Let $G$ be a permutation group on the set $A$, let $\sigma \in G$ and let $a \in A$. Prove that $\sigma G_A\sigma^{-1} = G_{\sigma(a)}$. Deduce that if $G$ acts transitively on $A$ then $\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = 1$.
So I solved the first part, which is just set theory. The second part I think I am on the right track. That is suppose $w$ is in the intersection, so it must be in all of the stabilizers by part (a) we solved first, so it must fix all elements $a \in A$, but how do we show then it must be equal to the identity ?
In this question, by using the words "permutation group", they mean that the action is faithful (so only the identity element fixes everything).
Any action of a group $H$ on a set $Y$ induces a permutation representation $H \to S_{|Y|}$. If $H$ acts faithfully, then the permutation representation $H \to S_{|Y|}$is injective. Because the map is injective, $H$ is naturally a subgroup of $S_{|Y|}$, the permutations of $Y$. Hence the terminology "permutation group". I personally find this terminology confusing, since it refers not to the group, but to the action. (Every group is a "permutation group" in a sense, by Cayley's Theorem.)