Zm × Zn is isomorphic to Zmn iff m and n are coprime

Dependencies:

$\mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn} \iff \gcd(m, n) = 1$

Proof

Proof of 'only if' part

$\mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn} \implies \mathbb{Z}_m \times \mathbb{Z}_n$ is cyclic.

Let $(a, b)$ be a generator of $\mathbb{Z}_m \times \mathbb{Z}_n$. Then $\operatorname{order}((a, b)) = mn$.

Let $o_a = \operatorname{order}(a)$ and $o_b = \operatorname{order}(b)$.

\begin{align} & o_a \mid m \wedge o_b \mid n \tag{$\mathbb{Z}_m$ and $\mathbb{Z}_n$ are cyclic} \\ &\Rightarrow o_a, o_b \mid \operatorname{lcm}(m, n) \\ &\Rightarrow \operatorname{lcm}(m, n) \textrm{ is a common multiple of } o_a \textrm{ and } o_b \\ &\Rightarrow \operatorname{lcm}(o_a, o_b) \le \operatorname{lcm}(m, n) \end{align}

$$\operatorname{order}((a, b)) = \operatorname{lcm}(o_a, o_b) \le \operatorname{lcm}(m, n) = \frac{mn}{\gcd(m, n)}$$

Therefore, $\gcd(m, n) = 1$.

Proof of 'if' part

Let $\gcd(m, n) = 1$.

Let $a$ and $b$ be generators of $\mathbb{Z}_m$ and $\mathbb{Z}_n$ respectively, $o_a = m$ and $o_b = n$.

\begin{align} & \operatorname{order}((a, b)) = \operatorname{lcm}(o_a, o_b) = \operatorname{lcm}(m, n) = \frac{mn}{\gcd(m, n)} = mn \\ &\Rightarrow \mathbb{Z}_m \times \mathbb{Z}_n = \langle (a, b) \rangle \wedge |\mathbb{Z}_m \times \mathbb{Z}_n| = mn \\ &\Rightarrow \mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn} \end{align}

Dependency for: None

Info:

Depth: 9
Number of transitive dependencies: 32