# 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.

- 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 $\Aut(N)$.
- The assumption that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime implies that $G=\ker(\psi)$.
- 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 $G$.

Add to solve later

Sponsored Links