# Homomorphism induced by group action

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 $|S_X|$?

• what is $S_X$ here? – SunShine Sep 13 at 10:33

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).
• @sequence: a transitive subgroup of $S_n$ is a subgroup which acts transitively on $\{ 1, 2, \dots n \}$. – Qiaochu Yuan Nov 21 '15 at 19:35
• @sequence: that's correct, but it doesn't follow that every element of this form lies in the kernel. Here is a more specific form of the exercise: show that $gHg^{-1}$ is the stabilizer of the coset $gH$. – Qiaochu Yuan Nov 21 '15 at 19:55
• @sequence: it's not hard to see this. What does it mean for an element $g' \in G$ to stabilize $gH$? It means that $g' g H = g H$. This means that $g^{-1} g' g H = H$, which is equivalent to saying that $g^{-1} g' g \in H$. – Qiaochu Yuan Nov 21 '15 at 20:10
• @sequence: no. It means $g^{-1} g' g = h$ for some $h \in H$, hence $g' = g h g^{-1}$, which is what I was trying to show. – Qiaochu Yuan Nov 21 '15 at 21:40
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.