# Multiplicative groups of nonzero reals and pos. reals

1. Dec 3, 2011

### Bachelier

WTS is that $\mathbb R^*/N \ \cong \ \mathbb R^{**}$ where $N = (-1, 1)$

then prove that $\mathbb R^*/\mathbb R^{**} \ is \ \cong \ to \mathbb Z/2\mathbb Z$

So the best answer in my opinion is to construct a surjection and use the first iso thm.

$f:\mathbb R^*\rightarrow\mathbb R^{**}$

$$f(x)=|x|,$$ is onto by construction. clearly a homomorphism

$Ker \ (f) = N$, hence $\mathbb R^*/N \ \cong \ \mathbb R^{**}$

part 2

$ψ:\mathbb R^*\rightarrow\ \ N$

$$ψ(x)=1 \ if \ x>0 \ and \ ψ(x)=-1 \ if \ x<0$$

by same thm, $\mathbb R^*/\mathbb R^{**} \cong \mathbb Z/2\mathbb Z$

because it has 2 elements one of each is the identity.