WTS is that [itex]\mathbb R^*/N \ \cong \ \mathbb R^{**}[/itex] where [itex]N = (-1, 1)[/itex](adsbygoogle = window.adsbygoogle || []).push({});

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

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

[itex]f:\mathbb R^*\rightarrow\mathbb R^{**}[/itex]

[tex]f(x)=|x|,[/tex] is onto by construction. clearly a homomorphism

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

part 2

[itex]ψ:\mathbb R^*\rightarrow\ \ N[/itex]

[tex]ψ(x)=1 \ if \ x>0 \ and \ ψ(x)=-1 \ if \ x<0[/tex]

by same thm, [itex]\mathbb R^*/\mathbb R^{**} \cong \mathbb Z/2\mathbb Z[/itex]

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Multiplicative groups of nonzero reals and pos. reals

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**