2.3.11

We already know that the composition of functions is an associative operation. To show that the composition of two isomorphisms is an isomorphism we first recall that the composition of two bijections is a bijection and note that $$(fg)(ab) = f(g(ab)) = f(g(a)g(b)) = f(g(a))f(g(b)) = (fg)(a)(fg)(b)$$, where $$f, g \in \text{Aut}(G), a,b \in G$$.

The identity element in $$\text{Aut}(G)$$ is the identity map $$i(x) = x$$, which is an isomorphism because $$i(ab) = ab = i(a)i(b).$$

Lastly, if $$f \in \text{Aut}(G)$$ then $$f^{-1} \in \text{Aut}(G)$$ as well because, as $$f$$ is a bijection on $$G$$ any two elements $$a, b \in G$$ can be written $$f(c), f(d)$$ (uniquely) and it follows that $$f^{-1}(ab) = f^{-1}(f(c)f(d)) = f^{-1}(f(cd)) = cd = f^{-1}(a)f^{-1}(b)$$, where the last equality holds because $$a = f(c) \iff f^{-1}(a) = c$$ and analogously for $$d$$.