Let $\mathbb{R}$ be the real number field and $\mathbb{R}^{\times}$ be the multiplicative group of it.
$\mathrm{Aut}(\mathbb{R}^{\times})$ denotes the group automorphism of $\mathbb{R}^{\times}$.

[My Question $(*)$]
Do all elements $\phi \in \mathrm{Aut}(\mathbb{R}^{\times})$ have the form $\mathbb{R}^{\times} \rightarrow \mathbb{R}^{\times} , x \mapsto x^r$ , where $r$ is a real number.

I'm trying to solve this question $(*)$. But I have no idea how to solve it.
Can you answer this question $(*)$ and explicitly write $\mathrm{Aut}(\mathbb{R}^{\times})$ ?


You must first bear in mind that exponentiation with general real exponents is only allowed for strictly positive bases (in other words, if you want to consider the quantity $x^y$, with $y \in \mathbb{R}$ arbitrary then you will have to restrict yourself to $x>0$).

The subgroups $\{\pm 1\}$ and $\mathbb{R}_{+}^{*}$ are such that they intersect trivially, commute element-wise (since the ambient group is abelian) and their product is the entire ambient group, hence you have the direct product decomposition:

$$\mathbb{R}^{*} \approx \{\pm 1\} \times \mathbb{R}_{+}^{*}\ (\mathrm{Gr})$$

Notice also that via any logarithm (in any base, that is), $(\mathbb{R}_{+}^{*}, \cdot)$ and $(\mathbb{R}, +)$ are isomorphic (as topological groups even).

It is also easy to see that for any arbitrary groups $G, G'$, both $\mathrm{Aut}_{\mathrm{Gr}}(G)$ and $\mathrm{Aut}_{\mathrm{Gr}}G'$ embed in $\mathrm{Aut}_{\mathrm{Gr}}(G \times G')$, so in your given case $\mathrm{Aut}_{\mathrm{Gr}} \mathbb{R}$ embeds in $\mathrm{Aut}_{\mathrm{Gr}} \mathbb{R}^{*}$.

However, it is easy to see that any group endomorphism of additive $\mathbb{R}$ is also a $\mathbb{Q}$-vector space endomorphism, in other words that:

$$\mathrm{Aut}_{\mathrm{Gr}} \mathbb{R}=\mathrm{Aut}_{\mathbb{Q}} \mathbb{R}$$

It is also relatively easy to establish that $\mathrm{dim}_{\mathbb{Q}} \mathbb{R}=|\mathbb{R}|$. In general, if $K$ is a field (not necessarily commutative), $V$ a $K$-vector space and $A$ a fixed basis of $V$ over $K$, then it is also straightforward that $\Sigma(A)$, the symmetric group over $A$ (i.e. the group consisting of all permutations of $A$ under composition of maps) embeds in $\mathrm{Aut}_{K} V$.

We combine this with the following:

for any infinite set $A$, it holds that $|\Sigma(A)|=|A|^{|A|}=2^{|A|}$

(the operation implicit in the above relation being cardinal exponentiation) in order to derive that:

$$2^{|\mathbb{R}|} \leqslant |\mathrm{Aut}_{\mathbb{Q}} \mathbb{R}| \leqslant |\mathrm{Aut}_{\mathrm{Gr}} \mathbb{R}^{*}|$$

On the other hand, according to a gross estimate:

$$|\mathrm{Aut}_{\mathrm{Gr}} \mathbb{R}^{*}| \leqslant |\mathbb{R}^{*}|^{|\mathbb{R}^{*}|}=|\mathbb{R}|^{|\mathbb{R}|}=2^{|\mathbb{R}|}$$

by which we conclude that the order of the automorphism group of $\mathbb{R}^{*}$ is actually $2^{|\mathbb{R}|}$.

The power function automorphisms of $\mathbb{R}^{*}$ that you were mentioning correspond naturally to homothecies over $\mathbb{R}$ of real coefficient (i.e. maps given by multiplication of the argument with a given constant, the coefficient of the homothecy), and as such only account for the real automorphisms of $\mathbb{R}$. This is a very narrow view of what is actually going on, as it ignores the presence of the rational automorphisms, which are as we have argued far more. As final words, there is a limit to how explicit a description of such automorphisms can be.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.