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 = (a_{1} a_{2}....a_{m})
then σ^{n} 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:
σ^{n} sends a_{1} to a_{1+n (mod k)}, and since n < k, n+1 ≤ k, so σ^{n}(a_{1}) = a_{n+1} ≠ a_{1}, so σ^{n} 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 σ^{1} is not a 1-cycle, since 1-cycles ARE the identity).