# Abelian Normal subgroup, Quotient Group, and Automorphism Group

## Problem 343

Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$.
Let $\Aut(N)$ be the group of automorphisms of $G$.

Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime.
Then prove that $N$ is contained in the center of $G$.

Contents

## Outline of the proof

Here is the outline of the proof.

1. Define a group homomorphism $\psi: G\to \Aut(N)$ by $\psi(g)(n)=gng^{-1}$ for all $g\in G$ and $n\in N$.
We need to check:
• The map $\psi(g)$ is an automorphism of $N$ for each $g\in G$.
• The map $\psi$ is in fact a group homomorphism from $G$ to $\Aut(N)$.
2. The assumption that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime implies that $G=\ker(\psi)$.
3. This implies that $N$ is in the center of $G$.

## Proof.

We define a group homomorphism $\psi: G \to \Aut(N)$ as follows.
For each $g\in G$, we first define an automorphism $\psi(g)$ of $N$.
Define $\psi(g): N \to N$ by
$\psi(g)(n)=gng^{-1}.$

Note that since $N$ is a normal subgroup of $G$, the output $\psi(g)(n)=gng^{-1}$ actually lies in $N$.

We prove that so defined $\psi(g)$ is a group homomorphism from $N$ to $N$ for each fixed $g\in G$.
For $n_1, n_2 \in N$, we have
\begin{align*}
\psi(g)(n_1n_2)&=g(n_1n_2)g^{-1} && \text{by definition of $\psi(g)$}\\
&=gn_1g^{-1}gn_2g^{-1} && \text{by inserting $e=g^{-1}g$}\\
&=\psi(g)(n_1) \psi(g)(n_2) && \text{by definition of $\psi(g)$}.
\end{align*}
It follows that $\psi(g)$ is a group homomorphism, and hence $\psi(g)\in \Aut(N)$.

We have defined a map $\psi:G\to \Aut(N)$. We now prove that $\psi$ is a group homomorphism.
For any $g_1, g_2$, and $n\in N$, we have
\begin{align*}
\psi(g_1 g_2)(n)&=(g_1g_2)n(g_1 g_2)^{-1}\\
&=g_1 g_2 n g_2^{-1} g_1^{-1}\\
&=g_1 \psi(g_2)(n) g_1^{-1}\\
&=\psi(g_1)\psi(g_2)(n).
\end{align*}

Thus, $\psi: G\to \Aut(N)$ is a group homomorphism.
By the first isomorphism theorem, we have
$G/\ker(\psi)\cong \im(\psi)< \Aut(N). \tag{*}$ Note that if $g\in N$, then $\psi(g)(n)=gng^{-1}=n$ since $N$ is abelian. It yields that the subgroup $N$ is in the kernel $\ker(\psi)$.

Then by the third isomorphism theorem, we have
\begin{align*}
G/\ker(\psi) \cong (G/N)/(\ker(\psi)/N). \tag{**}
\end{align*}

It follows from (*) and (**) that the order of $G/\ker(\psi)$ divides both the order of $\Aut(N)$ and the order of $G/N$. Since the orders of the latter two groups are relatively prime by assumption, the order of $G/\ker(\psi)$ must be $1$. Thus the quotient group is trivial and we have
$G=\ker(\psi).$

This means that for any $g\in G$, the automorphism $\psi(g)$ is the identity automorphism of $N$.
Thus, for any $g\in G$ and $n\in N$, we have $\psi(g)(n)=n$, and thus $gng^{-1}=n$.
As a result, the subgroup $N$ is contained in the center of $G$.

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