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 $h \in H$, $ghg^{-1} = gg^{-1}h = h \in H$." $\endgroup$ Commented Apr 22, 2014 at 3:47
  • $\begingroup$ Maybe I'm just missing it, but I don't see how $H$ is a subgroup by hypothesis. It might be extremely easy to prove, but no hypothesis was stated which shows $H$ is a subgroup. $\endgroup$
    – IAAW
    Commented Feb 6, 2023 at 6:10

2 Answers 2


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.


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
    Commented Apr 22, 2014 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
    Commented Apr 22, 2014 at 3:41
  • $\begingroup$ Note the phrase: "Then we can see that R^* is abelian, since..." $\endgroup$
    – Guest
    Commented Apr 22, 2014 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
    Commented Apr 22, 2014 at 3:43

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