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.

  • 2
    $\begingroup$ Don’t understand your first question. Second question: the cyclic group of order $p$ has $p-1$ automorphisms, including the identity autom. $\endgroup$
    – Lubin
    Oct 11 '15 at 5:08
  • 1
    $\begingroup$ 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. $\endgroup$
    – pjs36
    Oct 11 '15 at 5:12
  • $\begingroup$ @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 $N$? $\endgroup$ Oct 11 '15 at 5:16
  • 2
    $\begingroup$ 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 $N$. $\endgroup$
    – This Is Me
    Oct 11 '15 at 5:17
  • $\begingroup$ It’s homomorphism to the automorphism group of your $N$ because you prove that. It’s easy. $\endgroup$
    – Lubin
    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 $m$.

  • $\begingroup$ 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? $\endgroup$ Oct 11 '15 at 5:24
  • $\begingroup$ 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 $p-1$. $\endgroup$ Oct 11 '15 at 5:31

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