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The latest question to be asked at the Group Pub Forum is a classic: can every group be realised as the automorphism group of a group? The answer is no, and the canonical answer is the infinite cyclic group. See this answer of mine for the details. In the back-and-forth mailings Arturo Magidin made the following point, which I would never have realised:

A infinite group of exponent two may not have an automorphism of order two. It depends on the Axiom of Choice.

That is, if we assume choice then your group is the direct sum of infinitely many copies of $C_2$ and we can simply choose a pair of $C_2$s in the direct sum decomposition and switch them. Indeed, the direct sum will have automorphism group containing $S_{\infty}$. My question is therefore the following.

Question: Do not assume choice. What might the automorphism group of an infinite group of exponent two look like?

If it cannot be infinite cyclic then the canonical counter-example to the motivating question holds. But I don't really care about the motivating question, I am just interested on what is going on here...

(As an aside, I am not that comfortable with axioms - is "not assuming choice" the same as "assuming choice does not hold"? Is there some middle ground which sits before I decide to use choice or not? What happens here?...Basically, in your answer make it clear what is going on with your axioms!)

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  • What does "A countably infinite group of exponent two" mean? – Asaf Karagila May 06 '14 at 09:32
  • So it's a countable group that satisfies $x+x=0$, for all $x$? – Asaf Karagila May 06 '14 at 09:39
  • @Asaf Yes, precisely. – user1729 May 06 '14 at 09:42
  • Are you sure the word "countable" was mentioned, and not "infinite"? – Asaf Karagila May 06 '14 at 09:44
  • @AsafKaragila There was no stipulation. I am just more comfortable with countable groups, but I will remove this now. What I say is Arturo's point is just a paraphrase. What he actually said was: I will note that the fact that an abelian group other than the cyclic group of order 2 always has an automorphism of order 2 at least depends on the Axiom of Choice for infinite abelian groups; in the absence of Choice, there could be an infinite abelian group of exponent 2 with trivial automorphism group. – user1729 May 06 '14 at 09:46
  • Okay, that's a different story altogether. It is consistent that there is an abelian group of exponent $2$ with a trivial automorphism group. But you can't have this group countable. – Asaf Karagila May 06 '14 at 09:47
  • @AsafKaragila Ahh, okay, I didn't think it would be an issue. I will change this question then. (Although this makes my question slightly hand-wavey - the other reason I put "countable" there was because I was sort of assuming the group would be unique. It is unique if we assume choice, unless I am missing some subtlety (because it is just the direct sum of infinitely many copies of $C_2$).) – user1729 May 06 '14 at 09:48
  • No no, it's not a direct sum anymore. That's the thing. :-) – Asaf Karagila May 06 '14 at 09:51
  • @Asaf Yes, sorry, that was what I was meaning - if we assume choice then there is a unique countable group of exponent two, while if we do not assume it then I was hoping it still would be unique and so my question would have a nice answer. If I let go of my cardinality stipulation then it cannot be unique. – user1729 May 06 '14 at 09:58
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    Would the downvoter care to comment? – user1729 Jun 27 '14 at 15:14

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So to your question, it's hard to tell. Not assuming choice means that it might be true, in which case you already know the theory; or it might be false but its failure does not affect this; and so on.

It is, however, consistent that there exists an infinite abelian group of exponent $2$, or simply put, a vector space over $\Bbb F_2$, which does not admit any automorphism. It is consistent for such example to have the property that every proper subgroup is finite.

Note that this means that the group is not the direct sum of copies of $C_2$. However it does embed every such finite sum. If $G=\bigoplus_{i\in I}C_2$ then we can easily come up with an automorphism of $G$ which has order $2$. Pick some $i\in I$ and consider the automorphism switching the $i$-th copy of $C_2$, but keeping the rest in place.

The example above can be generalized, and we can have such group without automorphisms that every proper subgroup is the direct sum of copies of $C_2$ (and more specifically, the index set is well-orderable), but the group itself - again - is not.

You can see the details of the construction in Axiom of choice and automorphisms of vector spaces, which was later generalized to my masters thesis (which can be found through my user profile) to the generalized case.

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