Theorem
Prove that the set of all automorphisms of a group (G,⋅) forms a group with respect to the composition of mappings.
Proof
Let Aut(G) denote the set of all automorphisms of the group
(G,⋅). We show that Aut(G) forms a group under the
composition of mappings by verifying the group axioms.
1. Closure
Let f,g∈Aut(G). Since each is an isomorphism, both are homomorphisms and bijective.
Consider the composition f∘g. For any a,b∈G:
(f∘g)(a⋅b)=f(g(a⋅b))=f(g(a)⋅g(b))(since g is a homomorphism)=f(g(a))⋅f(g(b))(since f is a homomorphism)=(f∘g)(a)⋅(f∘g)(b)
Thus, f∘g f \circ g f∘g is a homomorphism. Being a composition of bijections, it is also bijective. Hence, f∘g f \circ g f∘g is an automorphism, showing closure.
2. Associativity
For any f,g,h∈Aut(G) f, g, h \in \operatorname{Aut}(G) f,g,h∈Aut(G), function composition is inherently associative:
(f∘g)∘h=f∘(g∘h)
(f \circ g) \circ h = f \circ (g \circ h)
(f∘g)∘h=f∘(g∘h)
3. Identity Element
The identity mapping IG:G→G \operatorname{I}_G : G \to G IG:G→G defined by IG(a)=a \operatorname{I}_G(a) = a IG(a)=a for all a∈G a \in G a∈G is an automorphism. It satisfies:
IG(a⋅b)=a⋅b=IG(a)⋅IG(b)
IG(a⋅b)=a⋅b=IG(a)⋅IG(b)\begin{align}
\operatorname{I}_G(a \cdot b) &= a \cdot b \nonumber\\
&= \operatorname{I}_G(a) \cdot \operatorname{I}_G(b)\nonumber
\end{align}IG(a⋅b)=a⋅b=IG(a)⋅IG(b)
IG(a⋅b)=a⋅b=IG(a)⋅IG(b)
Therefore, IG \operatorname{I}_G IG acts as the identity element in Aut(G) \operatorname{Aut}(G) Aut(G).
4. Inverses
Let f∈Aut(G) f \in \operatorname{Aut}(G) f∈Aut(G). Since f f f is a bijective homomorphism, its inverse f−1 f^{-1} f−1 exists.
For all a,b∈G a, b \in G a,b∈G, using the fact that f f f preserves the group operation, we have:
f−1(a⋅b)=f−1(a)⋅f−1(b)
f^{-1}(a \cdot b) = f^{-1}(a) \cdot f^{-1}(b)
f−1(a⋅b)=f−1(a)⋅f−1(b)
Hence, f−1 f^{-1} f−1 is also an automorphism, providing the inverse for every element in Aut(G) \operatorname{Aut}(G) Aut(G).