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?
Best Answer
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.