Final answer:
Aut(G) is proven to be a group by verifying it satisfies group axioms. Aut(Z) is isomorphic to Z_2 because it contains only two automorphisms: identity and negation. Aut(Z_5) has 4 elements corresponding to multiplication by each non-zero element of Z_5 and is not isomorphic to Z_4 because their operations differ. Aut(Z_p) has p-1 elements for a prime p.
Step-by-step explanation:
Aut(G) as a Group
To prove that Aut(G) is a group under composition, we must show that it satisfies the group axioms: closure, associativity, identity, and invertibility. Let φ and ψ be elements of Aut(G), then:
- Closure: The composition of two automorphisms φ ∘ ψ is another automorphism, thus inside Aut(G).
- Associativity: Composition of functions is associative.
- Identity: The identity map, id:G → G, is clearly an automorphism.
- Invertibility: If φ is an automorphism, then it has an inverse φ⁻¹ which is also an automorphism.
Aut(Z) ≅ Z₂
The only automorphisms of Z are the identity and the negation map. Thus, Aut(Z) is isomorphic to Z₂ since they have the same structure and order.
Elements of Aut(Z₅)
Since Z₅ is a finite field, every non-zero element has a multiplicative inverse. An automorphism must map 1 to another generator of the group. There are 4 such maps corresponding to multiplication by each non-zero element of Z₅. These 4 maps form a group, but since their operation is multiplication, not addition, this group is not isomorphic to Z₄.
Elements in Aut(Zᵗₚ)
For a prime p, there are p−1 non-zero elements in Zᵗₚ, each of which can serve as a possible image for 1 under an automorphism. Thus, there are p−1 automorphisms in Aut(Zᵗₚ).