 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-3/5d/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

A_{n} = ker (sgn : S_{n} \to {1, - 1}) .

l

(a_{1} a_{2} \dots a_{l}) = (a_{1} a_{l}) (a_{1} a_{l - 1}) \dots (a_{1} a_{2}),

using rightmost-first composition. Its sign of a permutation is therefore

(- 1)^{l - 1}

. Multiplicativity of the sign homomorphism gives

sgn (g) = (- 1)^{\sum_{j} (l_{j} - 1)} .

Thus

g \in A_{n} ⟺ the number of even-length cycles is even .

Equivalently,

n - t

is even when

t

counts all cycles, including fixed points. Odd-length cycles contribute positive sign and impose no further restriction.

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