Let $m, n \in \mathbb{N}, m, n > 1$ and we have the short exact sequence of $\mathbb{Z}$ modules (as is easily seen) $0 \rightarrow \mathbb{Z}_{m} \xrightarrow{\alpha} \mathbb{Z}_{mn} \xrightarrow {\beta} \mathbb{Z}_n \rightarrow 0$, where $\alpha (\bar{k}) = \hat{nk}, \beta (\hat{l}) = \bar{\bar{l}}$. I'm trying to prove that it is split iff $(m,n)=1$.
If $(m,n) = 1$ it follows easily that $\mathbb{Z}_{mn} = n\mathbb{Z}_{mn} \oplus m\mathbb{Z}_{mn}$ and I'm done since $ker(\beta) = Im (\alpha) = n \mathbb{Z}_{mn}$. I'm having trouble with the other implication:
Supposing $\mathbb{Z}_{mn} = n\mathbb{Z}_{mn} \oplus X$ for $X$ a subgroup of $\mathbb{Z}_{mn}$, then, writing $X = a\mathbb{Z}_{mn}$ with $0 < a < mn$ and using that $\hat{1} \in \mathbb{Z}_{mn} = n\mathbb{Z}_{mn} + a \mathbb{Z}_{mn}$, I get that $(n,a) = 1$.
So I am left to see that $a = m$, and obviously I should use here that $a\mathbb{Z}_{mn} \cap n\mathbb{Z}_{mn} = \hat{0}$, but I don't know how. I'm trying to use things like $\hat{m} = \hat{ak} + \hat{nl}$ for some $\hat{k}, \hat{l}$ or like $\widehat{1 + m} = \hat{ak} + \hat{nl}$ for some other $\hat{k}, \hat{l}$, but these don't seem to help.
$\require{AMScd}$
Suppose $(n,m)=1$. Then it is easily checked that the following is an isomorphism of exact sequences: \begin{CD} 0@>>>\mathbb{Z}_m @>{\alpha}>> \mathbb Z_{nm} @>{\beta}>> \mathbb Z_{n} @>>> 0\\ & @V{n}VV @V{f}VV @V{\text{id}}VV\\ 0@>>>\mathbb{Z}_m @>{p}>> \mathbb Z_{n}\oplus\mathbb Z_m @>{q}>> \mathbb Z_{n} @>>> 0 \end{CD}
Here, $p$ and $q$ are the canonical homomorphisms and $f:\mathbb Z_{nm}\to\mathbb Z_n\oplus\mathbb Z_m:\overline a\mapsto(\overline a,\overline a)$. Coprimeness is used to say $n:\mathbb Z_m\to\mathbb Z_m$ is an isomorphism.
Conversely, if the exact sequence is split, then we need an isomorphism $\mathbb Z_{nm}\cong\mathbb Z_n\oplus\mathbb Z_m$, which means $(n,m)=1$.