I was trying to solve the problem to show that there is a one-to-one correspondence between the set of left cosets of H in G and the set of right cosets. I attempted to prove it by creating a function mapping aH to Ha as follows: $f(ah) = ha$. I reasoned that is bijective as $h'a = ha$ implies $ah = ah'$, and given $ha$, we know that $f(ah) = ha$. However all of the proofs I found on the internet mapped ah to $ha^{-1}$. Can anyone explain why it is necessary to use the inverse?
[Math] Need for inverse in $1-1$ correspondence between left coset and right coset of a group
abstract-algebragroup-theory
Best Answer
For some fixed $a \in G$ the map $aH \to Ha, ah \mapsto ha$ is bijective, yes. But this only deals with two specific cosets. It is a completely different statement that there is a $1:1$ correspondence between the set of all left cosets $G ~/~ H$ and the set of all right cosets $H \setminus G$. Observe that the map $G ~/~ H \to H \setminus G$, $aH \mapsto Ha$ is not well defined. In fact, we have $aH = bH \Leftrightarrow (aH)^{-1} = (bH)^{-1} \Leftrightarrow H a^{-1} = H b^{-1}$. So instead, we have to take $G ~/~ H \to H \setminus G$, $aH \mapsto H a^{-1}$. This is a well-defined bijection.