Normal subgroup, quotient group, isomorphism.

Question

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.

Answer 1

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.

Answer 2

Notation: I'll use $\mathbb R^\ast$ and $\mathbb R^+$ for the nonzero and positive real numbers.

About part (a), I don't think the way you prove that the group is abelian makes sense. You're showing that all commutators vanish and to do this you switch the order of multiplication, but that's what abelian means: that you can switch the order of multiplication.

Personally I don't really think you need to say anything other than "multiplication of real numbers is commutative so $\mathbb R^\ast$ is abelian".

As for (b), try looking at the absolute value of a real number, it gives a map $\mathbb R^\ast \to \mathbb R^+$. Is it a homomorphism? What is its kernel?