I'm reading a proof and I don't quite understand one step of the proof.
We want to deduce that if G acts transitively on A then $ \bigcap_{\sigma \in G} \sigma G_{a} \sigma^{-1} = 1$. (Where $G_{a}$ is the stabilizer of the group $G$.)
The proof goes something like this:
If G acts transitively on A, then we know from a prior exercise that the kernel of the action is $\bigcap_{\sigma \in G} \sigma G_{a} \sigma^{-1}$.
Here is the part I don't understand: Moreover, because $G \leq S_A$, the homomorphism $\iota : G \rightarrow S_A$ producing this action is injective, and thus has a trivial kernel. Thus $\bigcap_{\sigma \in G} \sigma G_{a} \sigma^{-1} = 1$.
I don't understand why a homomorphism from a subgroup to the group implies the homomorphism is injective. Also, why does that imply that the kernel is trivial?
I know that the homomorphism is our permutation representation associated to the action. I don't understand why $G$ being a subgroup of $S_A$ forces the homomorphism to be injective. For a mapping to be injective we have to have that whenever $x \neq y$, then $f(x) \neq f(y)$.
Please help.