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 ?

If $N$ is normal in $G$, then $G$ acts by conjugation on $N$, which gives a group homomorphism $\def\Aut{\operatorname{Aut}}f:G\to \Aut N$. But since $|N|=17$ is prime, $N$ is cyclic and $\Aut N\cong (\Bbb Z/17\Bbb Z)^\times$, which has $16$ elements (it is actually cyclic, but that is not needed here). Now the image of $f$, which has odd order because $|G|$ is odd, must be the trivial subgroup because $1$ is the only odd divisor of $16$, and this means that $N$ is central in $G$.
