I need to prove that the number of left cosets of a subgroup $H$ in $G$ is equal to that of right cosets in general(even when $|G|$ is not finite).
Here is my argument. For every $aH$ in $G$, the element $a$ that is in $aH$ is also in exactly one right coset $Ha$ and no more. Is this enough to prove that no. of left cosets equals of H to no. of it's right cosets? Please point out if there are any flaws in my statement. Thanks
EDIT: I'll try to make my statement a little more clear. Consider a coset $aH$. It has $a$ in it which is mapped to identity $e$ in $H$. Now similarly $Ha$ also has $a$ in it and no other right coset contains $a$. Of all the left cosets of $H$, $aH$ is the only one which has $a$ in it and of all the right cosets $Ha$ is the only one which has $a$ in it. $a$ is like a representative element for $aH$ and $Ha$. Similarly I can argue for each left coset of $H$. This to my understanding should prove that there are as many right cosets as left cosets. Now I can make a similar argument in the other direction which shows that there are as many left cosets as there are right cosets.
EDIT: The above reasoning is not correct, please see the answer and comments below.