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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-2/8/b/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 8 / b / Solution by

Codex  0 2026-10-18

Use group axioms from two-sided local inverse identities. Choose any

a

in the nonempty semigroup and put

e_{R} = a_{L}^{- 1} a

,

e_{L} = a a_{R}^{- 1}

. The given identities say

b e_{R} = b

and

e_{L} b = b

for every

b

. Therefore

e_{L} e_{R} = e_{L}

by the right-identity property, and

e_{L} e_{R} = e_{R}

by the left-identity property. Thus

e_{L} = e_{R} = e

is a two-sided identity element.

For an arbitrary

a

b = a_{L}^{- 1}

and

c = a_{R}^{- 1}

. Inserting

e

into the original identities gives

ba = e

and

a c = e

. By associativity,

b = b e = b (a c) = (ba) c = ec = c .

Thus every element has a two-sided inverse element. Closure and associativity were assumed, so all the group axioms hold. No cancellation or identity was used before it had been established.

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