Show that as rings $\Bbb Z_{mn} \ncong \Bbb Z_m \times \Bbb Z_n$ if $\gcd(m,n) \neq 1$

Question

So I just got done showing explicitly that an isomorphism exists between these two rings if the $\gcd(m, n) = 1$, and I did not have much trouble with that. For some reason I'm having a lot harder of a time showing that the result is not true if $m$ and $n$ are not relatively prime. Can somebody help me out here? Thanks.

Answer 1

If $m$ and $n$ are not relatively prime then they have a least common multiple $L$ which is less than $mn$. (Namely $\frac {mn}{\gcd(n,m)}$)

And if you take any element of $(a,b)\in \mathbb Z_m\times \mathbb Z_n$ and add it to itself $L$ times you get the identity[1]. So there is no element with order $mn$. So it can not be isomorphic to $\mathbb Z_{mn}$ as $1\in \mathbb Z_{mn}$ has order $mn$.

.....

[1] If $a \in \mathbb Z_m$ then $(km)*a = 0$ for all integers $k$ and if $b \in \mathbb Z_n$ then $(kn)*b = 0$ for all integers $k$. So if $k = L = m*\frac n{\gcd(n,m)} = n*\frac m{\gcd(m,n)}$ then $L*(a,b) = (0,0)$.

Answer 2

Hint Show that $\mbox{lcm} (m,n) \cdot (x,y) =(0,0)$ for all elements $(x,y) \in \mathbb Z_m \times \mathbb Z_n$.