Let $G$ be a group (not necessarily finite) such every element of $G$ has order 2. Every such group is abelian [1].

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.

Or not?


1 Answer 1


Write $G$ as a direct sum of copies of $\mathbb{Z}/2\mathbb{Z}$ over some index set $X \cup \{e\}$. Then it is isomorphic as an abelian group to the Boolean algebra of all finite or cofinite subsets of $X$ (where the presence or absence of the basis element corresponding to $e$ tells you whether a given element of $G$ corresponds to a finite or cofinite subset).

  • $\begingroup$ Signed (very) long int, eh? I like it. $\endgroup$
    – Asaf Karagila
    May 26, 2012 at 20:45

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