Final answer:
The order of permutation α is 20, and it belongs to the alternating group A9. The compositions α∘β and β∘α in S100 and their orders can be found by applying one permutation after the other and analyzing the resulting cycle structure.
Step-by-step explanation:
Let's examine the permutation α=(1 9 2 4)(1 7 6 5 9)(1 2 3 8) in S9, the symmetric group on 9 elements. To find the order of α, we consider the lengths of its disjoint cycles. α has one 4-cycle, one 5-cycle, and one 4-cycle, which are disjoint. The order of α is the least common multiple (LCM) of the lengths of these cycles: LCM(4, 5, 4) = 20. Therefore, the order of α is 20.
Next, we determine if α is an even permutation and thus belongs to A9, the alternating group on 9 elements. A permutation is even if it can be expressed as a product of an even number of transpositions. Since we have a 4-cycle, which is an odd permutation, and a 5-cycle, which is also an odd permutation (each cycle can be expressed as 3 and 4 transpositions respectively), and another 4-cycle, the product of an odd number of odd permutations is even. Hence, α ∈ A9.
Examining the cycles α=(1,2,3,…,100), and β=(1,50,100) in S100, we find their compositions α∘β and β∘α which represent the result of applying β first then α, and vice versa. The orders of α∘β and β∘α are determined by looking at the cycles in the resulting permutations and taking the LCM of their lengths, after simplifications due to their specific forms.