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Let $\mathbb R^*$ be the group of all non zero real number under multiplication and $\,G^*$ be the subgroup of $\mathbb R^*$ consisting of all squares of reals.

What is the order of the quotient group $\;\dfrac{\mathbb R^*}{G^*}$?

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Square what? Like, any number that can be written as a square of a real number? That seems like a fancy way of saying "positive reals". Maybe squares of rational numbers? – rschwieb Mar 22 '13 at 14:06
$G^*={{x^2/x in R^*}}$ @rschwieb – prasad Mar 22 '13 at 14:11
up vote 4 down vote accepted

Write $\;G^* = \{g: g = x^2\mid x \in \mathbb R^*\}$

For each $g\in G^*$, there are two $x_i \in \mathbb R^*$ that solve $g = x_i^2,\;$ exactly one of which is also in $G^*$.

How does this relate to the number of cosets of $\,G^* \in \mathbb R^*$, recalling that the cosets of $\,G^* \leq \mathbb R^*\,$ necessarily partition $\,\mathbb R^*\,?\;$ Then recall $$\large\left[\mathbb R^*: \,G^*\right] = \left| \frac{\mathbb R^*}{G^*}\right|$$

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user59489: Does this help clear things up? – amWhy Mar 22 '13 at 14:26
two cosets right? so G* is a normal subgroup right? – prasad Mar 22 '13 at 14:27
thank you @amWhy – prasad Mar 22 '13 at 14:29
Yes, indeed, and you're welcome. – amWhy Mar 22 '13 at 17:02

Hint: What numbers are not squares in $\mathbb{R}^\ast$? Show that if $x\in\mathbb{R}^\ast$, then either $x$ or $-x$ is a square.

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Consider $\phi: \mathbb{R}^\ast \to \mathbb{R}^\ast$ given by $\phi(x)=x/|x|$. Then $\phi$ is a homomorphism. We have $\ker \phi=G^\ast$ because a real number is a square iff it is positive. We also have $\text{im}\, \phi=\{-1,1\}$, which means $G^\ast$ has index 2.

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+1, definitely the proof I prefer, as the quotient group in the question suggests the use of the first isomorphism theorem. – Andreas Caranti Mar 22 '13 at 20:01
+1 for looking elegant. – Babak S. Apr 3 '13 at 7:30

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