Thread: permutation

How to prove the following theorem. It is based on permutation (advanced abstract algebra)

Let n>=2 and sigma belongs to S_n be a cycle. Show that sigma is a K-cycle if and only if order of sigma is K.

Please give a proof.

suppose that |σ| = k, and that σ = (a₁ a₂....a_m)

then σⁿ sends a_j to a_{j+n (mod m)}

since σ^k = id, we have:

a_j = a_{j+k (mod m)}, for all j. in particular, 1 = 1+k (mod m) which means that k is a multiple of m.

since k is the least positive power of σ that equals the identity, 1+k is the smallest number > 1 such that 1 = 1+k (mod m), which shows that k = m, so σ is a k-cycle.

if, on the other hand, σ is a k-cycle, then for n < k:

σⁿ sends a₁ to a_{1+n (mod k)}, and since n < k, n+1 ≤ k, so σⁿ(a₁) = a_n+1 ≠ a₁, so σⁿ is not the identity.

however, σ^k(a_j) = a_{j+k (mod k)} = a_j, so σ^k = id, thus |σ| = k.

(we need k ≥ 2 so that σ¹ is not a 1-cycle, since 1-cycles ARE the identity).