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

Question

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?

Answer 1

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 $b.$

Answer 2

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.