Homomorphism induced by group action

Question

If $G$ is a group acting on $X$, which is a set of all left cosets of $H\leq G$ (such that $|G:H|=k$), by left multiplication, then this group action induces a homomorphism $\phi: G\to S_X$, such that $\phi(g)=\pi_g$, where $\pi_g(aH)=gaH$.

I would like to analyze this homomorphism, and I would appreciate your involvement (and help). It is in my understanding that this homomorphism is injective because $\ker(\phi)=e$. But I've also been trying to understand if it's surjective. Take an arbitrary permutation $\pi_{g'}$, then $\phi(g')=\pi_{g'}$, which implies that $\exists g\in G$ for each $\pi_g\in S_n$, and this $g$ is unique. We can also find an inverse homomorphism $\psi:S_X\to G$ defined by $\psi(\pi_g)=g$. Does this mean that the homomorphism $\phi$ is in fact an isomorphism, and that $|G|$ must be equal $|S_X|$?

Answer 1

The image of $\phi$ is a transitive subgroup of $S_{|G/H|}$, but that's all you can say about it; any transitive subgroup can appear (exercise). The kernel of $\phi$ is the intersection $\bigcap_{g \in G} gHg^{-1}$ of all of the conjugates of $H$ (exercise).

Answer 2

For arbitrary subgroup $H$ of $G$, the homomorphism $\phi\colon G\rightarrow S_X=S_{G/H}$ is not necessarily injective or surjective.

Consider $G$ with $|G|=9$ and $|H|=3$. We will get a homomorphism $\phi\colon G \rightarrow S_{G/H}\cong S_3$. Comparing orders, you will see that it is neither injective nor surjective.