 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-i/7/a/i/solution

Write

g \cdot x = gx g^{- 1}

. If

e

G

, then

e \cdot x = e x e^{- 1} = x

. Also, for

g, h, x \in G

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

The middle equality uses associativity and the inverse element rule

(g h)^{- 1} = h^{- 1} g^{- 1}

. Thus this is a left group action, the conjugation action of

G

on itself.

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