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