So I've been working on abstract algebra out of John B. Fraleigh's 3rd edition text. In the exercises of chapter 11, I came upon a question which I cannot even begin to solve.

"Show that there are the same number of left as right cosets of a subgroup H of a group G, that is, exhibit a one-to-one map of the collection of left cosets onto the collection of right cosets. (Note that this result is obvious by counting for finite groups. Your proof must hold for any group.)"

The only idea that I had was using a map $\phi : coset_{left} \rightarrow coset_{right}$ by $aH\phi = Ha$, but this seems far too easy. What thought process is wrong here? And how is this accomplished?

  • $\begingroup$ I'm working on this problem now. Is it possible to come up with one bijection $\phi: H \to aH$ with $\phi(h)=ah,$ and another bijection $\psi: H \to Hb$ with $\psi(h)=hb$, and then assert that $\phi\circ\psi^{-1}$ is a bijection mapping $aH$ to $Hb$? $\endgroup$ May 13, 2022 at 7:06

3 Answers 3


The map you propose is not well-defined.

To be well-defined, it must be true that if $aH=bH$, then $\phi(aH)=\phi(bH)$, or $Ha=Hb$.

Here's a counterexample: Take $G=S_3, H=\langle (1\ 2)\rangle, a=(2\ 3), b=(1\ 3\ 2)$.

Then $aH=\{(2\ 3), (1\ 3\ 2)\}=bH$, but we have $$Ha = \{(2\ 3),(1\ 2\ 3)\}\neq \{(1\ 3\ 2), (1\ 3)\} = Hb$$

Now, for your proof consider the function from the left to the right cosets defined by $$\varphi(aH)=Ha^{-1}$$

To show it is a bijection, you must check it is well-defined, injective, and surjective.

  • 1
    $\begingroup$ That hint is pretty clever. $+1$ $\endgroup$ Aug 4, 2015 at 4:06

Either the number of left cosets or the number of right cosets of $H$ in $G$ is equal to the number of members of $G$ divided by the number of members of $H$.

To see this, notice that

  • Every member $a\in G$ is a member of some right coset of $H$ since it is a member of $Ha$, and similarly for left cosets; and
  • Two distinct right cosets cannot intersect each other; nor can two left cosets. To see this, suppose $c\in Ha\cap Hb$. then $c=h_1 a = h_2 b$ for some $h_1,h_2\in H$. Consequently $a=h_1^{-1} h_2 b$ so $a\in Hb$ and similarly $b\in Ha$. Every member of $Hb$ belongs to $Ha$ and every member of $Ha$ belongs to $Hb$, so they are the same set.
  • Every coset $Ha$ has just as many members as $H$ itself. To show that, it is enough to show that $h\mapsto ha$ is a one-to-one and onto function. That it is onto follows instantly from the definition of $Ha$. To see that it is one-to-one, suppose $h_1 a=h_2 a$. Then $(h_1 a)a^{-1} = (h_1 a) a^{-1}$, so $h_1=h_2$.

Thus the right cosets of $H$ all have the same number of members and do not intersect each other and fill up all of $G$. So $|G|/|H|$ is the number of right cosets, and the same argument shows that that is the number of left cosets.

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    $\begingroup$ Does this argument apply to infinite groups? $\endgroup$
    – coldnumber
    Aug 4, 2015 at 3:59
  • $\begingroup$ @coldnumber : Yes, except there you can't speak of division. You must instead find a bijection between the set of all left cosets and the set of all right cosets. Now that I think of it, I'm not sure whether the axiom of choice is needed for that. But no axiom of choice is needed to show that the right cosets form a pairwise disjoint and exhaustive collection of subsets, as do the left cosets, and all cosets, either right or left, have the same cardinality as $H$. ${}\qquad{}$ $\endgroup$ Aug 4, 2015 at 4:08
  • $\begingroup$ Right, the disjointness follows from cosets being equivalence classes, and the rest from your map. You're saying that the axiom of choice is used to pick a representative element for each coset, or how? (I don't know much set theory) $\endgroup$
    – coldnumber
    Aug 4, 2015 at 4:11
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    $\begingroup$ @coldnumber : I'd have said the fact that they're equivalence classes follows from the fact that they're exhaustive and pairwise disjoint. Did you have in mind some way of showing that they're equivalence classes that does not prove from scratch that they are pairwise disjoint? ${}\qquad{}$ $\endgroup$ Aug 4, 2015 at 4:14
  • $\begingroup$ I was thinking of showing that the relation $aRb \iff a^{-1}b \in H$ is an equivalence relation whose equivalence classes are the left cosets of $H$. $\endgroup$
    – coldnumber
    Aug 4, 2015 at 4:15

First you would need to show that the function you define is well defined (it's not, but let's try to prove it anyway). Observe that $aH = bH$ implies that $b^{-1}aH = H$, so $b^{-1}a \in H$, so $H b^{-1} a = H$, so $H b^{-1} = H a^{-1}$. Okay, so we have proved a similar function is well defined. Why not try to go through a proof for that function... we need to prove it is injective and surjective.

It is certainly surjective (why?). If the number of cosets is finite, you can stop here by invoking the pigeon hole principle. Otherwise, note that $Ha = Hb$ implies that $Hab^{-1} = H$, so $ab^{-1} \in H$, so $ab^{-1} H = H$, so $a^{-1} H = b^{-1} H$.


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