abstract algebra - What is the automorphism group of $mathbb{R}$ under addition?

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?

Answer

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 $\operatorname{Aut}(\mathbb{R},+)$?