The factor group $\mathbb{R}^{*}/\{-1,1\}$ is isomorphic to $\mathbb{Z}_{2}$. True or False.

Question

I have been told that the answer for this question is False. And I'm trying to understand why.

For what I understand so far, $\mathbb{Z} _2$ is an abelian group. I also understand that for a group isomorphism of two groups, say:

$\phi: G\rightarrow H$

*If $G$ is abelian, then $H$ is abelian.

*If $G$ is cyclic, then $H$ is cyclic.

*If $G$ has a subgroup of order $n$, then $H$ has a subgroup of order $n$.

But I'm not entirely sure if it might be related to the question or not. However, any hint would be really helpful. Thank you.

Answer 1

Hint: $\mathbb{Z}_2 $ is finite, while $\mathbb{R}^*/\{-1,1\}$ is infinite.

Answer 2

One issue to think about is , as user 1337 pointed out, that the cardinalities must agree: Computing the cosets we get (use = for equivalence): $$a=b \Leftrightarrow ab^{-1}\subset\{-1,1\}$$ This means $a$=$b$ iff $a=b$ or $a=-b$. So the cosets are all the pairs $(a,-a)$, which is uncountably-infinite. You can see too, that moding out an infinite group by a finite group will not give you a finite quotient group. Also, as an uncountable group, it is not, cannot be, cyclic. Once you choose a representative from $(-a,a)$, say a, your group is isomorphic to the positive Reals.