# Group Homomorphism from $\Z/n\Z$ to $\Z/m\Z$ When $m$ Divides $n$

## Problem 613

Let $m$ and $n$ be positive integers such that $m \mid n$.

**(a)** Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined.

**(b)** Prove that $\phi$ is a group homomorphism.

**(c)** Prove that $\phi$ is surjective.

**(d)** Determine the group structure of the kernel of $\phi$.

Contents

## Proof.

### (a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined.

To show that $\phi$ is well-defined, we need show that the value of $\phi$ does not depends on the choice of representative $a$.

So suppose that $a+n\Z=a’+n\Z$ so that $a$ and $a’$ are two representatives for the same element.

This yields that $a-a’$ is divisible by $n$.

Now, $a+n\Z$ is mapped to $a+m\Z$ by $\phi$. On the other hand, $a’+n\Z$ is mapped to $a+m\Z$ by $\phi$.

Since $a-a’$ is divisible by $n$ and $m \mid n$, it follows that $a-a’$ is divisible by $m$.

This implies that $a+m\Z=a’+m\Z$.

This prove that $\phi$ does not depend on the choice of the representative, and hence $\phi$ is well-defined.

### (b) Prove that $\phi$ is a group homomorphism.

Let $a+n\Z$, $b+n\Z$ be two elements in $\Zmod{n}$. Then we have

\begin{align*}

&\phi\left(\, (a+n\Z)+(b+n\Z) \,\right)\\

&=\phi\left(\, (a+b)+n\Z) \,\right) &&\text{by addition in $\Zmod{n}$}\\

&=(a+b)+m\Z &&\text{by definition of $\phi$}\\

&=(a+m\Z)+(b+m\Z)&&\text{by addition in $\Zmod{m}$}\\

&=\phi(a+n\Z)+\phi(b+n\Z) &&\text{by definition of $\phi$}.

\end{align*}

Hence $\phi$ is a group homomorphism.

### (c) Prove that $\phi$ is surjective.

For any $c+m\Z \in \Zmod{m}$, we pick $c+n\Z\in \Zmod{n}$.

Then as $\phi(c+n\Z)=c+m\Z$, we see that $\phi$ is surjective.

### (d) Determine the group structure of the kernel of $\phi$.

If $a+n\Z\in \ker(\phi)$, then we have $0+m\Z=\phi(a+n\Z)=a+m\Z$.

This implies that $m\mid a$.

On the other hand, if $m\mid a$, then $\phi(a+n\Z)=a+m\Z=0+m\Z$ and $a+n\Z\in \ker(\phi)$.

It follows that

\[\ker(\phi)=\{mk+n\Z \mid k=0, 1, \dots, l-1\},\]
where $l$ is an integer such that $n=ml$.

Thus, $\ker(\phi)$ is a group of order $l$.

Since $\ker(\phi)$ is a subgroup of the cyclic group $\Zmod{n}$, we know that $\ker(\phi)$ is also cyclic.

Thus

\[\ker(\phi)\cong \Zmod{l}.\]

#### Another approach

Here is a more direct proof of this result.

Define a map $\psi:\Z\to \ker(\phi)$ by sending $k\in \Z$ to $mk+n\Z$.

It is straightforward to verify that $\psi$ is a surjective group homomorphism and the kernel of $\psi$ is $\ker(\psi)=l\Z$.

It follows from the first isomorphism theorem that

\[\Zmod{l}= \Z/\ker(\psi) \cong \im(\psi)=\ker(\phi). \]

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