 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-i/5/a/iv/solution

Take

x = h (1, 0, 0) = I_{3} + E_{12}

and

y = h (0, 0, 1) = I_{3} + E_{23}

x y = 100110111, y x = 100110011 .

They differ in the

(1, 3)

entry because

E_{12} E_{23} = E_{13}

whereas

E_{23} E_{12} = 0

. Therefore

x y \neq = y x

, and

H is a non-abelian group

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