Let $G$ be a group of odd order and $N$ a normal subgroup of order 17. Prove that $N\subset Z(G)=\{z\in G: \forall_{g\in G}: gzg^{-1}=z\}$.
I was given the hint to look at what $\operatorname{Aut}(N)$ looks like, and then apply the fact that you have a group homomorphism $f:G\to \operatorname{Aut}(N)$ with $f(g)=\phi_g|_N$, with $\phi_g(x)=gxg^{-1}$, restricted to $N$.
I'm not sure about my following approach, because it worked for me only using the second hint. This makes me doubt the correctness as hints are usually not given for no reason; what I have:
$\#N=17\implies N$ is cyclic and thus abelian. We then have $f(n)=\phi_n|_N=id_N$ for all $n\in N$. After all: $\phi_n|_N(x)=nxn^{-1}=nn^{-1}x=x$ for all $x\in N$. From this it follows that $N\subset\ker(f)$.
Now: $g\in \ker(f) \iff \phi_g|_N=id_N \iff$ $\forall_{n\in N}:\phi_g|_N(n)=n \iff \forall_{n\in N}: gng^{-1}=n \iff g\in Z(N)\subset Z(G)$.
So: $N\subset \ker(f)=Z(N)\subset Z(G)$ so that $N\subset Z(G)$.
This is where my mistake was: $Z(N)\not\subset Z(G)$.
Does anyone have any tips using my approach? Or is it not useful to look at the kernel of this homomorphism? What about the first hint? I'm not sure what else to say about $\operatorname{Aut}(N)$ other than finding what it is isomorphic to (Maybe something like $\mathbb{Z}/17\mathbb{Z}$?)
Let $N = \langle n \rangle$. We need to show that $gng^{-1} = n$ for all $g \in G$, i.e. $f(g) = \operatorname{id}_N$.
Now, note that $\operatorname{ord}(f(g)) \mid 16$, as $|\operatorname{Aut}(N)| = |\Bbb Z/16\Bbb Z| = 16$.
Also, since $f$ is a homomorphism, $\operatorname{ord}(f(g)) \mid \operatorname{ord}(g)$.
By Lagrange theorem, $\operatorname{ord}(g) \mid G$.
But $G$ is odd, so $\operatorname{ord}(g)$ is odd, so $\operatorname{ord}(f(g))$ is odd, so $\operatorname{ord}(f(g)) = 1$, so $f(g) = \operatorname{id}_N$ as required.