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 ?
 A: 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 $N\leqslant Z(G)$.
A: 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 $G$.
A: I might be wrong but I think this question has more or less the same idea as given in the link


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*http://crazyproject.wordpress.com/2010/05/20/if-a-subgroup-of-order-17-is-normal-in-a-group-of-order-3825-then-it-is-central/
