 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-2/9/c/solution

Z_{6}

is determined by the image

t

of its generator

1

, with the consistency condition

6 t = 0

in the target. Every element of either target satisfies this condition. Therefore

Z_{6} \to Z_{2} : f_{t} ([n]_{6}) = [t n]_{2}, t = 0, 1,

and

Z_{6} \to Z_{3} : f_{t} ([n]_{6}) = [t n]_{3}, t = 0, 1, 2.

These formulas are well-defined because changing

n

by six changes

t n

by a target multiple of its modulus. They list respectively two and three maps.

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