 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/b/b2/a/solution

e * x = e x e^{- 1} = x

. For

g, h \in G

(g h) * x = g h x h^{- 1} g^{- 1} = g * (h * x) .

Both action axioms hold, so conjugation is a group action on $G$ .

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