Show that ${\mathbb{R^\times}} \cong C_2 \times (\mathbb{R},+)$.

I know that the group $C_2$ is the cyclic group of order 2, and that $(\mathbb{R},+)$ is the group of reals under addition. However, I'm stumped on how to construct an isomorphism between the reals under multiplication and the the cyclic group times the reals under addition. Anyone who can give a rigorous argument would be highly appreciated.

  • $\begingroup$ Here's a hint. Call $G = C_2 \times (\mathbb{R},+)$. Recognize that the $C_2$ is effectively going to make two copies of $(\mathbb{R},+)$. And then, if we denote $C_2 = \{-1,1\}$ with operation as multiplication in $\mathbb{Z}$, we have that for any $g \in G$, taking $(-1,0) \circ g$ (where $\circ$ is the group operation), we move to the "other copy" of $(\mathbb{R},+)$. This is just like the positive and negative non-zero reals under multiplication. $\endgroup$ – Joe Apr 19 '18 at 20:55
  • $\begingroup$ The map is $(\pm 1,t)\to \pm \exp t$ in one direction. Here, we realize better $C_2$ as the multiplicative subgroup of $\mathbb R^\times$ with the elements $\pm 1$. It is also a map respecting the topologies (and differential structures). $\endgroup$ – dan_fulea Apr 19 '18 at 20:58

This is mainly a pattern recognition exercise. If you write out the definitions, among the things you are being asked to do is to find an inverse pair of functions $f$ and $g$ such that

  • $f(xy) = f(x) + f(y)$
  • $g(x+y) = g(x) g(y)$

You're expected to realize that you're familiar with two specific functions having these properties. When you try to build the isomorphism using them, the exercise follows rather straightforwardly.



Can you see how to construct an isomorphism between the group of POSITIVE reals under multiplication and $(\mathbb{R},+)$?

  • $\begingroup$ That's what I'm having trouble formulating. If you could show that, I could apply that to the idea above and have an isomorphism between $\mathbb {C_2}$ as well. $\endgroup$ – Mathmaniac Apr 19 '18 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.