TITLE: Number of left cosets equals number of right cosets?
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: 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 $a^{-1} H = b^{-1} H$.