If a normal subgroup $N$ of order $p$($p$ prime) is contained in a group $G$ of order $p^n$,then $N$ is in the center of $G$.
I want to use induction to prove this:
It is trivial when $G=p(n=1)$, assume $G=p^n$, $N$ is a normal subgroup in $G$ with order $p$. Since $G$ is p-subgroup, $G$ has a normal subgroup $H$ with order $p^{n-1}$,according to assumption, $N \subset C(H)$. But I don't know how to go on, is $C(H)=C(G)$?Can induction work in this case?
Notice that $N$ being normal implies a homomorphism from $G \to Aut(N)$ (sending $g$ to the function $\phi_{g} : x \to gxg^{-1}$ and the kernel is the centraliser of $N$ in $G$. Notice that $Aut(N)$ has order $p-1$ and so the two groups $G$ and $Aut N$ have relatively prime orders and hence any homomorphism is trivial. Therefore, the kernel must be the whole group, ie $G$ centralised $N$, and $N$ is in the center. Alternatively, you can prove that if $H$ is a non trivial normal subgroup of $G$, then $H \cap Z(G) \not = \lbrace 1 \rbrace$, and hence the result will immediately follow.
There is a well-known generalization here: let $N \unlhd G$, with $|N|=p$, the smallest prime dividing $|G|$, then $N \subseteq Z(G)$.
The proof runs along the same lines as explained by mich95: there is an injective homomorphism $G/C_G(N) \hookrightarrow Aut(N) \cong C_{p-1}$, hence $G=C_G(N)$, that is, $N$ is central.
