Q:

Abelian $2$-groups

Is every abelian group $A$ where every element has order two isomorphic to a direct product of cyclic groups of order two, $A\cong C_2\times C_2\times\ldots$?
I ask because I used this "fact" in one of my old answers here (which is relevant to some work I am doing), and have just realised that this is not obvious, and so perhaps not true.
Am I perhaps just not seeing something which I thought was obvious at the time? Or is there something more subtle going on?
(Note that there is no assumption that $A$ is finitely generated.)

A:

An abelian group is a $\mathbb{Z}$-module. An abelian group of exponent dividing $n$ is a $\mathbb{Z}/n\mathbb{Z}$ module. In your case, $A$ is a $\mathbb{Z}/2\mathbb{Z}$-module, so a vector space over the field $\mathbb{Z}/2\mathbb{Z}$. By standard axioms such as the axiom of choice, $A$ has a basis, and so is the direct sum of one dimensional subspaces. In other words, $A$ is the restricted direct product or direct sum of copies of $C_2$.
$\mathbb{Z}/n\mathbb{Z}$ is a self-injective ring if $n$ is nonzero, and this often gives you nice decompositions. See "DSC" group.
$A$ need not be a direct product. For instance if $A$ is countably infinite, then it is not a direct product of copies of $C_2$, since finitely many $C_2$s produces finite cardinality, and infinitely many $C_2$s produces at least a continuum cardinality.

