I was solving some exercises about automorphisms. I was able to show that $\operatorname{Aut}(\mathbb{Q},+)$ is isomorphic to $\mathbb{Q}^{\times}$. The isomorphism is given by $\Psi(f)=f(1)$, but when I try to do the same thing with $\operatorname{Aut}(\mathbb{R},+)$ I got stuck.

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

I would appreciate your help.

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    $\begingroup$ It's very big, assuming the Axiom of Choice: If $\{x_i\}_{i\in I}$ is any $\mathbb{Q}$-basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space, then any permutation of $I$ induces an automorphism of $\mathbb{R}$ as a $\mathbb{Q}$-vector space, and in particular as a group. That gives you at least $2^{\mathfrak{c}}$ automorphisms. Since the number of functions from $\mathbb{R}$ to itself is also $2^{\mathfrak{c}}$, that's how big this object is... $\endgroup$ Commented Mar 1, 2012 at 23:21
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    $\begingroup$ I am sure that equivalent questions have been asked several times but I can't track any of them down at the moment. Anyway, the relevant keyword here is "Cauchy functional equation" (en.wikipedia.org/wiki/Cauchy's_functional_equation). $\endgroup$ Commented Mar 1, 2012 at 23:30
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    $\begingroup$ @Qiaochu: Something like this perhaps? I'm not sure if this is a duplicate per se, though. $\endgroup$
    – Asaf Karagila
    Commented Mar 1, 2012 at 23:36
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    $\begingroup$ Or this, this, among others. $\endgroup$ Commented Mar 2, 2012 at 4:06
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    $\begingroup$ @spohreis: You should conclude by the comments that assuming the axiom of choice the automorphism group of $(\mathbb R,+)$ is equinumerous with the power set of the reals, so it cannot be $\mathbb R^\times$. Without the axiom of choice it is possible to have cases where it is exactly $\mathbb R^\times$. $\endgroup$
    – Asaf Karagila
    Commented Mar 2, 2012 at 13:27

1 Answer 1


First note that for every $\alpha\in\mathbb R^\times$ we have that $x\mapsto\alpha\cdot x$ is an automorphism of $(\mathbb R,+)$.

Also note that if $f(x+y)=f(x)+f(y)$ then $f(2)=f(1)+f(1)$, and by induction $f(n)=n\cdot f(1)$ for $n\in\mathbb N$, equally $f(k)=k\cdot f(1)$ for $k\in\mathbb Z$. This carries to rationals as well, so $f\left(\frac{p}{q}\right)=\frac{p}{q}\cdot f(1)$.

Now, if $f$ is continuous then for every $x\in\mathbb R$ we have $f(x)=x\cdot f(1)$. So setting $\alpha=f(1)$ gives us that the continuous solutions are the solutions defined by $\mathbb R^\times$, therefore $\mathrm{Aut}(\mathbb R,+)\cap\{f\in\mathbb R^\mathbb R\mid f\text{ continuous}\}\cong(\mathbb R^\times,\cdot\ )$.

The existence of non-continuous solutions requires some axiom of choice, since such solutions generate Lebesgue non-measurable sets. So if we assume that every set is Lebesgue measurable (e.g. Solovay's model of models of Determinacy) then indeed there are no other solutions.

However, assuming the axiom of choice we can generate a basis for the vector space $\mathbb R$ over $\mathbb Q$. Note that every permutation of this basis can be extended to an automorphism of the vector space, namely $(\mathbb R,+)$.

The cardinality of such basis (known as Hamel basis) is $2^{\aleph_0}$, we have $2^{2^{\aleph_0}}$ many non-continuous solutions if we assume that such basis exists.

For further reading:

  1. Is there a non-trivial example of a $\mathbb Q-$endomorphism of $\mathbb R$?
  2. Horst Herrlich, The Axiom of Choice. Springer, 2006. (In particular section 5.1)
  • $\begingroup$ (+1) for simple argument I can understand. Interesting to contrast that with complicated answers. Also, I forgot what the word endomorphism means, but will try to commit the term to memory. $\endgroup$ Commented Oct 22, 2018 at 8:58

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