Solution: We are given that $n \mid m$.
-
Let us prove that the map is well defined. That is, if $a + m\mathbb{Z} = b + m\mathbb{Z} $, then $a+n\mathbb{Z} = b + n\mathbb{Z} $. Equivalently, if $m \mid (b-a)$ then $n \mid (b-a)$. Since
\[
m \mid (b-a) \text{ and } n \mid m \implies n \mid (b-a).
\]
Thus, the map is well defined.
We now prove that $\varphi $ is a homomorphism. Before that we recall that the operation on the quotient group $\mathbb{Z} /n \mathbb{Z} $ is
\[
(a + n\mathbb{Z}) + (b + n\mathbb{Z}) = (a + b) + n\mathbb{Z} .
\]
Consider
\begin{align*}
\varphi \left( (a + m\mathbb{Z}) + (b + m\mathbb{Z} ) \right) & = \varphi \left( (a+b) + m \mathbb{Z} \right) \\
& = (a + b) + n\mathbb{Z} \\
& = (a + n\mathbb{Z} ) + (b + n\mathbb{Z} ) \\
& = \varphi (a + m\mathbb{Z} ) + \varphi (b + m\mathbb{Z} ).
\end{align*}
Thus, the map is well defined.
Now let us prove that $\varphi $ is surjective. Let $y + n\mathbb{Z} \in \mathbb{Z} /n\mathbb{Z} $. Then $y + m\mathbb{Z} \in \mathbb{Z} /m\mathbb{Z} $ and
\[
\varphi (y + m\mathbb{Z} ) = y + n\mathbb{Z} .
\]
-
Let us first find the kernel of the homomorphism.
\begin{align*}
\ker \varphi & = \left\{ a + m\mathbb{Z} : \varphi (a + m\mathbb{Z} ) = 0 \right\} \\
& = \left\{ a + m\mathbb{Z} : a + n \mathbb{Z} = 0 \right\} \\
& = \left\{ a + m \mathbb{Z} : n \mid a \right\}.
\end{align*}
Let us simplify the kernel that we obtained. If $a + m\mathbb{Z} \in \ker \varphi $, then $n \mid a$, that is, $a = nk$ for some $k \in \mathbb{Z} $. So, $a + m\mathbb{Z} = nk + m\mathbb{Z}$. As $n \mid m$, we can also write $m = nl$ for some $l \in \mathbb{Z} $. Thus we obtained
\[
a + m\mathbb{Z} = nk + nl \mathbb{Z} .
\]
Thus,
\[
\ker \varphi = \left\{ nk + nl \mathbb{Z} : k = 0,1,\dots, l-1 \right\}.
\]
So, the order of $\ker \varphi $ is $l$ and being a subgroup of a cyclic group it must be cyclic. Thus,
\[
\ker \varphi \cong \mathbb{Z} /l\mathbb{Z} .
\]