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Let $H$ be the subgroup {-1,1} in $G=\mathbb{R}^{*}$. Is $G/H$ isomorphic to $G$?

I know the answer is "no," but I am not really sure why. Is it because $a$ and $-a$ generate the same coset? If that's the case, if we just looked at the positive reals, would we then get an isomorphism?

Thanks in advance

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If you just looked at the positive reals, $\{-1,1\}$ would not be a subset. –  Jonas Meyer Feb 8 '12 at 4:17

2 Answers 2

up vote 2 down vote accepted

Each coset $\{a,-a\}$, with $a> 0$, has a square root, $\{\sqrt a,-\sqrt a\}$, and this is not true in $G$. Also, squaring is injective in $G/H$, but not in $G$. Note that $x\mapsto x^2$ is a surjective homomorphism from $G$ to $\mathbb R_{>0}$ with kernel $H$, so $G/H\cong \mathbb R_{>0}$. (You could also use $x\mapsto |x|$, or define a map explicitly from the quotient by $\{a,-a\}\mapsto a$ with $a>0$.)

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+1 too. Nice Answer, beat me to it! –  user21436 Feb 8 '12 at 4:25
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And as Nicky Hekster points out, the second point, that squaring is not injective, has the advantage that it would also apply if we replaced $\mathbb R^*$ with $\mathbb Q^*$, whereas the existence of square roots argument doesn't apply for $\mathbb Q^*/\{\pm1\}$. For the same reason, using absolute value rather than squaring would be better for establishing that $\mathbb Q^*/\{\pm 1\}\cong\mathbb Q_{>0}$. –  Jonas Meyer Feb 8 '12 at 15:19

$\mathbf R^*$ contains a non-identity element whose square is $1$. Is the same true of your quotient?

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A Quickie! +1 $$ –  user21436 Feb 8 '12 at 4:25
    
And that argument applies also to $\mathbb Q^*$, where square roots (see Jonas' remarks above) do not exist in general –  Nicky Hekster Feb 8 '12 at 14:03

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