 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-i/6/a/iii/solution

The hypothesis implies

x^{- 1} = x

for every

x \in G

, including

e

. For

a, b \in G

ab = (ab)^{- 1} = b^{- 1} a^{- 1} = ba .

Thus $G$ is an abelian group. This is the group of exponent two is abelian argument.

The converse is false: the cyclic group

C_{3}

is an abelian group, but its two nonidentity elements have order of a group element three.

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