 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-i/6/b/i/solution

ϕ : G \to G

ϕ (x y) = ϕ (x) ϕ (y) (x, y \in G) .

Such a map from a group to itself is an group automorphism. Its inverse is also a group homomorphism: if

u = ϕ (x)

and

v = ϕ (y)

, then

ϕ^{- 1} (uv) = x y = ϕ^{- 1} (u) ϕ^{- 1} (v)

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