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I know this is a duplicate, but I don't know ring theory or the Chinese remainder theorem, so I couldn't understand the answers of the duplicate.

I tried to prove it by constructing the isomorphism $\phi: x \mapsto (x \mod m, \ x \mod n)$. I already proved that this maps $U_{mn}$ to $U_m \times U_n$ and that it is injective, but I'm stuck on proving that it is surjective, or that $|U_{mn}| = |U_m \times U_n|$.

user388557
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    What is $R_m$? Do you mean $\mathbb Z_m$? – rschwieb Sep 28 '17 at 16:10
  • @rschwieb no, $R_m$ is the set of all integers between 1 and m that are coprime with $m$ under multiplication modulo m – user388557 Sep 28 '17 at 16:12
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    In that case, a more standard notation would be $U_m$ or $(\Bbb Z/m\Bbb Z)^\times$ – Prasun Biswas Sep 28 '17 at 16:14
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    I suppose this is the duplicate you're alluding to. Let's make a note here that you're specifically saying you're stuck on surjectivity, so that people notice it's not a duplicate. – rschwieb Sep 28 '17 at 16:15
  • An alternative indirect way to do it would be to prove $\mathbb Z_{mn}\cong \mathbb Z_m\times \mathbb Z_n$ are isomorphic rings (when $m$ and $n$ are coprime), and then automatically their groups of units are isomorphic. – rschwieb Sep 28 '17 at 16:23
  • This was asked a long time ago, but I hope this helps as it is the simplest way:

    You know that:

    $|U(m)\times U(n)| = |\phi(m)\times\phi(n)| = |\phi(mn)| = |U(mn)|$

    Note that $\phi$ is the totient function here and not the name of your function. The above equation means *order of the domain = order of co-domain*

    You've proven that the mapping is injective, which means *order of domain = order of the range*.

    So you know *order of range = order of co-domain*, which is the definition of surjectivity.

     – sanketalekar Jul 21 '22 at 23:41

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Hint for surjectivity: Remember (or look up) Bezout's lemma.

Lukas Heger
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  • I don't see how we can apply it to anything else than $m$ and $n$, so there exist integers $x, y$ such that $mx + ny = 1$. Then we want to write an element $(a, b) \in R_m \times R_n$ as $(z \mod m, z \mod n)$. Multiplying both sides of Bezout's lemma by $a$ gives $amx + any = a$. Either $x$ or $y$ is greater than 0, assume it's $x$. Then $amx \mod n = a$. So $n | (amx - a)$. I don't see how to proceed. Can I get another hint? – user388557 Sep 28 '17 at 16:36
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    When you have $mx+ny=1$ and you're given $(a,b) \in R_m \times R_n$, consider the integer $bmx+any$. – Lukas Heger Sep 28 '17 at 17:26
  • So something like this?: let $z = bmx + any$. Then $z \mod m = any \mod m$. But $any - a = a(ny - 1) = a(mx)$, so $m | any - a$, so $a = any \mod m = z \mod m$. In the same manner $b = z \mod n$. Suppose $z$ is not coprime with $mn$, then it has a factor $d$ that either divides $m$ or $n$ but not both since $m, n$ are coprime. Assume $d | m$ without loss of generality. Then $d | any$, so it divides $ay$. I still don't get how this integer is coprime with $mn$. – user388557 Sep 28 '17 at 18:41
  • I'm not sure how to prove this directly, but you can use the fact that $\gcd(z,m)=1$ and $\gcd(z,n)=1$ implies that $\gcd(z,mn)=1$. Do you see why $z$ is coprime to $n$ and $m$? – Lukas Heger Sep 28 '17 at 19:16