# 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 $G$.

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?

• In general, $C(H) \neq C(G)$. Consider $G$ the quaternion group, then every proper subgroup of $G$ is abelian, but $C(G) = \{1,-1\}$. I don't know if induction works for this, but considering the automorphisms of $N$ given by conjugation with elements of $G$ works. – Daniel Fischer Jun 12 '15 at 14:37
• For any nontrivial normal subgroup $N$ of $G$ we have that $H\cap Z(G)$ is nontrivial - see here. – Dietrich Burde Jun 12 '15 at 14:46
• Thanks for your example.@DanielFischer – haigang hu Jun 12 '15 at 14:57
• Thanks for your link.@DietrichBurde – haigang hu Jun 12 '15 at 14:59

## 2 Answers

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.

• The last statement has been proved at MSE (see link). – Dietrich Burde Jun 12 '15 at 14:46
• Thanks for your solution.@mich95 – haigang hu Jun 12 '15 at 14:56

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.