I'm having a problem trying to find the simplest way of proving this, which has most probably been solved a hundred of times but I am unable to find a good reference.

I have two groups, $(\mathbb{R},+)$ and $(\mathbb{R}_{>0},\times)$. I am trying to prove that the only class of isomorphisms between them is the class $F = \{f: f(x) = \exp(\alpha x), $ for all $\alpha \in \mathbb{R}_{>0}\}$. Existence is easy to prove: what I'm having trouble with is a clean algebraic uniqueness proof.

Does anyone know the proof or a reference containing this proof?

Thanks in advance!


marked as duplicate by Mikko Korhonen, Paul, Did, Dan Rust, Old John Dec 27 '13 at 11:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You can find some facts about the functional equation $f(x+y)=f(xy)$ in this post. $\endgroup$ – Martin Sleziak Dec 27 '13 at 10:21

Wait, it may not be true.

Consider $(\Bbb R,+)$ as an infinite (continuum) dimension vector space over $\Bbb Q$, and fix a basis (Hamel basis).

Then, any automorphism of this vector space (for example permuting the basis) will be an automorphism of $(\Bbb R,+)$, and you can compose this with any exponential.

  • $\begingroup$ I see... so, do you think the only way to make this statement true would be to impose the preservation of the strict total order? $\endgroup$ – peortega Jan 27 '13 at 22:39
  • 1
    $\begingroup$ Yes, or equivalently the continuity. $\endgroup$ – Martin Brandenburg Jan 27 '13 at 22:57
  • $\begingroup$ Great! Thanks a lot! $\endgroup$ – peortega Jan 27 '13 at 23:09

Not the answer you're looking for? Browse other questions tagged or ask your own question.