In lecture was defined for groups H and its cosets:
$ a \sim_L b $, if $ aH = bH$ and $ a \sim_R b $, if $ Ha = Hb $
For finite groups one can find $a,b \in H$ such that $a \neq b$, $aH = bH \neq H$
But I wonder now, were this also possible for some not finite groups? Beause spontanously I don't see how.
As always thanks in advance for any comments, hints or answers.
Let the group in question be $\Bbb Z$, which is infinite, and let the subgroup be $2\Bbb Z$. Then $1 + 2\Bbb Z = 5 + 2\Bbb Z\neq 2\Bbb Z$, but $1 \neq 5$, so $a = 1, b = 5$ works.
Also note that you cannot find such $a, b$ for all finite groups. Specifically, for any prime $p$, the cyclic group with $p$ elements has no subgroup $H$ that allows you to find such $a, b$.