If $H$ is a normal subgroup of $G$, then does conjugation, $g \cdot xH = g(xH)g^{-1}$, define a group action on $G/H$?
Here is my attempt at a proof:
First, the group action is well-defined since $g(xh)g^{-1} = gxg^{-1} ghg^{-1} = (gxg^{-1}) h'$ for some $h' \in H$ (following from $H \triangleleft G$). So $g(xH)g^{-1} \in G/H$, meaning we can also write the group action as $g \cdot xH = (gxg^{-1})H$. It is clear that $1 \cdot xH = xH$. For $a,b \in G$, we have $a \cdot (b \cdot xH) = a \cdot (bxb^{-1})H = (abxb^{-1}a^{-1})H = (ab) \cdot xH$.