Necessary and Sufficient Condition for Existence of Isomorphism between $\mathbb{Z}_{nm}$ and $\mathbb{Z}_n\bigoplus \mathbb{Z}_m$

Question

It is known that when $\gcd(n,m) = 1$, there exists an isomorphism between $\mathbb{Z}_{nm}$ and $\mathbb{Z}_n\bigoplus \mathbb{Z}_m$. This is the Chinese remainder theorem.

However, I am required to give a necessary and sufficient condition such that such an isomorphism exists. I conjectured that this is true if and only if $\gcd(n,m) = 1$, but I am not able to prove this. Is my claim correct?

Yes, you are correct. Prove it by contradiction by examining orders of elements. — Randall, Aug 24 '17 at 12:45
@Randall By orders of elements do you mean x^k for some integer k? — rr01, Aug 24 '17 at 12:53

Davood · Accepted Answer · 2017-08-26T19:38:11.960

Let $d:=\gcd(m,n)$ and let $D:=\dfrac{mn}{d}=\text{lcd}(m,n)$; notice that $m \mid D$ and $m \mid D$.

Then for every element $a=(b,c) \in \mathbb{Z}_n \bigoplus \mathbb{Z}_m$ we have: $$ \underbrace{(b,c)+ (b,c) + ... + (b,c)}_{D\text{-times}} = D (b,c)= (Db,Dc) = (0,0) ; $$ so every element of $ \ \ \mathbb{Z}_n \bigoplus \mathbb{Z}_m \ \ $ has order at most equal to $D:=\dfrac{mn}{d}=\text{lcd}(m,n)$.

On the otherhand we know that $ 1 \in \mathbb{Z}_{nm}$ has the order equal to $nm$.

Now suppose that $\mathbb{Z}_n \bigoplus \mathbb{Z}_m$ and $\mathbb{Z}_{nm}$ are isomorphic;
then $\mathbb{Z}_n \bigoplus \mathbb{Z}_m$ must have an element of order $mn$;
but we know that the order of every element of $\mathbb{Z}_n \bigoplus \mathbb{Z}_m$ is less than or equal to $$D:=\dfrac{mn}{d}=\text{lcd}(m,n) ; $$ which implies that $D=mn$; equivalently $d=1$.

Necessary and Sufficient Condition for Existence of Isomorphism between $\mathbb{Z}_{nm}$ and $\mathbb{Z}_n\bigoplus \mathbb{Z}_m$

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