Proof that composition on $Aut(G)$ is associative Would it be correct to prove it the way below? I'm feeling somewhat ambiguous as to the correctness of my proof. I would appreciate your input.
Proof:
We need to prove that for $\phi$, $\psi$, $\rho \in Aut(G)$, $\phi(\psi\rho)=(\phi\psi)\rho$. Suppose $g\in G$, then:
(i) $[(\phi\psi)\rho](g)=\phi[\psi\rho(g)]=\phi[\psi(\rho(g))]=\phi(\psi(g_\rho))=\phi(g_{\rho\psi})=g_{\rho\psi\phi}$.
(ii)$[\phi(\psi\rho)](g)=\phi[(\psi\rho)(g)]=\phi[\psi(\rho(g))]=\phi(\psi(g_\rho))=\phi(g_{\rho\psi})=g_{\rho\psi\phi}$.
Thus $Aut(G)$ is associative.

It should be well known that composition of functions is always associative. If not, here is the proof:
Suppose $f:A\to B$, $g:B\to C$ and $h:C\to D$ are functions. For any $a\in A$
$$
((h\circ g)\circ f)(a)=(h\circ g)(f(a))=h(g(f(a)))
$$
and
$$
(h\circ(g\circ f))(a)=h((g\circ f)(a))=h(g(f(a))).
$$
The two expressions are equal.
