# Automorphisms of $\mathbb{R^*}$ as a group

I was just thinking randomly about groups and cardinality of sets then I thought of the problem

What will be the cardinality of the group of all automorphisms of the multiplicative group $$\mathbb{R^*}$$?

My Attempt: At first I thought that there's only the identity automorphism and I also got some knowledge about it from few questions over here but then I realised that the questions here was about Ring automorphisms mainly. Then I thought if $$f(x) = x^{2n+1}$$ where $$n \in \mathbb{N}$$. Then $$f$$ is automorphism.

So I could say that $$Aut(\mathbb{R^*})$$ is at least countably Infinite but then I couldn't go any further.

Can anyone give me some ideas about it? I'd prefer if it involves simple group theory approaches. It's okay if it's not being possible to do so.

Edit : I was also looking for functions $$f$$ satisfying $$f(xy)=f(x)f(y)$$. But as I can't say anything about $$f$$ rather than it's bijection and preserves group structure, I could not find any form of $$f$$.

• Do you want continuous automorphisms only? Commented May 14, 2021 at 18:02
• No no just automorphisms. Commented May 14, 2021 at 18:03
• I'm looking for a general overview of what the automorphisms will be and what the $Aut(\mathbb{R^*})$ will be Commented May 14, 2021 at 18:04
• I recommend looking up the Cauchy functional equation. The analysis there should be transportable to this situation. Commented May 14, 2021 at 18:05
• I looked it up but it seems that it is beyond my current knowledge. I haven't dealt with functional equations yet. Commented May 14, 2021 at 18:08

First, note that the structure $$(\mathbb{R}_{>0};\times)$$ is isomorphic to the structure $$(\mathbb{R};+)$$ (think e.g. about the map $$x\mapsto ln(x)$$). The latter is a bit easier to think about. In particular, if you consider it as a vector space over $$\mathbb{Q}$$ it's easy to see that its dimension is $$2^{\aleph_0}$$; any permutation of a basis extends to an automorphism of the whole, and so $$Aut(\mathbb{R};+)$$ has cardinality at least $$2^{2^{\aleph_0}}$$. Since that's also an obvious upper bound (that's the number of functions from $$\mathbb{R}$$ to $$\mathbb{R}$$), we get $$\vert Aut(\mathbb{R}_{>0};\times)\vert=2^{2^{\aleph_0}}.$$

What about $$(\mathbb{R};\times)$$? It turns out that there is no real difference between $$(\mathbb{R}_{>0};\times)$$ and $$(\mathbb{R};\times)$$ as far as automorphisms are concerned:

Every automorphism of $$(\mathbb{R}_{>0};\times)$$ extends uniquely to an automorphism of $$(\mathbb{R};\times)$$, and every automorphism of $$(\mathbb{R};\times)$$ restricts to a (trivially) unique automorphism of $$(\mathbb{R}_{>0};\times)$$.

The key point is the following: a real number $$r$$ is non-negative iff it has a square root. Since this is a purely multiplicative property, it can't be affected by automorphisms - every automorphism of $$(\mathbb{R};\times)$$ must send positive reals to positive reals and negative reals to negative reals. Thinking a bit more about this gives the result above.

This gives the desired cardinality result: $$\vert Aut(\mathbb{R};\times)\vert=2^{2^{\aleph_0}}.$$

• help me understand some things.... considering $\mathbb{R}$ as vector space over $\mathbb{Q}$ how does this translate to finding the $Aut(\mathbb{R})$ when $(\mathbb{R},+)$ is a group structure. I understand that In order to be a vector space the set has to be an abelian group under the addition operation but I still cannot align with the concept that you used. Commented May 15, 2021 at 2:37
• I also understand the concept of finding the all possible basis permutations but I can't relate this idea when I'm considering it as group only, because there's no generating set in the group $(\mathbb{R},+)$. Commented May 15, 2021 at 2:44
• @Pritam You could show as an exercise that if $V_1$ and $V_2$ are two vector spaces over $\mathbb{Q}$, then any morphism of abelian groups $f : V_1 \to V_2$ is necessarily $\mathbb{Q}$-linear. Next, you need some theory about bases of infinite-dimensional vector spaces, like $\mathbb{R}$ over $\mathbb{Q}$ here, and specifically the lemma here. My argument also uses this theorem. Commented May 15, 2021 at 21:29