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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.

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

$\mathbb{R^+}$(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.

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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?

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$(\mathbb{R}^+,.)$ will be the 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

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 $H=\mathbb{R}_{>0}$.

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