# A finite normal subgroup is a subset of the center

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 $\mathbb{Z}/17\mathbb{Z}$?)

• Why $Z(N) \subset Z(G)$? May 20 '18 at 11:42
• @KennyLau I actually just looked at how I wrote it down and I think I see the mistake; I said $Z(N)=\{z\in N: gzg^{-1}=g\}$ for all $g\in G$ but that of course must be $nzn^{-1}$ for all $n\in N$ so they're not actually subsets.
– Marc
May 20 '18 at 11:45

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.

• Thank you for your answer. There's a couple of things I'm not completely getting; - I'm not familiar with the Lagrange theorem, what is it about in general? - How do you know $\operatorname{Aut} (N) \cong \mathbb{Z}/16\mathbb{Z}$? - How does it follow that $\#f(g)=1 \implies f(g)=id_N$?
– Marc
May 20 '18 at 11:56
• 1. Lagrange theorem states that the size of a subgroup divides the order of a group (how are you doing centralizers if Lagrange theorem hasn't been taught?) 2. $N$ is cyclic, so see this. 3. The order of $f(g)$ is by definition the least positive integer $r$ such that $f(g)^r = e$, where $e = \operatorname{id}_N$ is the identity of the group. Here, $r = 1$, so $f(g) = e$. May 20 '18 at 12:07
• Thank you for your references. I'm only a first-years mathematics student so often I find it hard to find sources to answer my questions without involving a lot of complex math. And actually, I might've had that theorem once but maybe it wasn't named
– Marc
May 20 '18 at 12:15
• Would you mind elaborating on $aut(N)\cong \mathbb{Z}/16\mathbb{Z}$ in this case? I understand the case with $\mathbb{Z}$ but I don't know how to use that when you don't have 'numbers' and 'primes' in $N$ as it is an arbitrary group with arbitrary operations
– Marc
May 20 '18 at 13:23
• every group of prime order is cyclic May 20 '18 at 14:07