I've to prove that $(\text{Aut}(G), \circ)$ is a subgroup of $(\text{Perm}(G),\circ)$
where:
$G$ is a group over the set $S$.
$\text{Perm}(G)$ denotes the set of all permutations in $S$
$\text{Aut}(G)$ denotes the set of automorphisms in $G$
so basically, one of the things I've to check is that if $f,g \in \text{Aut}(G)$ , then $f \circ g \in \text{Aut}(G)$. To prove that $f \circ g$ is an automorphism first I've to check that $f \circ g$ is an homomorphism.
So, the statement I'm stuck with proving is: $(f \circ g)(xy) = f(x) g(y)$ $\forall x,y \in G$,
I don't know where to start, all I know is $(f \circ g)(xy) = f(g(xy))$ but what property would let me "separate" the number $xy$ if I don't know how $f$ and $g$ behave?
NOTE: $f,g$ represent a permutation i.e. bijective mappings $f,g: S \rightarrow S $.
I appreciate your help.