Is $e^x$ the only isomorphism between the groups $(\mathbb{R},+)$ and $(\mathbb{R}_{> 0},*)$?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

$f:\mathbb{R}\rightarrow\mathbb{R}^{+} \text{ satisfying } f(x+y)=f(x)f(y)$, $f$ 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.

Answer 4

$e^{2x}$ is another one. In general if $\phi:\mathbb{R}\rightarrow\mathbb{R}$ is a group isomorphism, then $e^{\phi(x)}$ is another one.

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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

Answer 5

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 $(\mathbb{R}, +)$.