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.
Best Answer
Observation: if $ah \in aH$, then $(ah)^{-1} = h^{-1}a^{-1} \in Ha^{-1}$.
This suggests the "mapping" $gH \mapsto Hg^{-1}$. Before we show it is bijective, we must actually show it is a FUNCTION, i.e., that it is "well-defined" (constant on cosets).
Now $gH = g'H \iff g^{-1}g' \in H$. Since $H$ is a subgroup, $(g^{-1}g')^{-1} = g'^{-1}g \in H$.
This says that $g'^{-1}(g^{-1})^{-1} \in H$, which happens if and only if $Hg^{-1} = Hg'^{-1}$.
Thus if $\phi(gH) = Hg^{-1}$, we see that for any coset $g'H = gH$, that:
$\phi(g'H) = Hg'^{-1} = Hg^{-1} = \phi(gH)$, that is, $\phi$ is constant on cosets (independent of the "representative element", $g$). So $\phi$ is a well-defined function on left cosets. Now we can show $\phi$ is bijective, which is straight-forward.