(February 2008) Recall that \(S_n\) denotes the group of permutations on \(n\) symbols. (a) Find an element of \(S_{10}\) of order \(21\). (b) Prove that no element of \(S_{10}\) has order \(11\).
Solution
Solution. (a) The order of σ is the lcm of the orders of each individual disjoint cycle comprising σ (and the order of an n-cycle is n). Since 21 = lcm(3 , 7), the element σ = (1 2 3)(4 5 6 7 8 9 10) has order 21. (b) Suppose S10 has a permutation σ of order 11. Then σ must have a disjoint cycle structure such that the lcm of the cycle lengths is 11. If the disjoint cycles making up σ have lengths𝓁1,…,𝓁 k, then 11 =o(σ) = lcm(𝓁1,…,𝓁 k). Thus each 𝓁i divides 11. Since 11 is prime, each 𝓁i = 1 or 11. But an 11-cycle requires 11 distinct symbols, which can’t happen since we are in S10. So 𝓁i = 1 for all i, which means that σ is the identity. Yet it has order 11. This contradiction shows that no element of S10 has order 11. Here is a more abstract problem.