Can we describe all group isomorphisms from $(\mathbb R ,+)$ to $(\mathbb R^+ , .)$ ? I have tried that if $f$ is such an isomorphism , then $f(x)>0$ , and $f(r)=(f(1))^r , \forall r \in \mathbb Q$ , but nothing else . Please Help
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Yes and no. As an additive (abelian) group is a bit like a vector space (more technically, a $\mathbb{Z}$-module, a module over a PID), we can consider the isomorphisms as one-to-one correspondances of free bases between these two. Perhaps a simpler way to think about it is to use one particular isomorphism from $\mathbb{R}$ with $+$ to $\mathbb{R}^+$ with $\cdot$, and compose that with all automorphisms of $\mathbb{R}$ with $+$. – hardmath Mar 03 '15 at 16:17
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1See Is the group isomorphism $\exp(\alpha x)$ from the group $(\mathbb{R},+)$ to $(\mathbb{R}_{>0},\times)$ unique? or Is $e^x$ the only isomorphism between the groups $(\mathbb{R},+)$ and $(\mathbb{R}_{> 0},*)$? – Martin Sleziak Dec 01 '16 at 03:37
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It's not true that they are all of that form: think of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ and take any automorphism $g$ of it$^1$, then $f(t) = e^{g(t)}$ is such an isomorphism.
On the other hand, it's not too hard to show that any continuous isomorphism must be of the form you mentioned. This follows, after taking logarithms, from the fact that the only continuous additive functions $h: \mathbb{R} \to \mathbb{R}$ are the ones of the form $h(x) = c x$.
$^1$ For example, take $\{x_1, \ldots, x_n\}$ linearly independent over $\mathbb{Q}$, complete that set to a basis $\beta$ of $\mathbb{R}$, pick any matrix $A$ in $GL(n,\mathbb{Q})$ and define $g$ so that it is given by $A$ on the span of $\{x_1, \ldots, x_n\}$), and fixes all other elements of $\beta$.
Omar Antolín-Camarena
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$\newcommand{\R}[0]{\mathbb R}$$\newcommand{\Z}[0]{\mathbb Z}$$\newcommand{\Q}[0]{\mathbb Q}$This must be a duplicate. Anyway, one of them is $f(x) = e^{x}$.
And then you obtain all others by composing with any automorphism of the group $(\R, +)$, that is the isomorphisms $g : (\R, +) \to (\R, +)$. And these are obtained, although in a highly non-constructive way, by taking a Hamel basis of $\R$, and considering all $\Q$-vector space isomorphisms $\R \to \R$.
This is because if $g : (\R, +) \to (\R, +)$ is a group isomorphism, then $g$ is also a $\Q$-vector space isomorphism, as for $p, q \in \Z$, with $q \ne 0$, we have for all $x \in \R$ $$q g\left(\frac{p}{q} x\right) = g(p x) = p g(x),$$ so that $$g\left(\frac{p}{q} x\right) = \frac{p}{q} g(x).$$
The reason being that given any isomorphism $f : (\R, +) \to (\R^{+}, \cdot)$, you obtain an isomorphism $\log \circ f : (\R, +) \to (\R, +)$.
Andreas Caranti
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