Let $G$ be a group and $H$ its subgroup. Let $n=[G:H]$ be a cardinal number. Let $C=\{aH\,|\,a\in G\}.$ We have $n=\operatorname{card}(C).$ We define for any $g\in G$ the map $\phi_g:C\to C$ by the formula $\phi_g(aH)=gaH.$ The maps $\phi_g$ are well-defined because if $aH=bH,$ then $gaH=gbH$ for any $g\in G.$
$\phi_g$ must be 1-1 because if $gaH=gbH,$ then $aH=g^{-1}(gaH)=g^{-1}(gbH)=bH.$ $\phi_g$ must be onto because for $a,g\in G$ we have $aH=g(g^{-1}aH)=\phi_g((g^{-1}a)H).$
Therefore, $\phi_g$ is a permutation of $C.$ We can say that $\{\phi_g\,|\,g\in G\}\subseteq \operatorname{Sym}(n).$
Let $f:G\to\operatorname{Sym}(n)$ be defined by the formula
$f(g)=\phi_g.$
$f$ is a homomorphism because for $g_1,g_2\in G$ and $aH\in C$ we have
$(f(g_1g_2))(aH)=\phi_{g_1g_2}(aH)=g_1g_2aH=(\phi_{g_1}\circ\phi_{g_2})(aH)=(f(g_1)\circ f(g_1))(aH).$
I've just noticed this. Is there a name for this homomorphism? Are there names for its kernel and image?
Edit: Let's rename $f$ to $f_l$ and define $f_r$ analogously, but with right cosets instead of left cosets. Can it be for some $G$ and $H$ that
$\ker f_l\neq \ker f_r?$
Edit: OK, I think it can't be. We have
$ \begin{eqnarray} \ker f_l&=&\{g\in G \,|\, \phi_g=\operatorname{id}\}\\&=&\{g\in G\,|\,(\forall a\in G) gaH=aH\}\\&=&\{g\in G\,|\,(\forall a\in G) a^{-1}gaH=H\}\\&=&\{g\in G\,|\,(\forall a\in G) a^{-1}ga\in H\} \end{eqnarray} $
Analogously,
$ \ker f_r = \{g\in G\,|\,(\forall a\in G) aga^{-1}\in H\} $
and $ (\forall a\in G) a^{-1}ga\in H\iff (\forall a\in G) aga^{-1}\in H $
So the kernels are equal.