Show that all automorphisms of a group G form a group under function composition

Proof:
First part: Let Aut(G) be the set of all automorphisms of a group G.
To show that Aut(G) is a group under function composition, we need to
verify the following three facts.
(1) id(x)=x is the identity of Aut(G). Because for any f in Aut(G) and
any x in G, we have id(f(x))=f(x)=f(id(x)). So id*f=f*id.
(2) Aut(G) satisfies associative law. For any f,g,h in Aut(G), any x in G,
we have f*(g*h)(x)=f(g*h(x))=f(g(h(x)))=(f*g)(h(x))
Thus f*(g*h)=(f*g)*h
(3) For any f in Aut(G), it has an inverse f^(-1) also in Aut(G).
This is obvious because f is one-to-one and onto and thus f^(-1) can
be well defined. Also we note
f(f^(-1)(xy))=xy=f(f^(-1)(x)*f^(-1)(y))=f(f^(-1)(x))f(f^(-1)(y))
Thus f^(-1)(xy)=f^(-1)(x)*f^(-1)(y)
So f^(-1) is in Aut(G).
There Aut(G) is a group.
Second part: Let Inn(G) be the set of all inner automorphisms of G. We
want to show that Inn(G) is normal in Aut(G).
We consider any f_a in Inn(G) such that f(x)=a^(-1)xa, for any g in Aut(G),
we have
g^(-1) f_a g(x)=g^(-1)(f_a(g(x)))=g^(-1)(a^(-1)g(x)a)
=g^(-1)(a^(-1))g^(-1)(g(x))g^(-1)(a)
=g^(-1)(a)^(-1)xg^(-1)(a)
Let b=g^(-1)a, then
g^(-1) f_a g(x)=b^(-1)xb=f_b, which is also in Inn(G).
Therefore, gInn(G)g=Inn(G). Thus Inn(G) is normal in Aut(G).