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Let $G$ be a group with transitive automorphisms on $G-\{e\}$. I.e. for any $a,b\neq e$ in $G$, there exists some $f \in \operatorname{Aut}(G)$ such that $f(a) = b$. Is it then necessarily the case, that $G$ is a vector space (over $\mathbb F_p$ or $\mathbb Q$)? If $G$ is finite, this is the case (cf. Groups with transitive automorphisms ) but I don‘t know how to prove it for infinite $G$ or how to construct a counterxample.

I can show that either $k=\mathbb F_p$ or $k=\mathbb Q$ embeds into $G$, but then I am stuck. I tried constructing a maximal embedding $k^I \to G$ of some vector space into $G$ via Zorn’s lemma, but given two embeddings $f,g \colon k \to G$ with disjoint image I can’t show whether $f \times g $ defines an embedding of $k^2$ into $G$ and this is the crucial step missing in applying Zorn’s lemma.

This argument would work, if we assumed $G$ to be abelian of course, but I think we don’t need that assumption.

Arturo Magidin
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Sven-Ole Behrend
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    There are infinite groups in which any two nontrivial elements are conjugate. You can start with any torsion free group and embed it into such a group using HNN extensions. The resulting group has the desired property, but is not abelian. See the sketch here. – Arturo Magidin Jan 15 '23 at 00:56
  • Thank you for the answer! Are you aware of any counterexample in the case of $p$-torsion? Also, if you leave your comment as an answer, I will gladly accept it. – Sven-Ole Behrend Jan 16 '23 at 13:37
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    I am not sure about the torsion case; while one can do an HNN extension to make any two elements of order $p$ conjugate, the stable letter will usually have infinite order, so more work would have to be done after. – Arturo Magidin Jan 16 '23 at 21:40

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