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If so, how might I be able to prove it?

EDIT: OK, thanks to many answers especially spin's and Micah's explanations. All of the answers were extremely helpful in understanding -- I have accepted Micah's because it seems the most complete, but all answers provide helpful additions/perspectives! I have tried to summarize:

$\phi$ is an isomorphism between the groups if and only if $\phi(x) = e^{f(x)}$ where $f$ is an isomorphism from $(\mathbb{R},+)$ back to $(\mathbb{R},+)$.

Of course there are lots of such $f$, especially when we take the Axiom of Choice.

However, it seems from the answers and Micah's link (Cauchy functional equation) that the only "nice" solutions are $f(x) = cx$ for a constant $c$. It seems that all others must be "highly pathological" (in fact $\{(x,f(x))\}$ must be dense in $\mathbb{R}^2$).

A remaining question is, how strong is the statement

All such isomorphisms have the form $e^{cx}$ for some $c \in \mathbb{R}$

or its negation? Or what is required for each to hold?

One answer seems to be that supposing the reals have the Baire property is sufficient to rule out other solutions (as is assuming every subset of the reals is measurable, assuming the Axiom of Determinacy, and it holds in Solovay's model). For more, see this question, this question, and this mathoverflow question.

Community
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usul
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    well you could aslo have $e^{nx}$ for positive $n$. Or $a^nx$. – Joe Tait Feb 13 '13 at 16:35
  • Ah, good point. I think the question is still interesting when we exclude that case though! – usul Feb 13 '13 at 16:37
  • As I just added, it also works for any number positive number instead of $e$. Still interesting excluding that though :( – Joe Tait Feb 13 '13 at 16:39
  • See @GEdgar's comment below. – copper.hat Feb 13 '13 at 16:41
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    I think the negation of the last one is only a consequence, because it is a statement about the subsets of $\mathbb R$. It cannot imply statements about arbitrary sets.... It follows from the fact that $\mathbb R$ has a $\mathbb Q$-basis. – N. S. Feb 13 '13 at 21:06
  • @N.S. oh, I see, thank you. I will edit the question. – usul Feb 13 '13 at 21:07

5 Answers5

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Once you have an isomorphism, there are plenty. Take two Hamel bases of $\mathbf{R}$, say $(b_i)$ and $(c_i)$, and define $f$ by $$ f(b_i) = e^{c_i}, $$ extending it by $\mathbf{Q}$-linearity to all of $\mathbf{R}$.

PS Of course I am implicitly using here the argument of @Micah and @spin.

Andreas Caranti
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Let $f:\mathbb{R} \to \mathbb{R}^+$ be an isomorphism. Then $g=\log f$ is an automorphism of $(\mathbb{R}, +)$: that is, it satisfies $$ g(x+y)=g(x)+g(y) \, . $$ This is Cauchy's functional equation. The only continuous (or even measurable) solutions are the trivial ones (which correspond to $f(x)=e^{cx}$), but there are also exotic solutions that require some version of the axiom of choice to construct — which would yield similarly exotic $f$s.

Micah
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  • Very helpful, thanks! My question now (last question?) is how "strong" is the axiom that such solutions exist, or the axiom that they don't? – usul Feb 13 '13 at 21:08
  • @usul: I'm sorry, I'm not enough of a set theorist to give a good answer to that question. (Though many people here are; you'd probably get good answers if you asked it as a separate question!) – Micah Feb 13 '13 at 21:15
  • no problem, a search turned up several relevant threads and I have linked to them in my edited question! – usul Feb 13 '13 at 21:20
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$e^{2x}$ is another one. In general if $\phi:\mathbb{R}\rightarrow\mathbb{R}$ is a group isomorphism, then $e^{\phi(x)}$ is another one.

One should also note that if $\phi$ is a group isomorphism between $(\mathbb{R},+)$ and $(\mathbb{R}_{>0},*)$, then $\log(\phi)$ is an automorphism of $(\mathbb{R},+)$. Now the question is to determine the group automorphisms of $\mathbb{R}$ as a group under addition

Amr
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$f:\mathbb{R}\rightarrow\mathbb{R}^{+} \text{ satisfying } f(x+y)=f(x)f(y)$, $f$ continuous, then $f(x)=e^{cx}$

Try to solve this problem Any Lie Group Homomorphism from $\mathbb{R}\rightarrow S^1$ is of the form $e^{iax}$ for some $a\in\mathbb{R}$ and every such homomorphism is smooth.

Myshkin
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Suppose that $G \cong H$ as groups and that $f: G \rightarrow H$ is an isomorphism. Then every isomorphism $G \rightarrow H$ is of the form $f \circ g$, where $g: G \rightarrow G$ is an isomorphism. So to determine every isomorphism $G \rightarrow H$, it suffices to find just one and then the rest are given by composing with automorphisms of $G$.

Because $x \mapsto e^x$ is an isomorphism $(\mathbb{R}, +) \rightarrow (\mathbb{R}_{>0}, \cdot)$, every isomorphism $(\mathbb{R}, +) \rightarrow (\mathbb{R}_{>0}, \cdot)$ is of the form $x \mapsto e^{\varphi(x)}$, where $\varphi$ is an automorphism of $(\mathbb{R}, +)$.

spin
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