Need for inverse in $1-1$ correspondence between left coset and right coset of a group 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? 
 A: 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.
A: 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.
A: 
Can anyone explain why it is necessary to use the inverse? 

You need a map from the group to itself that reverses the left-right order of multiplication ( $f(xy)=f(y)f(x)$, an anti-homomorphism), sends $H$ to $H$, and is a 1-1 correspondence (is "bijective").  In some situations there are functions other than the inverse that accomplish this, but in the generality of all group/subgroup pairs, inverse is the only option. 
In semigroups, algebraic structures that satisfy the group axioms except for the existence of inverses, the concepts of sub-semigroup and cosets with respect to a subsemigroup make perfect sense and can be defined in the same way as for groups.  There the theorem is not true, the number of left and right cosets can be different.  This indicates that inverses have to be used somewhere in the proof.  
