(January 1997) Suppose that \(\sigma\) is a permutation in the alternating group \(A_{10}\) given by \[ \begin{pmatrix} 1&2&3&4&5&6&7&8&9&10\\ 4&7&2&6&10&1&5&?&?&3 \end{pmatrix}, \] where the images of \(8\) and \(9\) have been lost. Determine the images of \(8\) and \(9\) under \(\sigma\). What is the order of \(\sigma\)?
Solution
Solution. The two missing outputs are also 8 and 9. If σ(8) = 9, then we must have σ(9) = 8, which would give σ = (1 4 6)(2 7 5 10 3)(8 9). However, since 3-cycles and 5-cycles are even, and 2-cycles are odd, that would make σ an odd permutation. But we were told σ∈A10, meaning that σ is even. Contradiction! Therefore σ is not the permutation above, and therefore σ(8)⁄= 9. So we must have σ(8) = 8, and hence σ(9) = 9. Thus, σ = (1 4 6)(2 7 5 10 3). The order of σ is the lcm of the orders of the individual (disjoint) cycles in the above decomposition. So the order of σ is o(σ) = lcm(3, 5) = 15.