If $(m,n)\neq 1$, prove $\mathbb{Z}_{mn} \not \cong \mathbb{Z}_{m} \times \mathbb{Z}_{n}$. I am really struggling with this problem. We just learned isomorphism in rings. We have not learned groups. From reading multiple sources, so far I have:

*

*Suppose $(m,n) = d > 1 \Longrightarrow \frac{m}{d}, \frac{n}{d}$ integers.

*Suppose $k = \textrm{lcm} (m,n)$, then $k = \frac{m n}{d}$ integer $\Longrightarrow m \vert k, n \vert k$.

How would I proceed from here?
 A: Hint: If $m,n$ are not relatively prime, then $k = \mathrm{lcm}(m,n)$ is strictly smaller than $mn$.

*

*Show that $k x= 0$ whenever $x \in \mathbb{Z}_m \times \mathbb{Z}_n$.

*Find an element $x \in \mathbb{Z}_{mn}$ such that $k x \neq 0$.

A: lets prove $mcd(m,n)=1 \Leftrightarrow \mathbb{Z}_{n} \times \mathbb{Z}_{m}  cyclic\Leftrightarrow\mathbb{Z}_{nm}  \cong \mathbb{Z}_{n} \times \mathbb{Z}_{m}$
(first part)($\Rightarrow$)
suppose $\mathbb{Z}_{n} \times \mathbb{Z}_{m}$  not cyclic  and $\mathbb{Z}_{nm}$ cyclic
for every morfism $$f:(\mathbb{Z}_{nm})\rightarrow\mathbb{Z}_{n} \times \mathbb{Z}_{m}$$
$$1 \rightarrow(g_n,g_m)$$
with $k=ord(g_n,g_m)$ $k\neq \rvert \mathbb{Z}_{n} \times \mathbb{Z}_{m}\rvert =mn$ but $f(k)=(kg_n,kg_m)=k(g_n,g_m)=e$
and $e\neq k$
so f is not inyective and $\mathbb{Z}_{nm} \not \cong \mathbb{Z}_{n} \times \mathbb{Z}_{m}$
(first part)($\Leftarrow$)
suppose $ord(\mathbb{Z}_{nm})= ord(\mathbb{Z}_{n} \times \mathbb{Z}_{m})=nm\Rightarrow \mathbb{Z}_{nm} \cong \mathbb{Z}_{n} \times \mathbb{Z}_{m}$
(second part)($\Rightarrow$)
$be (g_n,g_m)$ the generator of $\mathbb{Z}_{n} \times \mathbb{Z}_{m} \Rightarrow <g_n>=\mathbb{Z}_{n}$ and $ <g_m>=\mathbb{Z}_{m}$
$ ord((g_n,g_m))= min_{k\in\mathbb N}\{kg_n=0_{\mathbb{Z}_{n}},kg_m=0_{\mathbb{Z}_{m}}\}=min_{k\in\mathbb N}\{k\rvert n,k\rvert m\}=mcd(n,m)$
and $ ord((g_n,g_m))=\rvert \mathbb{Z}_{n} \times \mathbb{Z}_{m}\rvert =nm$
so $mcd(n,m)=\frac{nm}{mcm(n,m)}=1\Rightarrow n,m$ are coprimes
(second part)($\Leftarrow$)
n,m coprimes so $ord(1_{\mathbb{Z}_{n}},1_{\mathbb{Z}_{m}})=mcm(n,m)=mn$
A: We know that ${\mathbb Z}_{mn}$ is cyclic. So to show that
$$ {\mathbb Z}_{mn} \not\cong {\mathbb Z}_m \times {\mathbb Z}_n $$
it would be enough to show that ${\mathbb Z}_m \times {\mathbb Z}_n$ is not cyclic when $\gcd(m,n) \ne 1$.
Let $\ell=\text{lcm}(m,n)=mn/g$, where $g=\gcd(m,n)$. Thus, $g>1$ if and only if $\ell<mn$.
We show that every $(a,b) \in {\mathbb Z}_m \times {\mathbb Z}_n$ has order dividing $\ell$ when $g=\gcd(m,n)>1$.
Let $(a,b) \in {\mathbb Z}_m \times {\mathbb Z}_n$. Since $m \mid \ell$ and $n \mid \ell$, we have
$$ \ell (a,b) = (\ell a,\ell b) = (0,0). $$
This completes our claim, and the proof of non-isomorphism between ${\mathbb Z}_{mn}$ and ${\mathbb Z}_m \times {\mathbb Z}_n$ when $\gcd(m,n)>1$. $\blacksquare$
