# permutation

• Feb 3rd 2013, 08:00 AM
permutation
How to prove the following theorem. It is based on permutation (advanced abstract algebra)

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

Please give a proof.
• Feb 3rd 2013, 10:24 AM
Deveno
Re: permutation
suppose that |σ| = k, and that σ = (a1 a2....am)

then σn sends aj to aj+n (mod m)

since σk = id, we have:

aj = aj+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 a1+n (mod k), and since n < k, n+1 ≤ k, so σn(a1) = an+1 ≠ a1, so σn is not the identity.

however, σk(aj) = aj+k (mod k) = aj, so σk = id, thus |σ| = k.

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