This question is inspired by questions $4.1.1$ and $4.1.2$ of Dummit and Foote. The hypothesis for the first question is formulated as: "Let $G$ act on the set $A$", and the hypothesis for the second question is formulated as "Let $G$ be a permutation group on $A$".
I know that $G$ acting on $A$ induces a permutation representation of $G$, which is a homomorphism of $G$ into $S_A$. This means that $G$ is homomorphic to some subgroup of $S_A$.
So I think the difference between the hypothesis that in the second hypothesis $G$ must not only be homomorphic, but isomorphic to a subgroup of $S_A$?
Please confirm this or explain what I'm missing here. Thanks
A permutation group is automatically a subset of $S_A$. Equivalently, it is an embedding $G\to S_A$, which means this group homomorphism must be injective (aka one-to-one). In general, a group action needn't be specified by an injective map $G\to S_A$. If the map is not $1$-to-$1$, then different elements of $G$ are sent to the same permutation, so they act the same way on $A$, whereas distinct permutations always act differently on $A$. The term for this is an unfaithful action.