Finding a normal subgroup $H$ of $\mathbb{Z}_{mn}$ of order $m$

Find a normal subgroup $H$ of $\mathbb{Z}_{mn}$ of order $m$ where $m$ and $n$ are positive integers. Show that $H$ is isomorphic to $\mathbb{Z}_{m}$.

I am honestly not even sure where to start. My initial thoughts were if $\mathbb{Z}_{mn}$ was isomorphic to $\mathbb{Z}_{m}\times \mathbb{Z}_{n}$ then I could find a subgroup $H$ from that group. However, I discovered that $\mathbb{Z}_{mn}$ is isomorphic to $\mathbb{Z}_{m}\times \mathbb{Z}_{n}$ but the converse is not true. Would $H$ look like a cyclic group of order $m$ with an arbitrary generator $a$?

Any help would be great.

• Suppose $x$ is a generator for $\Bbb Z_{mn}$. What can you say about the order of $nx$ (Here, $nx = x + x + \cdots + x$ ($n$ times))? – David Wheeler Mar 4 '15 at 7:10

$\mathbb Z_{mn}$ is cyclic has an element of order $mn$ say $x$ Then $nx$ has order $m$.Let $H=\langle nx\rangle$