What is cycle?

Example 1.20 (Cycle type as a census)

How many permutations of $\mathfrak{S}_9$ have the cycle type $(4, 3, 2)$ — one $4$-cycle, one $3$-cycle, one transposition? Choose the supports and the cyclic orders:

list the nine symbols in a row ($9!$ ways), bracket the first four, next three, last two into cycles, and divide by the rotations inside each bracket ($4$, $3$ and $2$ of them) which give the same permutation. (Distinct cycle lengths here, so no further division; equal lengths would also require dividing by the permutations of the equal brackets.) Every such permutation has order $\operatorname{lcm}(4,3,2) = 12$ and signature $(-1)^3(-1)^2(-1)^1 = +1$ (Theorem 1.19 and the signature theorem below). One partition of $9$, one conjugacy class, one census — the combinatorics of $\mathfrak{S}_n$ is the arithmetic of partitions.

Example 1.23 (Three roads to one sign)

Let $\sigma \in \mathfrak{S}_5$ send $1, 2, 3, 4, 5$ to $3, 5, 4, 1, 2$. Via cycles: $1 \mapsto 3 \mapsto 4 \mapsto 1$ and $2 \mapsto 5 \mapsto 2$, so $\sigma = (1\,3\,4)(2\,5)$ and $\varepsilon(\sigma) = (-1)^{2}(-1)^{1} = -1$. Via inversions: in the value list $3, 5, 4, 1, 2$ the out-of-order pairs are $(3,1)$, $(3,2)$, $(5,4)$, $(5,1)$, $(5,2)$, $(4,1)$, $(4,2)$: seven of them, and $(-1)^7 = -1$. Via transpositions: $\sigma = (1\,4)(1\,3)(2\,5)$, three factors, $(-1)^3 = -1$. Three computations, one parity: the uniqueness in Theorem 1.21 guarantees that no bookkeeping scheme can ever make them disagree — which is exactly what makes $\varepsilon$ usable as an invariant (see the weekend problem).