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Define $\mathrm{E}(\mathbb{Z}_{i})$ to be the group of invertible elements of the ring with unity $\mathbb{Z}_{i}$.

Show that $\mathrm{E}(\mathbb{Z}_{mn})$ is isomorphic to $\mathrm{E}(\mathbb{Z}_{m}) \times \mathrm{E}(\mathbb{Z}_{n})$ if and only if $m$ and $n$ are relatively prime.

I see that it is relatively easy to prove that $\mathbb{Z}_{mn}$ is isomorphic to $\mathbb{Z}_{m}\times \mathbb{Z}_{n}$ iff $m$ and $n$ are relatively prime using the Chinese Remainder Theorem. However, is it possible to make an extension of this theorem to prove the above statement?

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    The theorem gives you one direction; for the other, have you tried counting the number of elements in $E(\mathbb{Z}_{mn})$ and in $E(\mathbb{Z}_m)\times E(\mathbb{Z}_n)$ when $m$ and $n$ are not relatively prime?2011-01-03
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    Should be $E(\mathbb{Z}_n)$ in the second line.2011-01-03
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    Thank you Arturo, I see they are different and I can use a short proof by contradiction to show this in the other direction. Is it possible to use this theorem directly to prove the theorem in the first direction?2011-01-03
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    Yes: just show that if $R$ and $S$ are rings, then the units of $R\times S$ are exactly the products of units of $R$ and units of $S$. Also, if you address the comment (start it with `@whoever`) then the recipient gets a notification. If it's your question or your answer you get notified anyway, but not otherwise. Finally, if you figure it out, consider writing it up as an answer; that way others can comment and make suggestions, and eventually you can accept your own answer.2011-01-03

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