Let $N$ be a normal subgroup of a finite group $G$ Let $N$ be a normal subgroup of a finite group $G$. Suppose $|N|=5$ and that $|G|$ is odd. Prove $N$ is contained in $Z(G)$ , the center of $G$
because the order of $N$ is prime, then N is cyclic, but
I'm more concerned with how I can derive for  $g\in G$ and $g^{-1}ng=n$ for all $n\in N$
 A: Consider $G$ acting on $N$ via conjugation. Every element $g \in G$ defines an homomorphism of groups $N \to N$. Now, it is known that $\operatorname{Aut}(C_5)=C_4$, so the above action defines a map $G \to C_4$. But, since $|G|$ is odd, every element of $G$ has odd order. 
It follows that the homomorphism defined by conjugation has to be the trivial one, so every element of $G$ acts trivially on $N$, that is, for any $x \in N, g \in G$ $$gxg^{-1}=x$$ which implies $$gx = xg$$ so $N \subseteq Z(G)$.
A: By using N/C Lemma, $G/C_G(N)$ is isomorphic to a subgroup of $Aut(N)$.
Note that $|Aut(N)|=4$.
Since $|G|/|C_G(N)|$ divides $|Aut(N)|$ and $|G|$ is odd, we have $G=C_G(N)$, that is $N\leq Z(G).$
A: In general: if $N \unlhd G$ with $|G|$ is odd and $|N|$ is a product of different Fermat prime numbers (there are $32$ possibilities here!), then $N \subseteq Z(G)$.
A: $gng^{-1}=n^k$ . This implies $g^lng^{-l}=n^{k^l} \forall l.$ Putting $l=G$ we have $n=n^{k^{G}}$ Thus we have 5 divides $k^{G}-1$. Since $G$ is odd we have therefore 5 divides $k-1$ and hence $n^k=n$ .This completes the proof.
