# 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)

• Hi, for finite groups this works. If all cosets have the same cardinality $k$, while the group has $n$ elements in total, then the number of left cosets must equal $n/k$ (because every element is in exactly one left coset) and the number of righ cosets must also equal $n/k$ because each element is in exactly one right coset. However for infinite groups the number $n$ does not exist and this approach does not work in that same way. Commented May 13, 2022 at 7:26
• The more general problem is that you now make a bijection between a single left coset and a single right coset and the thing that you feed into the function (and get out of it) are still group elements. You want a bijection between the set of left cosets and the set of right cosets, where the thing you put in the function is a coset. This is indeed not so easy to find. I will think about it. Commented May 13, 2022 at 7:28
• Ah, that makes sense. I was afraid I was only mapping to a single coset. Could I express the set $L$ of left cosets of $H$ as $L=\{gH:g\in G\}$ and the set $R$ of right cosets of $H$ as $R=\{Hg:g\in G\}$? Then create a function $\phi: L \to R$ where $gH \mapsto Hg^{-1}$? Commented May 13, 2022 at 7:39
• Yes! Yes! I wrote that as an answer and only later saw that you commented, otherwise I just would have commented that indeed that is what you should do. Commented May 13, 2022 at 7:41

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!

• Thank you! Although I did write your function $\phi$ in the comments it was worth it to consider that I need to show it is well defined. Commented May 13, 2022 at 7:43
• Does the well defined part basically mean that some $gH$ maps to exactly one right coset, I.e. $\phi$ is indeed a function? Commented May 13, 2022 at 7:46
• Thank you! For surjectivity, would this work? Suppose $Hg\in R$. Then $g\in G,$ meaning $g^{-1}\in G$ as well, so there exists $g^{-1}H\in L.$ Since $\phi(g^{-1}H)=Hg,$ $\phi$ is surjective. Commented May 13, 2022 at 8:00
• Yes that works! Commented May 13, 2022 at 8:01
• There is a mistake in this line 'By definition of right cosets, for all $h \in H$, $ha^{-1} = hb^{-1}$'. This is not true. What the definition right coset gives you is that there exists a $h \in H$ such that $a^{-1} = hb^{-1}$ or, more generally, for each $h_1 \in H$ there exists a $h_2 \in H$ such that $h_1a^{-1} = h_2b^{-1}$. But these $h_1, h_2$ need not be equal Commented May 13, 2022 at 8:51

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

• @Vincent ops, there should've been "$Hg^{-1}$". Fixed. With that it works for non-abelian as well. Commented May 13, 2022 at 8:01