In this question, it is asked whether the additive group of a ring can be isomorphic to its group of units. The answer is yes: the trivial ring is the easiest example. However, the top-voted answer asserts that $\mathbb R\times \mathbb Z/2\mathbb Z$ (under addition) is isomorphic to $(\mathbb R\times \mathbb Z/2\mathbb Z)^\times$. Why is this the case?
I can verify that the units of the ring $\mathbb R\times \mathbb Z/2\mathbb Z$ are of the form $(x,1)$, where $x\neq0$, so that there is an isomorphism between $(\mathbb R\times\mathbb Z/2\mathbb Z)^\times$ and $\mathbb R^\times$ which takes $(x,1)$ to $x$, but I cannot find an isomorphism $\mathbb R^\times\to\mathbb R\times \mathbb Z/2\mathbb Z$.
The isomorphism $\mathbb{R}\times\mathbb{Z}_2\to\mathbb{R}^\times$ is given by $(x,s)\mapsto(-1)^s e^x$.
The "parity bits" of $\mathbb{Z}_2$ corresponds to two connnected components of $\mathbb{R}^\times$ (the positive or negative numbers), and exponentials famously turn addition into multiplication, i.e. $e^{x+y}=e^x\cdot e^y$.
The inverse isomorphism $\mathbb{R}^\times\to\mathbb{R}\times\mathbb{Z}_2$ is then $r\mapsto(\ln|r|,\,\mathrm{sgn}\,r)$.