I've been noodling around with a family of subdirect products recently, and it seems elementary enough that I suspect there might be something written about them already. Here is the idea.
Let $\Gamma$ be a simple graph on the set $\{1, \ldots, n\}$ and let $G$ be a group. We will define a subgroup of $G^n$. For each $g \in G$ and $i \in \{1, \ldots, n\}$ we define $g_i$ to be the element of $G^n$ with a $g$ in position $i$, and also in every position $j$ such that $j$ is adjacent to $i$. Then we define $G^\Gamma$ to be the subgroup of $G^n$ generated by all $g_i$.
For example, suppose that $\Gamma$ is a path of length 2, with edges connecting 1 to 2 and 2 to 3. Then $G^\Gamma$ is generated by elements of the form $(g, g, 1), (g, g, g)$, and $(1, g, g)$. It's not hard to show that in this case we get all of $G^3$.
For another simple example, if $\Gamma$ is a complete graph, then $G^\Gamma$ is the diagonal subgroup of $G^n$.
Has anyone seen this family of subdirect products before?
This is not a complete answer but perhaps an interesting observation.
Let $G$ be a group and let $A$ be an $m\times n$ matrix with integer entries.
For every row $R$ of $A$ and $g\in G$, let $g^R$ be the element of $G^n$ such that the entry in the $i$th position is $g^{R_i}$.
(So for example, $g^{[1,0,2,-1]}=(g,1,g^2,g^{-1})$.)
Now, let $G^A=\langle g^R\rangle\leq G^n$, where $g^R$ runs over every row $R$ and every $g\in G$.
(Note that $G^A$ generalises $G^\Gamma$, just take $A$ to be the adjacency matrix of $A$ plus the identity matrix.)
Now, we have the following observations:
Applying invertible row operations (over the integers) to $A$ does not affect $G^A$. (Permuting rows or multiplying a row by $-1$ doesn't change the generating set, adding a row to another does change the generating set but not the generated group, because $g^{R+S}=g^Rg^S$.)
Permuting columns or multiplying a column by $-1$ does change $G^A$, but not its isomorphism type.
Adding a column to another can change the isomorphism type of $G^A$.
The last one is unfortunate, because otherwise, we could just reduce $A$ to Smith normal form and this would give us the isomorphism type of $G^A$.
Still, this is still helpful in simplifying the problem and can solve it in some cases.
For example, let $\Gamma$ be a path graph on $n$ vertices (so of length $n-1$) and $A$ be the adjacency matrix of $\Gamma$ plus the identity matrix. One can check that, if $n\not\equiv 2 \pmod 3$, then $A$ can be reduced to the identity matrix using just the operations allowed above (without adding a column to another), and so in that case $G^\Gamma=G^n$.