Idea for proving there are the same number of left cosets as right cosets of $H\leq G$ (for finite and infinite groups)? I'm trying to prove that there are the same number of left and right cosets of a subgroup $H$ of any group $G$ (finite or infinite) by finding a bijection mapping the left cosets of $H$ onto the right cosets of $H$. It seems a lot of people have asked about this problem, too. I'm thinking I could define $\phi:H\to aH$ by $h\mapsto ah$ and $\psi:H\to Hb$ by $h\mapsto hb$. Then after proving $\phi$ and $\psi$ are bijections, that would mean $\phi^{-1}$ exists and is a bijection, so then the composition $\psi\circ\phi^{-1}$ is a bijection from $aH$ to $Hb$, thus proving the collections of cosets have the same cardinality. Would this work?
(PS: I'm aware the function $\phi:aH\to Ha$ where $ah\mapsto ha^{-1}$ is a possibility)


I'm trying to prove that there are the same number of left and right cosets of a subgroup $H$ of any group $G$ (finite or infinite) by finding a bijection mapping the left cosets of $H$ onto the right cosets of $H$.

Well, either you want to show that left/right cosets have the same cardinality or the collections of left/right cosets have the same cardinality. These are different things.
You've correctly shown that $gH$ is equinumerous to $Hg'$ for any $g,g'\in G$. But this doesn't show that $G\backslash H=\{gH\ |\ g\in G\}$ is equinumerous to $G/H=\{Hg\ |\ g\in G\}$. For that you would consider the following function
$$f:G\backslash H\to G/H$$
$$f(gH)=Hg^{-1}$$
The most important thing you need to show is that it is well defined, i.e. $gH=g'H$ if and only if $Hg^{-1}=Hg'^{-1}$. Which I leave as an exercise. With that the inverse is simply given by $Hg\mapsto g^{-1}H$.
