Let $G$ a finite group of odd order. Let $N \triangleleft G$ and $|N|=17$. Prove that $N \subset Z(G)$. 
Let $G$ a finite group of odd order. Let $N \triangleleft G$ and
  $|N|=17$. Prove that $N \subset Z(G)$.

Here is what I tried:
I need to prove that: $∀n\in N,∀g \in G :gn=ng$.
Let $g \in G, n\in N$.
Since $|N|=17$, then $N$ cyclic, and then $N$ is an abelian subgroup.
Since $N\triangleleft G$, then $gng^{-1}=n' \in N$.
Now I need to prove that $n'=n$.
I now that this is true for $g\in N$, as $N$ is abelian, but I can't prove that this is true for all $g\in G$.
Any hints ?

Hint. Notice that the automorphism group of $N\cong \mathbb{Z}_{17}$ has order $16$.  Then use the fact $G$ has odd order to prove that no element of $G$ can induce a nontrivial homomorphism of $N$.
Note that by the same proof we may show that whenever a group of odd order has a subgroup $N\unlhd G$ whose order is a Mersenne prime, we may conclude that $N\leqslant Z(G)$.
