TITLE: $|G-H|<\infty$ so $|G|<\infty$
QUESTION [8 upvotes]: 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.

REPLY [11 votes]: 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.