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.
Best Answer
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 $m$.