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

Solution 1:

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 $G$.

Solution 2:

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 $N\leqslant Z(G)$.