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I do not know much about infinite groups. $\mathbb{R}$ is especially different from others I have worked before - it does not seem to have any generator like $\mathbb{Z}$ does or we could say that its every non-trivial element generates a subgroup isomorphic to $\mathbb{Z}$.

I attempted to find the automorphism group of $\mathbb{R}$. There are only three kinds of automorphism operations we can perform on $\mathbb{R}$:

  1. identity: $\psi_1: x \mapsto x$, $\psi_1 = \mathrm{id}$
  2. reflection: $\psi_2:x \mapsto -x$, $\psi_2 \circ \psi_2 = \mathrm{id}$
  3. translation: $\phi_r:x \mapsto x+r$, $r \in (-\infty, \infty) = \mathbb{R}$

Therefore the $\mathrm{Aut}(\mathbb{R}) = \mathbb{R} * \mathbb{Z}_2$.

Is my reasoning correct?

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marked as duplicate by Seirios, Davide Giraudo, Jim, Norbert, Ben Millwood Feb 23 '13 at 19:48

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    $\begingroup$ Is translation an additive map? You'll want to view $\bf R$ as a countably infinite dimensional vector space over $\bf Q$; check that any additive homomorphism of rational vector spaces is in fact a $\bf Q$-linear map (this sort of thing generalizes to modules over integral domains and their localizations). $\endgroup$ – anon Feb 23 '13 at 18:06
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    $\begingroup$ @anon $\dim_{\mathbb Q}\mathbb R>\aleph_0$. $\endgroup$ – Hagen von Eitzen Feb 23 '13 at 18:12
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    $\begingroup$ Oops, continuum-dimensional. :slip-up: $\endgroup$ – anon Feb 23 '13 at 18:13
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Note that $f(x)=x+r$ is an automorphism means that $f(1)+f(-1)=f(0)=0$ therefore $1+r+(-1)+r = 2r = 0$ so $r=0$. Therefore no actual translation is an automorphism of the additive group of $\Bbb R$. But it is true that translations are automorphisms of the ordered set $\Bbb R$.

Let $f\colon\Bbb R\to\Bbb R$ be an automorphism of the additive group.

By induction one can show that if $f(1)=r$ then $f(q)=r\cdot q$ for every rational number $q$.

If we require that $f$ is continuous then this is enough to show that $f(x)=r\cdot x$ for every $x\in\Bbb R$. But if we don't require this then we can generate a lot more automorphisms using the axiom of choice.

The method is simple, consider $\Bbb R$ as a vector space over $\Bbb Q$ and using the axiom of choice let $H$ be a Hamel basis for $\Bbb R$ over $\Bbb Q$, any permutation of $H$ induces an automorphism of $\Bbb R$ as a vector space, which is also an automorphism of the additive group.

Related: What is $\operatorname{Aut}(\mathbb{R},+)$?

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No, it is not correct; first of all, translation is not a group automorphism of $\mathbb{R}$ (perhaps you meant to say dilation?) But there are many more automorphisms than these; in fact, thinking only additively, $\mathbb{R}$ is a $\mathbb{Q}$-vector space whose dimension is the continuum, and it should be clear that there are many ways of constructing $\mathbb{Q}$-linear maps from $\mathbb{R}$ to $\mathbb{R}$ itself (which will, in particular, be group homomorphisms).

However, I believe that if you're only interested in continuous automorphisms, then dilation (and reflection and the identity, which are cases of dilation) are the only ones.

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  • $\begingroup$ In fact if you are interested in measurable/Baire automorphisms then the continuous ones are the only ones. $\endgroup$ – Asaf Karagila Feb 23 '13 at 18:16
  • $\begingroup$ I needed this result for calculation of the automorphism (deck) group of $\mathbb{R}$ covering $\mathbb{R}$ by an identity covering map which is continuous. Hence, using only the dilation, in my understanding $Aut(\mathbb{R}, \mathbb{R})$ would be $\mathbb R^\times$. Is that correct? $\endgroup$ – Dávid Natingga Feb 23 '13 at 19:08
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    $\begingroup$ @David: Yes. If you only consider continuous automorphisms of $(\Bbb R,+)$ you get $(\Bbb R^\times,\cdot)$. $\endgroup$ – Asaf Karagila Feb 23 '13 at 20:30
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No, your reasoning is not correct. For one thing, translation is not an automorphism: it doesn't even map $0$ to $0$. For another, you haven't even tried to explain why these should be all the automorphisms. As it happens, there are far far more, but even if there weren't this would be a gaping hole in your argument.

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