 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-i/7/a/solution

a : G \times X \to X

satisfying

a (e, x) = x, a (g h, x) = a (g, a (h, x)) (g, h \in G, x \in X) .

ρ_{a} (g) : X \to X

ρ_{a} (g) (x) = a (g, x)

ρ_{a} (g) \circ ρ_{a} (g^{- 1}) = ρ_{a} (e) = id_{X} = ρ_{a} (g^{- 1}) \circ ρ_{a} (g) .

Hence each

ρ_{a} (g)

is a bijection, with inverse

ρ_{a} (g^{- 1})

, and belongs to the symmetric group

Sym (X)

. Finally,

ρ_{a} (g h) (x) = a (g h, x) = a (g, a (h, x)) = (ρ_{a} (g) \circ ρ_{a} (h)) (x),

ρ_{a} : G \to Sym (X)

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