Tor one of two cyclic abelian groups is cyclic of gcd order

By [L1] with $M=\mathbb{Z}\mathrm{/}n$, the balanced Tor group is canonically the kernel of multiplication by $m$ on $\mathbb{Z}\mathrm{/}n$. Both cyclic modules have the explicit two-term projective resolutions supplied by [L1], so its supplied-resolution hypotheses are met. AC is used here through its comparison clause, not in the following arithmetic.

Put $g=\gcd (m,n)$, $m=ga$, $n=gb$, with $g,b>0$. By [L2], $\gcd (a,b)=1$ and $mx\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$ is equivalent to $b\mid ax$. By [L3] choose integers $u,v$ with $ua+vb=1$. If $b\mid ax$, then $x=uax+vbx$ is divisible by $b$; the converse is immediate. Hence the kernel in step 1.1 consists precisely of the multiples of $b$ modulo $n=gb$.

The map $\mathbb{Z}\mathrm{/}g\to \mathbb{Z}\mathrm{/}n$, $[t{]}_g\mapsto [bt{]}_n$, is well defined and additive since $bg=n$. Its image is the kernel from step 2.1. It is injective because $gb\mid bt$ holds exactly when $g\mid t$, by cancellation of the nonzero integer $b$. Thus it is an isomorphism onto that kernel, including $g=1$ when both groups in this identification are trivial. Composing with step 1.1 proves the claim.