Let $R^{*}$ be the group of nonzero real numbers under multiplication and let $R^{+}$ be the group of positive numbers under multiplication.
Prove
(a) $\{-1,1\}$ is a normal subgroup of $R^{*}$.
(b) The quotient group $R^{*}/\{-1,1\}$ is isomorphic to $R^{+}$.
Attept: Please check my proof.
Proof (a): Let $H = \{-1,1\}$. Then by hypothesis $H$ is a subgroup of $R^*$. We know that if a group is abelian, then all of its subgroups are normal. Then we can see that $R^{*}$ is abelian, since let g be an an element of $R^{*}$ not equal to zero, and $h$ be an element of $H$. Then $ghg^{-1} = hgg^{-1} = h$, which is an element of $H$. So since $R^*$ is abelian, $H$ is a normal subgroup.
In addition we can see every left coset is also a right coset.
proof(b): Can anyone please help me? I don't know how to start part (b).
Any help would be really appreciated.
Thank you.
Best Answer
On b, I would use the surjective homomorphism $abs$ that takes $R^{*}$ to $R^+$ by absolute value. Since $|rs| = |r|\cdot|s|$, then this map is indeed a homomorphism that is pretty clearly surjective.
By the first isomorphism theorem, since $\ker(abs)$ is $\{-1,1\}$, then we have the needed isomorphism.