$\text{order }a\otimes b=\gcd\left(\text{order }a,\text{order }b\right)$? Let $A,B$ be abelian groups with $a\in A$ having order $m$ and
$b\in B$ having order $n$. 
Then $m\left(a\otimes b\right)=\left(ma\right)\otimes b=0\otimes b=0$
and likewise $n\left(a\otimes b\right)=a\otimes\left(nb\right)=a\otimes0=0$.
This shows that $a\otimes b$ has finite order $k$ that divides $m$
and $n$, or equivalently $k$ divides $\gcd\left(m,n\right)$.
My question is: 

Can it also be shown that $k$ equalizes $\gcd\left(m,n\right)$?

 A: The key result is: If $\sum_i{a_i}\otimes{b_i}=0$ in $A\otimes{B}$, there exist finitely generated abelian groups $A_0\subseteq{A}$ and $B_0\subseteq{B}$ such that: $a_i\in{A_0}$, $b_i\in{B_0}$ and $\sum_i{a_i}\otimes{b_i}=0$ in $A_0\otimes{B_0}$. Hence you may apply the structure theorems of finitely generated abelian groups to prove that $k=\mathrm{gcd}\{m,n\}$.
A: We can generalize this result by proving that
$\mathbb{Z}/m\mathbb{Z}\bigotimes_{\mathbb{Z}} \mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}/d\mathbb{Z}$ where $d$ is gcd($m,n$)
Observe first that 
$a\bigotimes b=a\bigotimes (b.1)=(ab)\bigotimes 1=ab(1\bigotimes 1)$. Consequently, $\mathbb{Z}/m\mathbb{Z}\bigotimes_{\mathbb{Z}} \mathbb{Z}/n\mathbb{Z}$ is a cyclic group with $1\bigotimes 1$ as generator.
Since $m(1\bigotimes 1)=m\bigotimes 1=0\bigotimes 1=0$  and $n(1\bigotimes 1)=1\bigotimes n=1\bigotimes 1=0$, then this cyclic group has order dividing $d$.
Consider the map $\varphi :\mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}\rightarrow \mathbb{Z}/d$ defined by $\varphi (a,b)=ab$ .this map is well-defined(easy to check). Evidenly $\varphi$ is $\mathbb{Z}$-linear. 
By the universal property of tensor, it induces a $R$-module homorphism $\Phi :\mathbb{Z}/m\mathbb{Z}\bigotimes \mathbb{Z}/n\mathbb{Z}\rightarrow \mathbb{Z}/d\mathbb{Z}$ where $\phi(a\bigotimes b)=ab$.
Consider a $R$-module homorphism map defined as follows
$\Psi :\mathbb{Z}/d\mathbb{Z}\rightarrow \mathbb{Z}/m\mathbb{Z}\bigotimes \mathbb{Z}/n\mathbb{Z}$
$\Psi(a)=a(1\bigotimes 1)$
Clearly, $\Phi \Psi =Id_{\mathbb{Z}/d\mathbb{Z}}$, and $\Psi \Phi =Id_{\mathbb{Z}/m\mathbb{Z}\bigotimes \mathbb{Z}/n\mathbb{Z}}$
Hence, $\mathbb{Z}/m\mathbb{Z}\bigotimes_{\mathbb{Z}} \mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}/d\mathbb{Z}$ where $d$ is gcd($m,n$) from which implies what you claim.
