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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 $f$.

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

1 Answer 1

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The following expands the comments above into an answer:

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: $$\vert Aut(\mathbb{R};\times)\vert=2^{2^{\aleph_0}}.$$

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  • $\begingroup$ 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. $\endgroup$
    – Itachi
    Commented May 15, 2021 at 2:37
  • $\begingroup$ 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 $(\mathbb{R},+)$. $\endgroup$
    – Itachi
    Commented May 15, 2021 at 2:44
  • $\begingroup$ @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. $\endgroup$ Commented May 15, 2021 at 21:29

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