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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.

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    $\begingroup$ 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. $\endgroup$ – Matt Samuel Feb 23 at 11:47
  • $\begingroup$ Are there automorphisms which are discontinuous? $\endgroup$ – Ajay Kumar Nair Feb 23 at 11:50
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    $\begingroup$ The only continuous automorphisms are multiplications by scalars in the usual topology. But yes, there are more discontinuous automorphisms than there are continuous ones. $\endgroup$ – Matt Samuel Feb 23 at 11:52
  • $\begingroup$ Thanks! Can we conclude from here that the ones mentioned above are all the continuous automorphisms of $ ( \mathbb{R}^{\times}, \cdot )$? $\endgroup$ – Ajay Kumar Nair Feb 23 at 11:55
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    $\begingroup$ Use this question and this one. $\endgroup$ – Dietrich Burde Feb 23 at 11:58
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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 $$f(r) = \begin{cases} r^a&\mbox{ if }r>0\\ -r^a&\mbox{ if }r<0 \end{cases}$$

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