# 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 $\phi$.

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

### More from my site

• Group of $p$-Power Roots of 1 is Isomorphic to a Proper Quotient of Itself Let $p$ be a prime number. Let $G=\{z\in \C \mid z^{p^n}=1\}$ be the group of $p$-power roots of $1$ in $\C$. Show that the map $\Psi:G\to G$ mapping $z$ to $z^p$ is a surjective homomorphism. Also deduce from this that $G$ is isomorphic to a proper quotient of $G$ […]
• A Group Homomorphism that Factors though Another Group Let $G, H, K$ be groups. Let $f:G\to K$ be a group homomorphism and let $\pi:G\to H$ be a surjective group homomorphism such that the kernel of $\pi$ is included in the kernel of $f$: $\ker(\pi) \subset \ker(f)$. Define a map $\bar{f}:H\to K$ as follows. For each […]
• Subgroup of Finite Index Contains a Normal Subgroup of Finite Index Let $G$ be a group and let $H$ be a subgroup of finite index. Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.   Proof. The group $G$ acts on the set of left cosets $G/H$ by left multiplication. Hence […]
• Surjective Group Homomorphism to $\Z$ and Direct Product of Abelian Groups Let $G$ be an abelian group and let $f: G\to \Z$ be a surjective group homomorphism. Prove that we have an isomorphism of groups: $G \cong \ker(f)\times \Z.$   Proof. Since $f:G\to \Z$ is surjective, there exists an element $a\in G$ such […]
• Group Homomorphisms From Group of Order 21 to Group of Order 49 Let $G$ be a finite group of order $21$ and let $K$ be a finite group of order $49$. Suppose that $G$ does not have a normal subgroup of order $3$. Then determine all group homomorphisms from $G$ to $K$.   Proof. Let $e$ be the identity element of the group […]
• Abelian Normal subgroup, Quotient Group, and Automorphism Group 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 […]
• Cyclic Group if and only if There Exists a Surjective Group Homomorphism From $\Z$ Show that a group $G$ is cyclic if and only if there exists a surjective group homomorphism from the additive group $\Z$ of integers to the group $G$.   Proof. $(\implies)$: If $G$ is cyclic, then there exists a surjective homomorhpism from $\Z$ Suppose that $G$ is […]
• A Group Homomorphism is Injective if and only if Monic Let $f:G\to G'$ be a group homomorphism. We say that $f$ is monic whenever we have $fg_1=fg_2$, where $g_1:K\to G$ and $g_2:K \to G$ are group homomorphisms for some group $K$, we have $g_1=g_2$. Then prove that a group homomorphism $f: G \to G'$ is injective if and only if it is […]

#### You may also like...

##### Example of an Infinite Group Whose Elements Have Finite Orders

Is it possible that each element of an infinite group has a finite order? If so, give an example. Otherwise,...

Close