Let $G$ be a group. Say what it means for a map $\varphi: G \rightarrow G$ to be an automorphism. Show that the set-theoretic composition $\varphi \psi = \varphi \circ \psi$ of any two automorphisms $\varphi, \psi$ is an automorphism. Prove that the set $\mathrm{Aut}(G)$ of all automorphisms of the group $G$ with the operation of taking the composition is a group.
I have said: A map is an automorphism of a group $G$ if it is an isomorphism to itself.
a) For the next bit, I want to show if $\varphi, \psi$ is bijective, then $\varphi \circ \psi$ is bijective: For two elements $a, b \in G$ we have
$$\varphi \circ \psi (ab) = \varphi(\psi(ab)) = \varphi(\psi(a)\psi(b))$$
as $\psi$ is an isomorphism. Also, as $\varphi$ is an isomorphism, we have
$$\varphi(\psi(a) \psi(b)) = \varphi \circ \psi(a) \varphi \circ \psi(b)$$
Showing $\varphi \circ \psi$ is an isomorphism iff $\varphi, \psi$ are isomorphisms.
For the group bit, we want to prove the 3 group axioms.
1) Associativity: $\varphi \circ (\psi \circ \zeta) = (\varphi \circ \psi) \circ \zeta$. So for some $x \in G$, we get:
$$\varphi \circ (\psi \circ \zeta)(x) = (\varphi \circ \psi) \zeta(x) = \varphi(\psi(\zeta(x))) $$
2) Identity: If we let the identity automorphism, $e: G \rightarrow G$, be the map $e(x) = x$, then clearly we get that $e \circ \psi = \psi \circ e = \psi$.
3) Inverse: As the automorphisms are bijective (already proved) then we know that by definition of a bijection, there is well defined inverse such that $\psi^{-1}: G \rightarrow G$ exists.
(Second edit to correct proof for inverse): For any two elements $a,b \in G$, we want to see if $\psi^{-1}(ab) = \psi^{-1}(a)\psi^{-1}(b)$. Apply $\psi$ to both sides gives us
$$\psi \circ \psi^{-1}(ab) = \psi(\psi^{-1}(ab)) = ab$$
Doing the same on RHS gives us $ab$ and so we have proved the inverse exists and is unique.
Is this right and enough to prove this?
EDIT: Actually, can I just say that by definition of two bijective maps, the composition is also bijective and this is enough?
Best Answer
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.