# To show number of left cosets equals number of right cosets

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.

• I'm not sure if I understand you correctly. Are you saying that $aH \mapsto Ha$ is a bijection between left and right cosets? If so, that is incorrect. In fact this mapping is not even well defined - $aH=bH$ does not imply $Ha=Hb$. You're on the right track though, something quite similar will work. – Dániel G. Mar 27 '15 at 15:18
• No, I don't mean a bijection or any kind of map between the sets $aH$ and $Ha$. – levitt Mar 27 '15 at 15:24
• @levitt: In the moment you are trying to prove your claim purely set-theoretical ("If the equivalence classes of two equivalence relations on a set have all a fixed cardinality, then the cardinalities of the two sets of equivalence classes (i.e., quotient sets) are both the same."). If you use a tiny bit of group theory, you can give a bijection between the left and the right cosets. (Hint: Do you know a self-inverse bijection on $G$ that reverses orders in products?) – j.p. Mar 27 '15 at 16:05
• Yes, given a left coset $\Gamma$ of $H$ and an element $a\in\Gamma$ there is exactly one right coset $Ha$. But as you vary which element $a\in\Gamma$ you choose to represent the left coset $aH=\Gamma$, you will in general obtain different right cosets $Ha$. It's important to understand a single left coset $\Gamma$ can be written as $aH$ for any choice of $a\in\Gamma$, and the corresponding right coset $Ha$ will depend on which $a$ you picked. So your argument does not set up a correspondence between left cosets and right cosets. – anon Mar 27 '15 at 17:39
• @levitt: In David's answer you can see the missing part of your proof (your map was not "well-defined", and only for normal subgroups you can prove that). Knowing the trick, one can also prove the statement more direct by defining $X^{-1} = \{x^{-1} | x\in X\}$ for subsets $X$ of $G$. Restricting this function (which is not depending on any representative/choice, hence well-defined) to the left cosets G/H you have to show (1) the image of a left coset is a right coset and (2) for every right coset there is a left coset mapped onto it. Try this proof (and look at the example $S_3$ from chat). – j.p. Mar 28 '15 at 16:06

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.

This is the right idea, but it only shows that there are at least as many right cosets as left cosets. You need to state the argument in the other direction as well.

• You are right. We need an argument in the other direction also. Thanks. – levitt Mar 27 '15 at 15:27
• You also need to be a bit more precise with the argument. As relep points out, the way you stated it is a bit confusing. Something like: each $a\in xH$ is in one an only one $Hy$ is a bit more clear. – Tim Raczkowski Mar 27 '15 at 15:41
• Thanks for pointing that out, I should have been a little precise. Your statement " each $a\in xH$ is in one an only one $Hy$ " isn't immediate to me. I'm trying to sink it in. I was talking about $a$ which is $a.1$. – levitt Mar 27 '15 at 15:56
• @TimRaczkowski: How do you think the current approach can be made into a working proof? (see my comments to the questions) If you don't want to solve levitt's homework now, please post it in a week. Thanks! – j.p. Mar 27 '15 at 16:18
• You raise a good point, and I must admit I haven't thought about the details. I'll get back to this. – Tim Raczkowski Mar 27 '15 at 16:25