# Is there a group $G$ for which $\mathrm{Aut}(G) \simeq (\mathbb{R},+)$?

I know the classic theorem that $$(\mathbb{Q},+)$$ cannot be expressed as an automorphism group, i.e. there is no group $$G$$ such that $$\mathrm{Aut}(G)\simeq (\mathbb{Q},+)$$.

Theorem A. If $$L$$ is a locally cyclic group with no element of order $$2$$, then $$L$$ cannot be expressed as an automorphism group.

But how about $$(\mathbb{R},+)$$?

I think the answer might be no, and the proof proceeds by showing that if $$\mathrm{Aut}(G) \simeq \mathbb{R}$$, then some subgroup or quotient $$H$$ of $$G$$ will satisfy $$\mathrm{Aut}(G) \simeq \mathbb{Q}$$, which contradicts Theorem A.

So how can we construct this $$H$$? Assuming the axiom of choice, we can say $$\mathbb{R} = \mathbb{Q}\oplus B$$ for some additive subgroup $$B$$ of $$\mathbb{R}$$ (just by picking a $$\mathbb{Q}$$-basis for $$\mathbb{R}$$). Now I'm tempted to do some Galois-type thing, where you use a "fixed" subgroup $$H=\mathrm{Fix}(B)=\{g\in G : b(g) = g \text{ for all } b\in B \},$$ and then try to say something about $$\mathrm{Aut}(H)$$ or $$\mathrm{Aut}(G/H)$$ (where, for the latter, maybe we take the normal closure of $$H$$). But I can't complete this line of reasoning. It seems key that $$\mathrm{Aut}(G)$$ splits as a direct sum --- that seems special.

Am I just totally off-base here? Is there some obvious group $$G$$ whose automorphism group is $$\mathbb{R}$$?

Related questions

If I can prove that $$\mathbb{R}$$ is not an automorphism group, then the same would hold for the isomorphic group $$\mathbb{R}_{>0}$$ of positive reals under multiplication. But how about $$\mathbb{R}^\times \simeq \mathbb{R}_{>0}\times C_2\simeq \mathbb{R}\times C_2$$ --- can this be achieved as an automorphism group? If $$\mathrm{Aut}(G)$$ is a direct product of abelian groups, what can be said about $$G$$?

A related observation is that if $$\mathrm{Aut}(G)$$ is abelian, then $$G$$ is nilpotent of rank $$\leq 2$$. So there is a sense in which $$G$$ is "almost" abelian.

I know $$\mathbb{Q}^\times$$ is the automorphism group of $$(\mathbb{Q},+)$$, and that $$\mathbb{R}^\times$$ constitutes the continuous automorphisms of $$(\mathbb{R},+)$$ ... Then there's multiplicative/additive groups of other rings/fields ...

I can't even resolve this question in the case of $$\mathrm{Aut}(G) \simeq \mathbb{Z}\times \mathbb{Z}$$. For this one I can show that $$G'$$ would necessarily by cyclic, but that's about it.

I'm very curious about which types of groups can be achieved as automorphism groups.

A quick observation

If $$\mathrm{Aut}(G) \simeq \mathbb{R}$$, then $$G$$ must be infinitely generated (or else its automorphism group is countable) and nilpotent of rank $$\leq 2$$ (this follows for any group whose automorphism group is abelian).

• It's certainly not the case 'generically' that if $H\subset Aut(G)$ then there's some subgroup $G'$ of G with $Aut(G')\equiv H$, even if $Aut(G)=H\times K$; consider the automorphism group of $C_p$ for $p$ prime. I don't see anything in your reasoning offhand that uses any properties of $\mathbb{R}$ that don't mirror to this case. Nov 27, 2020 at 16:08
• Yes, it would have to be some special property of $\mathbb{R}$ that comes up here. Subgroup or quotient would be OK here --- just as long as we can use the fact that $\mathbb{Q}$ is a direct summand of $\mathbb{R}$ to reduce to the case of $\mathbb{Q}$ (which is a known exercise). Nov 27, 2020 at 16:10
• @Ehsaan: re: your related questions, it's consistent with ZF that every endomorphism of $\mathbb{R}$ is measurable, hence continuous; if this is true then $\text{Aut}(\mathbb{R}) \cong \mathbb{R}^{\times} \cong \mathbb{R} \times C_2$. (With the axiom of choice $\text{Aut}(\mathbb{R})$ is much larger and in particular nonabelian.) Dec 6, 2020 at 17:56
• Partial answer: $G$ cannot be abelian. If $G$ is abelian, then $x\mapsto x^{-1}$ is an automorphism of $G$ of order $\leq 2$. $\mathbb{R}$ has no nonidentity elements of finite order, so $x=x^{-1}$ for all $x\in G$. This implies that $G$ is the underlying abelian group of an $\mathbb{F}^2$-vector space $V$, and we have $\operatorname{Aut}(G)=\operatorname{Aut}_{\mathbb{F}_2}(V)$. We must have $\dim V\geq 2$, but this means $\operatorname{Aut}(G)$ contains $\operatorname{GL}_2(\mathbb{F}_2)$ as a subgroup. This would make $\operatorname{Aut}(G)$ nonabelian, contradiction! Jan 7, 2021 at 5:56