Number of left cosets equals number of right cosets?

Question

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?

Answer 1

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.

Answer 2

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 $$Ha = \{(2\ 3),(1\ 2\ 3)\}\neq \{(1\ 3\ 2), (1\ 3)\} = Hb$$

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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.

Answer 3

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 $a^{-1} H = b^{-1} H$.