Showing one to one correspondence 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 of $H$ in $G$.
What is the basic technique/principle for showing one to one correspondence?
 A: As noted above, you must find a bijection $f$ between $G/H$ (the set of left cosets) and $H\backslash G$ (the set of right cosets).
The standard bijection is such that $f(gH) = Hg^{-1}$. The first step is to show that such a mapping indeed exists, that is that whenever $gH = g'H$ then $Hg^{-1} = Hg'^{-1}$. Thus you can define $f(X)$ as $Hg^{-1}$ for any $g \in X$ without the result depending on the choice of such a $g$.
This step is essential. Note that in it lies the reason for not defining $f$ simply by $f(gH) = Hg$; if $gH = g'H$, it is not generally the case that $Hg = Hg'$.
Then you must show that $f$ is onto and that it is one to one.
A: You have that the set of right cosets is given by {$gH: g \in G$}, and the set of left cosets is given by {$Hg: g \in G$} , although different elements$g,g'$ of $G$ may generate the same coset. This happens precisely when $gg'^{-1} \in H$. This is independent of whether you have a right coset or a left coset.
The "standard" bijection is given by $gH \rightarrow Hg^{-1}$ . Can you see this is a bijection (Hint: the coset H itself is the "zero coset", so you can compute the "kernel " of the map, now you just need to show the map is onto, i.e., every Hg appears listed somewhere in the map; use $(a^{-1})^{-1}= a$? Like Tony Jacobs said, one way of showing correspondence is by constructing a bijection.( In the case of infinite sets, there are related methods like Constructing injections in both directions, i.e., Cantor-Schroeder-Bernstein theorem. Don't worry about that for now).
Now, to show the map is onto, say you want to see if the coset Hg is listed , or "hit" by the map. Then remember that $g=(g^{-1})^{-1}$. To see about injection, remember that, by definition, two elements  $g,g'$ in G generate the same coset iff $gg'^{-1} \in H$
A: In general, one shows a one-to-one correspondence by constructing a function from one set to the other, and then proving that it is one-to-one and onto.
Does this help?
A: Let $H$ be a subgroup of $G$ and consider $f$  be a mapping from
$f:Ha \to Hb$ defined by:
$f(ha) = hb$ where $h \in H$
now To prove that $f$ is one-one and onto
$f(h_1a) = f(h_2a) \implies h_1b = h_2b$ hence $h_1 = h_2$ by the cancellation property. Hence $f$ is one-one
In $h_ia$ in $Ha$ is different since $h_ia = h_ja$ for $i \ne j \implies h_i = h_j$ hence $o(Ha) = o(H)$ for any $a \in G$
So $o(Ha) = o(Hb)$ so $Ha$ is onto.
$f$ is one-one and onto.
