Multiplicative groups of nonzero reals and pos. reals

[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.

 

[fresh_42]

As long as you do not tell us how you imbed $\mathbb{R}^{**}$ in $\mathbb{R}^*$ as a normal subgroup we cannot say anything. We need an embedding, not a surjection.