# A Homomorphism from the Additive Group of Integers to Itself ## Problem 163

Let $\Z$ be the additive group of integers. Let $f: \Z \to \Z$ be a group homomorphism.
Then show that there exists an integer $a$ such that
$f(n)=an$ for any integer $n$. Add to solve later

## Hint.

Let us first recall the definition of a group homomorphism.
A group homomorphism from a group $G$ to a group $H$ is a map $f:G \to H$ such that we have
$f(gg’)=f(g)f(g’)$ for any elements $g, g\in G$.

If the group operations for groups $G$ and $H$ are written additively, then a group homomorphism $f:G\to H$ is a map such that
$f(g+g’)=f(g)+f(g’)$ for any elements $g, g’ \in G$.

Here is a hint for the problem.
For any integer $n$, write it as
$n=1+1+\cdots+1$ and compute $f(n)$ using the property of a homomorphism.

## Proof.

Let us put $a:=f(1)\in \Z$. Then for any integer $n$, writing
$n=1+1+\cdots+1,$ we have
\begin{align*}
f(n)&=f(1+1+\cdots+1)\\
&=f(1)+f(1)+\cdots+f(1) \quad \text{ since } f \text{ is a homomorphism}\\
&=a+a+\cdots+a\\
&=an.
\end{align*}
Thus we have $f(n)=an$ with $a=f(1)\in \Z$ as required. Add to solve later

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