I'm trying to prove that if $m,n$ are relatively prime, $\mathbb{Z}/m \times \mathbb{Z}/n \cong \mathbb{Z}/mn$ as groups under addition. Here is my attempt. I'm going to use multiplicative notation, but I acknowledge that the groups are abelian and therefore it's more natural to use additive notation. I'm going to use the below two lemmas without proof.
Lemma 1: Any two finite cyclic groups of order $k$ are isomorphic.
Lemma 2: If $a \in G$ is an element of order $n$ and $a^k = e$, then $n \mid k$.
Consider the element $(1,1) \in \mathbb{Z}/m \times \mathbb{Z}/n$. If $(1,1)^k = (0,0)$, then $1^k = 0$ in $\mathbb{Z}/m$ and $1^k = 0$ in $\mathbb{Z}/n$; in $\mathbb{Z}/m$, $|1| = m$ and in $\mathbb{Z}/n$, $|1| = n$. By Lemma 2, $m \mid k$ and $n \mid k$. As $m$ and $n$ are relatively prime, we have $mn \mid k$.
I'm trying to show that $|(1,1)| = mn$, which would require that $k \mid mn$. I'm not sure how to deduce this from the fact that $m$ and $n$ are relatively prime. If I prove this, then $(1,1)$ has the same order as $\mathbb{Z}/m \times \mathbb{Z}/n$, so it generates the group. But $\mathbb{Z}/mn$ is also cyclic of order $mn$ generated by $1$, so by the lemma, these two cyclic groups are isomorphic.
My questions are:
How do I prove that $mn = k$?
Is there a different, more natural proof strategy for this fact, or is this the standard proof?
