# Abelian Groups and Surjective Group Homomorphism

## Problem 167

Let $G, G’$ be groups. Suppose that we have a surjective group homomorphism $f:G\to G’$.
Show that if $G$ is an abelian group, then so is $G’$.

Contents

## Definitions.

Recall the relevant definitions.

• A group homomorphism $f:G\to G’$ is a map from $G$ to $G’$ satisfying
$f(xy)=f(x)f(y)$ for any $x, y \in G$.
• A map $f:G \to G’$ is called surjective if for any $a\in G’$, there exists $x\in G$ such that
$f(x)=a.$
• A surjective group homomorphism is a group homomorphism which is surjective.

## Proof.

Let $a, b\in G’$ be arbitrary two elements in $G’$. Our goal is to show that $ab=ba$.
Since the group homomorphism $f$ is surjective, there exists $x, y \in G$ such that
$f(x)=a, f(y)=b.$

Now we have
\begin{align*}
ab&=f(x) f(y)\\
&=f(xy) \text{ since } f \text{ is a group homomorphism}\\
&=f(yx) \text{ since } G \text{ is an abelian group}\\
&=f(y)f(x) \text{ since } f \text{ is a group homomorphism}\\
&=ba.
\end{align*}
Therefore, we obtain $ab=ba$ for any two elements in $G’$, thus $G’$ is an abelian group.

Let $\Z$ be the additive group of integers. Let $f: \Z \to \Z$ be a group homomorphism. Then show that...