Proving that (group) automorphisms form a subgroup of the permutation group on the set 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.

That is not the statement you want to be proving. Recall that a function $p:G\rightarrow G$ is a homomorphism when 
$$p(xy)=p(x)p(y)\text{ for all }x,y\in G.$$
You want to prove that, given two homomorphisms $f:G\rightarrow G$ and $g:G\rightarrow G$, then their composition $(f\circ g):G\rightarrow G$ is also a homomorphism. The composition $f\circ g$ is defined by
$$(f\circ g)(x)=f(g(x))\text{ for all }x\in G.$$
So, do you see what the right statement you want to prove is? Also, remember that in your argument, somewhere you should use the hypotheses: namely, that $f$ and $g$ are themselves homomorphisms. 
