TITLE: Need for inverse in $1-1$ correspondence between left coset and right coset of a group
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: At first you might be inclined to try the map $\phi(aH)=Ha$.  The trouble is that this map isn't well defined - you can find examples where two elements in the coset $aH$ map to different right cosets.  In other words, $aH=bH$ does not imply that $Ha=Hb$.
The solution to this problem is to instead use $\phi(aH)=H{a^{-1}}$ as the internet solutions suggested.  The naive approach is to notice that $(aH)^{-1}=H^{-1}a^{-1}=Ha^{-1}$, so we can expect that $aH=bH$ implies that $Ha^{-1}=Hb^{-1}$, which proves that $\phi(aH)=Ha^{-1}$ is well defined.
(Disclaimer. Be careful here.  When I write $(aH)^{-1}$ I am refering to the set $\{x^{-1}:x\in aH\}$.  Later, in quotient groups, $(aH)^{-1}$ will refer to the inverse of the element $aH$ in the quotient group $G/H$, which is an entirely different thing.)
I'd also like to add that it is important to note that your theorem makes no assumption that the group is finite.  In a finite group, in fact, Lagrange's theorem makes it quite easy to prove that the number of left and right cosets are the same.  To prove this for infinite groups, however, we need your lemma.