Best Answer: Let Aut(G) be the set of all automorphisms φ: G --> G. In order to show that this is a group under the operation of composition, we must verify:
(1) Is the set is closed under composition? Yes! If you are given isomorphisms φ, ψ: G --> G, then it is not too tough to show that ψ∘φ and φ∘ψ are isomorphisms. I can expand on this in more detail if you like, but you have probably seen a proof before that a composition of bijective functions is bijective. If a and b are elements of the group, ψ∘φ(ab) = ψ(φ(ab)) = ψ(φ(a)φ(b)), because φ is an isomorphism. Since ψ is also an isomorphism, ψ(φ(a)φ(b)) = ψ∘φ(a)ψ∘φ(b), so the composition ψ∘φ preserves products. Thus, ψ∘φ is an isomorphism if ψ and φ are.
(2) Is the set associative? Yes! All you need to do is show that, for any three isomorphisms φ, ψ and ξ, φ∘(ψ∘ξ) = (φ∘ψ)∘ξ. To do that, just show that for each x in G, φ∘(ψ∘ξ)(x) = (φ∘ψ)∘ξ(x) = φ(ψ(ξ(x))). It's just pushing around definitions.
(3) Does the set contain an identity element? Yes! Let the identity automorphism e: G --> G be the map e(x) = x. Clearly, e∘φ = φ∘e = φ.
(4) Does each element of the set have an inverse under ∘? Yes! Since each isomorphism φ: G --> G is bijective, there is a well-defined inverse map φ^(-1): G --> G. You may have already seen a proof that the inverse of an isomorphism is an isomorphism. If not, it isn't too difficult to prove: I'll leave it to you, but I can expand on it if you need me to. Further, the composition φ^(-1) ∘ φ = φ ∘ φ^(-1) = e.
Since Aut(G) satisfies all the group axioms, it forms a group under ∘, as needed.
The identity element would just be the identity function $i_{D}:T\to T$ in this case, since given any bijection $f:T\to T$, and $t\in T$, we have $(f\circ i_{D})(t)=f(i_{D}(t))=f(t)=i_{D}(f(t))=(i_{D}\circ f)(t)$.
Best Answer
A function $f$ is injective iff $f(a)=f(b)$ implies $a=b$, for every pair of elements $a$ and $b$. Now suppose $(f\circ g)(a)=(f\circ g)(b)$. By definition of $\circ$, you have $(f\circ g)(a)=(f\circ g)(b)$ implies $f(g(a))=f(g(b))$ which implies (use injectivity of $f$) that $g(a)=g(b)$ which implies (now use injectivity of $g$) that $a=b$. Since composition is associative for an arbitrary function, it is associative also for the subset of functions given by injective ones. For the identity, note that the identity function is trivially injective.
For the inverse, you need also surjectivity (an injection of a set in itself is not in general also surjective). You want to show that $f^{-1}$ is also one-to -one. So suppose $f^{-1}(a)=f^{-1}(b)$. Apply $f$ to both sides and deduce $a=b$. Here you use injectivity to guarantee that $f^{-1}$ is defined.