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Let $G$ is a group and $H<G$ such that $|G-H|<\infty$. Prove that $|G|<\infty$.

Truthfully, there is a hint for it:

$H$ cannot be an infinite subgroup.

It is clear if $|H|<\infty$, since $|G-H|<\infty$ then $|G|<\infty$ and problem will be solved. But cannot understand why "$H$ cannot be an infinite subgroup". Thanks for your help.

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Look at the size of the cosets of $H$. –  Arthur Sep 6 '12 at 21:20
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2 Answers 2

up vote 11 down vote accepted

We assume that $H\neq G$ (otherwise take $G$ infinite to get a counter-example). Let $x_0\in G\setminus H$. Then $x_0 H\cap H=\emptyset $ as $H$ is a subgroup, and since $|G\setminus H|$ is finite, so is $x_0 H$ (as the map $a\mapsto x_0 a$ is a bijection). We deduce that $H$ is finite.

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why "and so is $x_0H$" ? can you please explain ? –  Belgi Sep 6 '12 at 21:25
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@Belgi There is a bijection between the elements of $H$ and the elements of $x_0H$ given by $x\rightarrow x_0x$. –  Arthur Sep 6 '12 at 21:27
    
@Belgi I've added the detail. –  Davide Giraudo Sep 6 '12 at 21:27
    
backslash is for the quotient or $-$ (minus) of sets ? –  Belgi Sep 6 '12 at 21:30
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I would suggest the following, which I believe is clearer: "[...], so is $x_0H$. As the map $a\mapsto x_0a$ is a bijection, we deduce that H is also finite." –  M Turgeon Sep 6 '12 at 22:13
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Either $|G-H|= 0$ or there exists an element $g \in G-H$. In the first case it is clear that $G=H$ and this would allow $G$ to be infinite. And in the second: Suppose H was not finite. Then $gH\cap H = \{1\}$ and hence $gH-\{1\} \subseteq G-H$ would be infinite contrary to your assumption. Hence $H$ is finite.

With $H$ being finite and $G-H$ being finite we can conclude that $G$ is also finite.

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At the first line, you shoul write $g\in G\setminus H$. Why would $1$ be in $gH$? –  Davide Giraudo Sep 6 '12 at 21:28
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