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.

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# Multiplicative groups of nonzero reals and pos. reals

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