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).