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.

  • 1
    $\begingroup$ 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))? $\endgroup$ – 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 $

Now use any two finite cyclic groups are isomorphic

  • $\begingroup$ I apologize for not being clear but the end goal is to use the fundamental homomorphism theorem or the first isomorphism theorem. So it would be a better idea to construct a function that is a bijective homomorphism. However, I do not know how to construct such a function for this case. $\endgroup$ – Rick Strut Mar 4 '15 at 16:21
  • $\begingroup$ It looks like I'm just working with numbers mod mn here. So if I put x=1 and think about the set of numbers {0,n,2n,3n,...,(m-1)n}. I can see a correspondence with {0,1,2,3,...,(m-1)} but do not know how to make the function. $\endgroup$ – Rick Strut Mar 4 '15 at 16:23

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