# Automorphism group of ($\mathbb{R}^{\times}, \cdot$)

Let ($$\mathbb{R}^{\times}, \cdot$$) be the multiplicative group of non-zero real numbers. I want to find the automorphism group of ($$\mathbb{R}^{\times}, \cdot$$). My guess is that it is the group of all rational numbers $$p/q$$ such that $$p, q$$ are odd integers which are co-prime.

We know that any automorphism of the additive group of real numbers ($$\mathbb{R}, +$$) is of the form $$x \mapsto \lambda x$$. Also, ($$\mathbb{R}_{>0},\cdot$$) is isomorphic to ($$\mathbb{R}, +$$), where ($$\mathbb{R}_{>0},\cdot$$) denotes the multiplicative group of positive reals. $$\require{AMScd}$$ $$\begin{CD} (\mathbb{R}_{>0}, \cdot) @>{\phi}>> (\mathbb{R}_{>0}, \cdot)\\ @V{\log(x)}VV @VV{\log(x)}V\\ (\mathbb{R},+) @>{\lambda x}>> (\mathbb{R}, +) \end{CD}$$

By the diagram above, any automorphism $$\phi$$ of $$(\mathbb{R}_{>0}, \cdot)$$ is of the form $$\phi(x) = \exp(\lambda \log(x)) = x^\lambda$$where $$\lambda \in \mathbb{R}^{\times}$$. As any automorphism $$\psi$$ of ($$\mathbb{R}^{\times}, \cdot$$) should take positive numbers to positive numbers, $$\psi$$ must restrict to an automorphism on ($$\mathbb{R}_{>0},\cdot$$). The only automorphisms of ($$\mathbb{R}_{>0},\cdot$$) which extend to the automorphisms of ($$\mathbb{R}^{\times}, \cdot$$) are $$x \mapsto x^\lambda$$ such that $$\lambda = \frac{p}{q}$$ such that $$p,q$$ are odd integers which are co-prime. This is because:

1. For irrational $$\lambda$$'s, the map is not well defined for negative reals.
2. For $$\lambda = \frac{p}{q}$$, and $$q$$ even, the map is not well defined for negative reals.
3. For $$\lambda = \frac{p}{q}$$, and $$p$$ even, the inverse of this map does not exist.

From here on, I don't know how to proceed in proving my guess, if it is right in the first place.

• It's not true that the automorphism group of the additive group of real numbers is only multiplication by scalars. There are also horrible $\mathbb{Q}$-vector space automorphisms that are also group automorphisms. – Matt Samuel Feb 23 at 11:47
• Are there automorphisms which are discontinuous? – Ajay Kumar Nair Feb 23 at 11:50
• The only continuous automorphisms are multiplications by scalars in the usual topology. But yes, there are more discontinuous automorphisms than there are continuous ones. – Matt Samuel Feb 23 at 11:52
• Thanks! Can we conclude from here that the ones mentioned above are all the continuous automorphisms of $( \mathbb{R}^{\times}, \cdot )$? – Ajay Kumar Nair Feb 23 at 11:55
• Use this question and this one. – Dietrich Burde Feb 23 at 11:58

## 1 Answer

We have that $$(\mathbb{R}^*,\cdot)\simeq \{-1,1\}\times (\mathbb{R}^+,\cdot)$$. For any automorphism $$f:(\mathbb{R}_{>0},\cdot)\to(\mathbb{R}_{>0},\cdot)$$ we can define $$f'(\epsilon,x) = (\epsilon,f(x))$$ and this is an automorphism of $$(\mathbb{R}^*,\cdot)$$. Does this define all automorphisms? Suppose conversely that $$g(\epsilon,x) = (g_1(\epsilon,x),g_2(\epsilon,x))$$ is an automorphism. If we compose with the absolute value, we obtain $$|g|:(\mathbb{R}^*,\cdot)\to (\mathbb{R}^+,\cdot)$$ given by $$|g|(\epsilon,x) = g_2(\epsilon,x)$$ The kernel is exactly $$\{g^{-1}(-1,1), 1\}$$, and since $$(-1,1)$$ is the only element of order $$2$$, $$g(-1,1) = (-1,1)$$, hence $$g_2(\epsilon,x)$$ is independent of $$\epsilon$$ and hence arises from an automorphism $$g_2:(\mathbb{R}_{>0},\cdot)\to (\mathbb{R}_{>0},\cdot)$$. Thus $$g(\epsilon,x) = (g_1(\epsilon,x),g_2(x))$$ If $$g_1(\epsilon,x) = (-1,g_2(x))$$, then $$g(\epsilon,x) = (-1,1)g(-\epsilon,x)$$. Since $$g(\epsilon,x)^2 = g(-\epsilon,x^2) = g(-\epsilon,x)^2$$, it follows that $$-\epsilon = 1$$. Hence $$g(\epsilon,x) = (\epsilon,g_2(x))$$ This characterizes the automorphisms as being automorphisms of $$(\mathbb{R}_{>0},\cdot)$$ linearly extended to be able to pull out minus signs. These will usually no longer be given by a single formula involving exponents, even if they are continuous. The continuous automorphisms end up being $$f(r) = \begin{cases} r^a&\mbox{ if }r>0\\ -r^a&\mbox{ if }r<0 \end{cases}$$