If a group G is isomorphic to H, prove that Aut(G) is isomorphic to Aut(H) Properties of Isomorphisms acting on groups: Suppose that is

If a group G is isomorphic to H, prove that Aut(G) is isomorphic to Aut(H)

Properties of Isomorphisms acting on groups:

Suppose that ϕϕ is an isomorphism from a group G onto a group H, then:
1. ϕ−1ϕ−1 is an isomorphism from H onto G.
2. G is Abelian if and only if H is Abelian
3. G is cyclic if and only if H is cyclic.
4. If K is a subgroup of G, the ϕϕ(K)={ϕϕ(k)|k∈∈K} is a subgroup of H.
5. If L is a subgroup of H, then ϕ−1ϕ−1L={g∈∈G|ϕϕ(g)∈∈L} is a subgroup of G.
6. ϕϕ(Z(G))=Z(H)

Steps to proving G is isomorphic to H:

1. Define a function ϕϕ from G to H
2. Prove that ϕϕ is one-to-one
3. Prove that ϕϕ is onto
4. Show that ϕ(ab)=ϕ(a)ϕ(b)ϕ(ab)=ϕ(a)ϕ(b) for all a,b∈Ga,b∈G

Proof:

Suppose that ϕ:G→Hϕ:G→H is an isomorphism.

Since G is a group, it has elements a,b∈Ga,bG such that ab∈Gab∈G.    H is a group as well.    Therefore, ϕ(a)=x∈Hϕ(a)=x∈H, ϕ(b)=y∈Hϕ(b)=y∈H, and ϕ(a)ϕ(b)=ϕ(ab)=xy∈Hϕ(a)ϕ(b)=ϕ(ab)=xy∈H. Thus ϕϕ is one-to-one and operation preserving.
For any element h∈Hh∈H, there is a g∈Gg∈G such that ϕ(g)=hϕ(g)=h, so ϕϕ is onto.
The isomorphism has an inverse operation ϕ−1:H→Gϕ−1:H→G such that ϕ−1(h)=g∈Gϕ−1(h)=g∈G.
Similarly, ψ(a)=r∈Gψ(a)=r∈G, ψ(b)=s∈Gψ(b)=s∈G, and ψ(a)ψ(b)=ψ(ab)=rs∈Gψ(a)ψ(b)=ψ(ab)=rs∈G. Thus ψψ is one-to-one and operation preserving.
For any element g¯∈Gg¯∈G, there is a g∈Gg∈G such that ψ(g)=g¯ψ(g)=g¯, so ψψ is onto.    So ψψ is an automorphism on G.
The automorphism θ(ψ):H→Hθ(ψ):H→H is defined by θ(ψ)=ϕ∘ψ∘ϕ−1θ(ψ)=ϕ∘ψ∘ϕ−1 and is an isomorphism.