# Normal subgroup of prime order in the center

Problem: If $$N$$ is a normal subgroup of order $$p$$ where $$p$$ is the smallest prime dividing the order of a finite group $$G$$, then $$N$$ is in the center of $$G$$.

Solution: Since $$N$$ is normal, we can choose for $$G$$ to act on $$N$$ by conjugation. This implies that there is a homomorphism from $$G$$ to the automorphism group of $$N$$, which has $$p - 1$$ elements. Thus the homomorphism is trivial and $$N$$ is in the center of $$G$$

My first question is why conjugation implies the automorphism group.

My second question is why the automorphism only has $$p-1$$ elements; i.e. why is conjugation by the identity excluded even though it's a valid conjugation.

• Don’t understand your first question. Second question: the cyclic group of order $p$ has $p-1$ automorphisms, including the identity autom. Oct 11 '15 at 5:08
• Indeed, the statement is purely about cyclic groups (if it helps psychologically, it happens to have prime order). Remember that isomorphisms are completely determined by where generators get sent. Oct 11 '15 at 5:12
• @Lubin I mean how do we know that there exists a homomorphism from $G$ to the automorphism group of $N$, and not just a regular non-homomorphism action/function from $G$ to the automorphism group of $N$? Oct 11 '15 at 5:16
• The fact that $N$ is normal is exactly the condition that conjugating an element of $N$ gives another element of $N$. You need that in order to have $G$ acting (well-defined) by conjugation on $N$. Oct 11 '15 at 5:17
• It’s homomorphism to the automorphism group of your $N$ because you prove that. It’s easy. Oct 11 '15 at 18:28

In general, an action of a group $G$ on a set $X$ is equivalent to a homomorphism $\varphi: G \to \text{Sym}(X)$, where $\text{Sym}(X)$ is the set of all permutations of $X$, i.e., bijections $X \to X$. (This is called a permutation representation; see here for more.) In this problem, the set $N$ (on which $G$ acts) has the structure of a group, and the bijections induced by elements of $G$ happen to also be automorphisms. Concretely, given $g \in G$, then the induced automorphism is just \begin{align*} \varphi_g : N &\to N\\ n &\mapsto g n g^{-1} \, . \end{align*} (This is called an inner automorphism.)
As pointed out in the comments, $\text{Aut}(\mathbb{Z}/m\mathbb{Z}) \cong (\mathbb{Z}/m\mathbb{Z})^\times$, the set of units, for any $m \in \mathbb{Z}_{>0}$. These are exactly the cosets that are represented by an element in $\{0, \ldots, m-1\}$ that is relatively prime to $m$.
• So using $\text{Sym}(X)$ instead, if $G$ acts on $N$ by conjugation, and conjugation of $N$ by an element of $G$ constitutes an automorphism of $N$, this action induces a homomorphism of $\phi: G \to \text{Sym}(N)$, and $\text{Sym}(N)$ has $p!$ elements. But it's actually supposed to be $(p-1)!$ elements I think (in order for the proof to work). What's going on in this case? Oct 11 '15 at 5:24
• As I mentioned in the answer, the elements of $\text{Sym}(N)$ induced by elements of $G$ are actually automorphisms, not mere bijections, so the image of the homomorphism $\varphi$ is contained in the subgroup $\text{Aut}(N) \leq \text{Sym}(N)$. By the argument in my second paragraph, the group $\text{Aut}(N)$ has order $p-1$. Oct 11 '15 at 5:31