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It is a fun exercise to prove that if $G$ is finitely generated, and $|\text{Aut}(G)| = p$ for some prime $p$, then $G$ is one of $\Bbb Z_3, \Bbb Z_6, \Bbb Z$, or $\Bbb Z \times \Bbb Z_2$. The finitely generated is essential (at least, in my proof), since the very end reduces to doing casework with the structure theorem on finitely generated abelian groups. But what about the more general case?

Are there any non-finitely generated groups $G$ with $|\text{Aut}(G)| = p$?

Some immediate comments: $G$ must still be abelian (because $\text{Inn}(G)$ is cyclic iff trivial); $p$ must be 2 (because being abelian means implies $a \mapsto a^{-1}$ is an automorphism, and this map is an involution, so $2\mid p$); and $G$ must be either indecomposable or an indecomposable times $\Bbb Z_2$ (because if $G = A \times B$, then $\text{Aut}(G) \times \text{Aut}(G)$ is a subgroup of $\text{Aut}(G)$; and the only groups with trivial automorphism group are the trivial group and $\Bbb Z_2$; and you can only stick one copy of $\Bbb Z_2$ in there, since otherwise there's an extra automorphism given by swapping two copies of $\Bbb Z_2$). But that's all I know.

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    $\begingroup$ This is relevant. Take the cross-product of the group there with one of the two you point out. $\endgroup$
    – user1729
    Commented Jul 2, 2014 at 6:30
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    $\begingroup$ The group of rational numbers with square-free denominator has automorphism group of order $2$. I can't readily think of an example for an odd prime $p$, though. $\endgroup$
    – James
    Commented Jul 2, 2014 at 6:53
  • $\begingroup$ @James: I think that is a pretty example. Please consider undeleting it as an answer. You may want to add a bit more detail, but I enjoyed validating it as it is :-) $\endgroup$
    – Jyrki Lahtonen
    Commented Jul 2, 2014 at 6:59
  • $\begingroup$ @James I agree with Jyrki! $\endgroup$
    – user98602
    Commented Jul 2, 2014 at 7:01
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    $\begingroup$ I think the claim of your first sentence is false. Clearly the infinite cyclic group $\mathbb{Z}$ is finitely generated, and $\operatorname{Aut}(\mathbb{Z})$ has order $2$. $\endgroup$
    – spin
    Commented Jul 2, 2014 at 10:25

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The group of rational numbers with square-free denominator has automorphism group of order $2$. (EDIT: As noted in the comments, there there aren't examples for odd $p$.)

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    $\begingroup$ There is no example for odd prime $p$. If $|Aut(G)|$ is prime, then $Inn(G) \cong G/Z(G)$ is cyclic, so $G$ must be abelian. Now if $G$ has elements of odd order $> 1$, then $g \mapsto g^{-1}$ is a nontrivial automorphism of order $2$, hence $|Aut(G)| = 2$. If $g^2 = 1$ for all $g$, then $G$ is elementary abelian and $|Aut(G)|$ is not prime. $\endgroup$
    – spin
    Commented Jul 2, 2014 at 10:28
  • $\begingroup$ @spin What you say is dependent on choice (because of your last sentence). I do not know what happens in the absence of choice though.. $\endgroup$
    – user1729
    Commented Jul 2, 2014 at 20:22
  • $\begingroup$ @spin, Yes, thank you. This was noted in an earlier comment too, but I was a bit rushed earlier, and failed to delete the last sentence. $\endgroup$
    – James
    Commented Jul 3, 2014 at 3:56
  • $\begingroup$ @user1729 Why does that depend on choice? One can reduce that to every set (with more than one element) having more than one bijection to itself; this depends on choice, I take it? $\endgroup$
    – user98602
    Commented Jul 3, 2014 at 18:57
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    $\begingroup$ @Mike Read the link in my comment to your question. $\endgroup$
    – user1729
    Commented Jul 3, 2014 at 19:01

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