How do I compute the isomorphism class of $A\otimes_\mathbb{Z} B$, where $A$ and $B$ are abelian of finite order?
I can do this for a few examples, but I am unsure of how to proceed in the general case.
Specifically, I am interested in the cases where...
$|A|$ and $|B|$ are coprime
$\pi(|A|)\cap\pi(|B|)=1$ (where $\pi(n)$ denotes the set of prime divisors of $n$)
all Sylow subgroups of $A$ and $B$ are elementary abelian
Is there a way of seeing such results intuitively?
On
Suppose $A=\Bbb{Z}/m\Bbb{Z}$ and $B=\Bbb{Z}/n\Bbb{Z}$. If $\gcd(m,n)=1$ then $n$ is a unit in $A$ and it annihilates $B$. For any pure tensor $a\otimes b\in A\otimes_{\Bbb{Z}}B$ we have $$a\otimes b =(nn^{-1}a)\otimes b=(n^{-1}a)\otimes(nb)=(n^{-1}a)\otimes0=0.$$ As all pure tensors are trivial and $A\otimes_{\Bbb{Z}}B$ is generated by them, it follows that $A\otimes_{\Bbb{Z}}B=0$. Now use the fundamental theorem of finite abelian groups to extend this to finite abelian groups of coprime order.
More generally it is true that $$(\Bbb{Z}/m\Bbb{Z})\otimes_{\Bbb{Z}}(\Bbb{Z}/n\Bbb{Z})\cong\Bbb{Z}/\gcd(m,n)\Bbb{Z},$$ by using an argument similar to the case where $\gcd(m,n)=1$, and considering the map $$(\Bbb{Z}/m\Bbb{Z})\times\Bbb{Z}/n\Bbb{Z})\ \longrightarrow\ \Bbb{Z}/gcd(m,n)\Bbb{Z}:\ (a,b)\ \longmapsto\ a\cdot b.$$ Again using the fundamental theorem of finite abelian groups, this result extends in a nice way to all finite abelian groups.
The main theorem on finite abelian groups is that they can all be written as direct sums of cyclic groups of prime power order (this is called the elementary divisor decomposition). So write $A=\oplus_{i=1}^m \mathbb{Z}/{p_i^{e_i}}$ and $B=\oplus_{j=1}^n \mathbb{Z}/{q_j^{f_j}}$ so that $A \otimes B = \oplus _{1 \leq i \leq m, 1 \leq j \leq n} (\mathbb{Z}/p_i^{e_i} \otimes \mathbb{Z}/q_j^{f_j})$. Now you have reduced to the case $A$ and $B$ cyclic of prime power order. Figure out what happens when $p$ and $q$ are distinct and when they are the same.