$N$ a normal subgroup of $G$ with $|G|$ odd and $|N|=5$ satisfies $N \subset Z(G)$

Question

Exercise

Let $N$ be a normal subgroup of $G$. Suppose that $|N|=5$ and that $|G|$ is odd. Show that $N \subset Z(G)$.

I am sorry for not writing any work of mine but I really don't know where to start. I would appreciate any hints that could help me to have an idea of what to do.

Answer 1

In general if $N$ is a normal subgroup of $G$, then there is an injective homomorphism $G/C_G(N) \hookrightarrow \text{Aut}(N)$ (here $C_G(N)=\{g\in G : gn=ng \text{ for all } n \in N\}$, the centralizer of $N$ in $G$). This can be seen by the conjugation action of $G$ on $N$. Since in this case $N\cong C_5$, $\text{Aut}(N)\cong C_4$ and $|G|$ is odd, it follows that $G=C_G(N)$, that is $N \subseteq Z(G)$.
The statement can be generalized to $|N|$ being a Fermat prime number (of the form $2^{2^{n}}+1$).

Answer 2

$G$ acts on $N$ by conjugation, hence permuting the non-identity elements of $N$. That action defines a homomorphism $$\rho:G\to S_4.$$ Since $G$ is has odd order, $\rho$ sends $G$ to one of the cyclic subgroups $C_3$ in $S_4$. It follows that $G$ fixes a non-identity element of $N$. Since any non-identity element of $N$ is a generator, $G$ fixes all elements of $N$.