Idea for proving there are the same number of left cosets as right cosets of $H\leq G$ (for finite and infinite groups)?

Question

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)

Answer 1

As I wrote in the comments you showed that all cosets (left and right) have the same cardinality, but your question is about something else: the set of all left cosets having the same cardinality as the set of all right cosets.

You are absolutely right in saying that what you want to do is create a bijection $\phi$ from the set of left cosets to the set of right cosets.

Here is one that works. Given a left coset $gH$ we define the right coset $\phi(gH)$ by:

$$\phi(gH) = Hg^{-1}$$

One thing that you should check is that it is well defined. That means:

if $gH$ and $qH$ are the same left coset but just written in a different way then $\phi(gH)$ and $\phi(qH)$ must obviously be the same right coset (possibly written in a different way).

So from the way I introduced the function one would have $\phi(qH) = Hq^{-1}$. So what we want to show is that

Whenever $gH = qH$ we have $Hg^{-1} = Hq^{-1}$.

The general principle is this (you will see this more often): we have a function that takes as input a set and gives as output an other thing (in this case also a set, but in other applications it could have been a number). Now in order to describe the function, or say how to compute it we write something like this 'take a random element $g$ in the input and use it in this and this way to create the output'. The question that then must be checked by the reader (the well definedness) is that the outcome does not depend on which random element of the set we had chosen, i.e. the answer stays the same if we would have taken $q \in gH$ in the role of $g$.

Only once we have verified that, we can say that we really have a map on the set of cosets. The next step is then of course to see that it is a bijection.

Good luck!

Answer 2

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 $Hg\mapsto g^{-1}H$.