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Here's my question

Using the Third Isomorphism Theorem, show that if m, n are positive integers then there is an isomorphism:

$\Bbb Z_m \cong \Bbb Z_{mn}/\Bbb Z_{n}$

I began this by assuming $m \ge n$ so I can apply the Third Isomorphism Theorem to see that:

  1. $\Bbb Z_m / \Bbb Z_n $ is a normal subgroup of $\Bbb Z_{mn} / \Bbb Z_n$
  2. $(\Bbb Z_{mn} / \Bbb Z_n)(\Bbb Z_m / \Bbb Z_n) $ is isomorphic to $\Bbb Z_{mn} / \Bbb Z_n$

As $\Bbb Z_n \subseteq \Bbb Z_m \subseteq \Bbb Z_{mn}$

But I don't even know where to begin proving this, any help would be greatly appreciated!

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    $\begingroup$ I think you have a typo. Do you mean $\mathbb{Z}_{m} \cong \mathbb{Z}_{mn}/\mathbb{Z}_{n}$? $\endgroup$ Commented Dec 5, 2014 at 1:08
  • $\begingroup$ I do, thank you! I've edited my post. $\endgroup$
    – tom982
    Commented Dec 5, 2014 at 2:01
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    $\begingroup$ Your first point seems fishy considering you don't even necessarily have $\mathbb{Z}_n \leq \mathbb{Z}_m$, let alone $\mathbb{Z}_n \unlhd \mathbb{Z}_m$. E.g., is $\mathbb{Z}_3 \leq \mathbb{Z}_5$? $\endgroup$
    – syusim
    Commented Dec 5, 2014 at 2:10
  • $\begingroup$ My mistake, I've removed that. Thank you for your reply! $\endgroup$
    – tom982
    Commented Dec 5, 2014 at 2:23

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