# Group isomorphism $h:(\mathbb R,+)\to (\mathbb R^+,\times)$ that is not an exponential function.

Let $$\mathbb{R}^+$$ denote the set of positive real numbers. Group isomorphism $$h_b:(\mathbb R,+)\to (\mathbb R^+,\times)$$ can be given by the exponential function: $$h_b(r)=b^r$$, where $$b$$ is a positive real number and is not $$1$$. Moreover, after we define the group automorphism $$A_x:(\mathbb R^+,\times)\to (\mathbb R,+)(r\mapsto xr)$$, the set of all exponential functions(with positive base) are related by $$h_k(r) = h_bA_rh_b^{-1}(k)$$, with $$A_r$$ the automorphism defined above.

I start to wonder that is this the only possible way to construct the isomorphism? Are there any isomorphisms $$i:(\mathbb R,+)\to (\mathbb R^+,\times)$$ different from the exponential function?(It needs not to be continuous) But what about the case if we are looking for a continuous isomorphism?

• Do you want the isomorphism to be continuous? – D. Brogan Mar 26 at 1:13
• Thank you for the comment. I've added that to my question. I would like to know both cases: when we are looking for a topological group isomorphism, and when we only require it to be a group isomorphism. – William Sun Mar 26 at 1:20
• I should point out this question too, which is very informative relating to this questions (and is essentially a duplicate modulo composition with $\exp$). – Dan Rust Mar 26 at 1:37

First we will construct an isomorphism $$(\mathbb R,+)\to(\mathbb R^+,\times)$$ which is not an exponential. Let $$V$$ be $$\mathbb R$$ as a vector space over $$\mathbb Q$$ and let $$\{v_\alpha\in\mathbb R : \alpha\in A\}$$ be a basis for $$V$$. Then any permutation $$\sigma$$ of $$A$$ induces linear operator $$T=T_\sigma$$ on $$V,$$ and so $$T:(\mathbb R,+)\to(\mathbb R,+)$$ is a group isomorphism. $$T$$ is not continuous (unless $$\sigma$$ is the identity), and so $$\exp\circ T:(\mathbb R,+)\to(\mathbb R^+,\times)$$ is an isomorphism which is not continuous, hence not an exponential.
Now suppose that $$\phi:(\mathbb R,+)\to(\mathbb R^+,\times)$$ is a continuous group isomorphism. Write $$b=\phi(1).$$ Then $$\phi(-1)=b^{-1}$$ and induction gives $$\phi(k)=b^k$$ for all $$k\in \mathbb Z.$$ Furthermore, for $$p/q\in\mathbb Q,$$ we have $$b^p = \phi(p) = \phi((p/q)\cdot q) = \phi(p/q)^q,$$ so that $$\phi(p/q)=b^{p/q}.$$ We then use continuity to extend $$\phi$$ to the reals, giving that $$\phi(x)=b^x$$ for all $$x,$$ so $$\phi$$ is an exponential with base $$b.$$
Let $$\beta = \{b_{\lambda} \in \mathbb{R} \mid \lambda \in \Lambda\}$$ be a $$\mathbb{Q}$$-basis for $$\mathbb{R}$$ and let $$\phi \colon \beta \to \beta$$ be a permutation of $$\beta$$. Then $$\phi$$ induces a group isomorphism $$h_\phi\colon \mathbb{R} \to \mathbb{R}$$ by linear extension. The composition $$\exp \circ h_\beta$$ is then a group isomorphism as well.