Let $G$ be a group (not necessarily finite) such every element of $G$ has order 2. Every such group is abelian .
Clearly, every Boolean algebra $B$ is a group of this type, when equipped with the symmetric difference operation.
Question: is there any group $G$ such that there does not exist a Boolean algebra $B$ equipped with the symmetric difference such that $B$ is isomorphic to $G$?
EDIT: It is clear to me, that every such group must be a direct sum of $\mathbb Z/2\mathbb Z$. This suggests a counterexample: take a countably infinite direct sum of $\mathbb Z/2\mathbb Z$. Clearly, we can model this group by all finite subsets of $\mathbb N$ equipped with symmetric difference, and this is not a Boolean algebra. But there is a mistake in this reasoning; the set system of question is not a Boolean algebra as a set system. It can still be a Boolean algebra as an abelian group.