Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Determine all subgroups of $\mathbb{R}^*$ that have finite index.

How can I able to solve this? Can anybody help me please? Thanks for your time.

share|improve this question
1  
Do you know any subgroups of $\mathbb{R}^*$? –  Zev Chonoles Apr 19 '13 at 6:04

3 Answers 3

Suppose that $G$ is a group of finite index in the multiplicative group $\mathbb R^\times$, then $G\cap(0,\infty)$is a finite index subgroup of $\mathbb R_{>0}$. Now $\mathbb R_{>0}$ is a divisible group, so $\mathbb R_{>0}/G\cap \mathbb R_{>0}$ is a finite divisible abelian group. This means this quotient is a trivial group, and therefore $G\cap \mathbb R_{>0}=\mathbb R_{>0}$.

Can you see what $G$ must be now?

share|improve this answer
    
$(\mathbb{R}^+,.)$ will be the answer.am I right? –  haltui Apr 19 '13 at 6:12
    
Well, there are two finite index subgroups. –  Mariano Suárez-Alvarez Apr 19 '13 at 6:15
    
what is the other? –  haltui Apr 19 '13 at 6:21
    
Well, think about this for a while. Longer. –  Mariano Suárez-Alvarez Apr 19 '13 at 6:23
    
ok.thanks for your help.still I cant find nothing but I shall try more. –  haltui Apr 19 '13 at 6:24

$\mathbb{R^+}$(set of all positive nonzero real numbers) is the only proper subgroup of $\mathbb{R^*}$ of finite index.

To prove first part let us assume that $\mathbb{R^*}$ has a proper subgroup $H \neq \mathbb{R^+}$ such that $[\mathbb{R^*} : H] = n$ is finite. Thus we have $(xH)^n = x^n H = H$ for each $x\in \mathbb{R^*}$. Thus $x^n \in H$ for each $x\in \mathbb{R^*}$. Now let $x \in \mathbb{R^+}$, then $ x = (\sqrt[n]{x})^n \in H$. Thus, $\mathbb{R^+} \subset H $. Since $H\neq \mathbb{R^+}$ and $ \mathbb{R^+} \subset H$, we may conclude that $H$ must contain a negative number say $-y$ for some $y\in \mathbb{R^+}$. Since $\frac{1}{y}\in \mathbb{R^+}\subset H $ and $-y\in H$, and since $H$ is closed under multiplication we conclude that $-y(\frac{1}{y}) = -1 \in H$. Since $H$ is closed and $ \mathbb{R^+} \subset H$, and $-1 \in H$.We conclude that $\mathbb{R^-} \subset H$, where $\mathbb{R^-}$ is the set of all nonzero negative real numbers. Since $\mathbb{R^+}\subset H $ and $\mathbb{R^-}\subset H $ , we conclude that $H =\mathbb{R^*} $, which is a contradiction since $H$ is a propser subgroup of $\mathbb{R^*} $. Hence $\mathbb{R^+}$ is the only proper subgroup of $\mathbb{R^*}$ of finite index.

share|improve this answer

First, notice that $|\cdot | : x \mapsto | x|$ and $\operatorname{sign} : x \mapsto \left\{ \begin{array}{cl} 1 & \text{if} \ x>0 \\ -1 & \text{if} \ x<0 \end{array} \right.$ are surjective homomorphisms from $\mathbb{R}^{\times}$ to $\mathbb{R}_{>0}$ and $\mathbb{Z}_2$ respectively, with $\text{ker}(\operatorname{sign})=\mathbb{R}_{>0}$. In particular, $\mathbb{R}^{\times} \simeq \mathbb{Z}_2 \times \mathbb{R}_{>0}$ (for the isomorphism, take $x \mapsto (\operatorname{sign}(x),|x|)$) and $\mathbb{R}_{>0}$ is of index two in $\mathbb{R}^{\times}$.

Let $H$ be a subgroup of finite index in $\mathbb{R}^{\times}$; then $|H|$ is of finite index in $\mathbb{R}_{>0}$. But $\mathbb{R}_{>0}$ is an infinite divisible abelian group, so $|H|=\mathbb{R}_{>0}$ hence $\mathbb{R}_{>0} \subset H$.

You deduce that $[\mathbb{R}^{\times}: H] \leq [\mathbb{R}^{\times}: \mathbb{R}_{>0}]=2$, that is $H= \mathbb{R}^{\times}$ or $H=\mathbb{R}_{>0}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.