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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.)

Community
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user1729
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  • direct sum, not direct product, it uses axiom of choice – Jack Schmidt Jul 15 '13 at 13:27
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    Such a group is a vector space over the field of $2$ elements. – Tobias Kildetoft Jul 15 '13 at 13:29
  • I find it quite obvious. And sure, if you relax the conditions on the group it no longer holds (if the elements just have order a power of $2$, the group need not even be abelian). – Tobias Kildetoft Jul 15 '13 at 13:33
  • @TobiasKildetoft I deleted my comment when I saw Jack Schmidt's answer! Also, the abelian assumption is given. – user1729 Jul 15 '13 at 13:34
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    A fine answer has been posted. However, the SE software will not let me accept an answer within six minutes. This is to let anyone else give a better answer. Therefore, I dare someone to give a better answer than Jack Schmidt's. Go on! Make it worth my while not accepting it as soon as I can... – user1729 Jul 15 '13 at 13:37
  • Given all the other data, the "abelian" thing is superfluous: any group of exponent two is abelian. – DonAntonio Jul 15 '13 at 16:16
  • @DonAntonio: In my now-deleted comment, I had wondered if it really was obvious what TobiasKildetoft said, giving the fact that relaxing the exponent would give you crazy abelian groups (for example, Prüfer quasi-cyclic groups). Thus his comment, and thus mine. Also, I suppose I was not just wondering about $2$-groups when I asked the question. – user1729 Jul 15 '13 at 16:51

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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.

Jack Schmidt
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  • So having a fixed finite exponent is the important part? – user1729 Jul 15 '13 at 13:42
  • Yes, exactly. If the exponent is 4, then you get a direct sum of $C_2$s and $C_4$s. If every element is a 2-element, but not bounded, then you can get $\mathbb{Z}[\tfrac12]/\mathbb{Z}$ which is directly indecomposable (and you can get other weirder groups too). – Jack Schmidt Jul 15 '13 at 13:44