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I'm trying an old problem here:

Suppose $n\mid m$, and I have a natural ring homomorphism $\varphi\colon \mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ defined by $\varphi(j)=j\mod{n}$. I can verify that this is a ring homomorphism, but why does it induce a surjective group homomorphism on the unit groups $(\mathbb{Z}/m\mathbb{Z})^\times\to(\mathbb{Z}/n\mathbb{Z})^\times$?

My question in particular is why is it surjective? I take $k\in(\mathbb{Z}/n\mathbb{Z})^\times$. Then there exist integers $s,t$ such that $sn+tk=1$. Since $n\mid m$, I can also write $m=na$. So multiplying through I get $sm+tka=a$. I'm lost here. How can I find a unit in $(\mathbb{Z}/m\mathbb{Z})^\times$ which maps to $k$ to show surjectivity?

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In general, even if $k$ is a unit modulo $n$ you need not have that $k$ is a unit modulo $m$; e.g., $m=6$, $n=3$, $k=2$. – Arturo Magidin Oct 7 '11 at 21:34
@Henning: Yes: the point is that Calvin seems to be trying to show that if $k$ is a unit modulo $n$, then it will have to be a unit modulo $m$ (at least, up to the point where he says he is "lost"), and that is false. Instead, he needs to show that there exists an $r$ such that $\gcd(k+nr,m)=1$; or else reduce to a case where it is true, like Rolando suggests. – Arturo Magidin Oct 7 '11 at 22:05
@Arturo, I think the part leading up to "I'm lost here" is just meant to be "here are some random deductions I've made that turn out to lead nowhere" -- i.e., showing that he's done some work, which is a good thing and should be encouraged. They don't seem to aim anywhere in particular, but are just the most immediate conclusions one can draw from the available assumptions. Trying that is certainly a good initial plan in general. In particular: given the explicit wish to "find a unit which maps to $k$", I don't see any implication that he expects $k$ itself to be that unit. – Henning Makholm Oct 7 '11 at 22:35
@Henning: I don't necessarily disagree; however, what my first comment is meant to show is that that line of attack cannot succeed, so that it was not merely that Calvin didn't get anywhere: that path doesn't lead anywhere. Perhaps I should have been clearer as to what my comment was meant to be doing (I was certainly not claiming the desired theorem is false). – Arturo Magidin Oct 8 '11 at 2:46
@Arturo: OK; perhaps I was too thin-skinned. Sorry. – Henning Makholm Oct 8 '11 at 2:50

First reduce to the case where $n$ and $m$ are both powers of a prime $p$, using the Chinese remainder theorem. After this reduction the result is clear since the $p$-adic valuation of $n$ is lesser than the $p$-adic valuation of $m$ for any prime $p$.

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If you want a "pedestrian" solution without knowledge in group theory or number theory, you have to note that when $k$ is your unit, then $k+ln$ are also units, so you have several candidates for the unit in the bigger group and you only have to plug them into your equation and show that one of them gives a coefficient of $n$ divisible by $a$.

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