Can the group-structure of my constraint reduce the corresponding assignment problem to polynomial time search complexity?

Question

Consider a finite Abelian group $G$ where each element is its own inverse. As an explicit example, let $G$ be the integers 0 to 31 under the bitwise XOR $\oplus$ operation. We have a collection of $n$ non-empty and possibly non-distinct EDIT: (subsets) of $G_i \leqslant G$ satisfying a cardinality constraint $\sum_{i=1}^n |G_i| \le |G|$. The task is to assign to each subset $G_i$ a unique element $x_i\in G$ respecting the following constraint:

For $G_i, G_j$ with $i\ne j$, $x_i\ne x_j$ and $x_i\oplus x_j \notin Q_{ij} = \{ z | x \oplus y = z$, For $x\in G_i$, $y\in G_j\}$.

As a specific example in case I have erred in my above notation, if $G_i = \{5\}$ and $G_j =\{1, 8\}$, Then $Q_{ij} = \{5\oplus1=4, 5\oplus8=13\}$, so $x_i \oplus x_j$ may not be 4 or 13. The problem can be fed into a CSP solver, or be naturally mapped onto a Maximum Independent Set formulation, but I'm wondering: does the group structure of the constraint lead to a reduction in the problem complexity, specifically allowing this to be solved in polynomial time?

Edit: Specifically the time to find any solution that satisfies the constraints or to prove it is unsatisfiable.

A possible alternative formulation of the above constraints is that $x_i\ne x_j$ for $i\ne j$ and $Q=\bigcup_{i=1}^n \{z|x_i \oplus g_i$ for $g_i\in G_i\}$ has $|Q|=\sum_{i=1}^n |G_i|$.

For example, the group structure embedded in the problem allows for a simplification of the search space. If $S_0=\{x_i\}$ is a valid solution, so is $S_x=\{z_i |z_i=x\oplus x_i$ for $x_i\in S_0\}$. This can be seen since the group structure + each element being its own inverse leads to the constraints being unchanged under this transformation. We can thus without loss of generality pin $x_1 =0$. I don't see obvious further reductions beyond this point.

• cpoole: @EmilJeřábek 0ドル\in Q_{ij}$ only if $G_i$ and $G_j$ share a common element, which is not guaranteed.
• Emil Jeřábek: (comment edited but grace period expired) The constraint $x_i\ne x_j$ is redundant: if $x_i=x_j,ドル then $x_i+x_j=0\in Q_{ij}$. The condition can ve restated so that you want to pick a coset of each $G_i$ such that the cosets are pairwise disjoint. Also note that what you have here (an abelian group of exponent 2ドル$) is actually a vector space over the field $\mathbb F_2,ドル and the $G_i$ are subspaces.
• Emil Jeřábek: All subgroups share the 0 element, so it is most certainly guaranteed.
• cpoole: The original question posed the $G_i$ as subgroups when the intention was just an arbitrary subset. Your comment makes sense to me when viewing them as subgroups.
• Emil Jeřábek: Oh. That changes the problem completely.