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.

  • 1
    $\begingroup$ Part (a) is pretty much correct, but could be cleaned up a bit. I would say something like "The group $R^*$ is abelian since multiplication of real numbers is commutative. Hence $H$ is normal since, for any $g \in R^*$ and $h \in H$, $ghg^{-1} = gg^{-1}h = h \in H$." $\endgroup$ – manthanomen Apr 22 '14 at 3:47

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?

  • $\begingroup$ The OP shows that the group is normal by using the fact that we commute. I see no error. $\endgroup$ – Vladhagen Apr 22 '14 at 3:39
  • $\begingroup$ @Vladhagen: No, read it more carefully. The OP states that if the group is abelian then all subgroups are normal, which is true. The OP then goes on to show the group is abelian in the awkward manner I described. $\endgroup$ – Guest Apr 22 '14 at 3:41
  • $\begingroup$ Note the phrase: "Then we can see that R^* is abelian, since..." $\endgroup$ – Guest Apr 22 '14 at 3:42
  • $\begingroup$ Sure. You win. I am not sure whether this is just because the OP is not perfect at English or whether/she actually thought this was necessary. The wording is awkward, I will give you that. $\endgroup$ – Vladhagen Apr 22 '14 at 3:43

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.


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