing $\;|G|\;$
