Richard Bellman (ed.)'s a collection of modern mathematical classics analysis PDF

By Richard Bellman (ed.)

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Consider the powers of α: α, α2 , . .. Since Sn is finite, at some point an element in this list will be repeated. Suppose that αt = αs , for some s < t. Then multiplying both sides by α−s , we see that αt−s = (1) . Let r be the smallest natural number such that αr = (1) . Set G = {(1), α, . . , αr−1 } ⊂ Sn . Now check that G is a permutation group: we have (αi )−1 = α−i = αr−i for any i, 1 ≤ i < r. 38 CHAPTER 3. PERMUTATION GROUPS and αi αj = αi+j = αk , where i + j ≡ k (mod r), 0 ≤ k < r . G is called the cyclic permutation group generated by α and will be denoted by ⟨α⟩.

8} are all accounted for now, and we see that α = α3 α2 α1 . The cycles are even disjoint, that is, no two have a number in common. 2. Any permutation is a product of disjoint cycles. To see this, let α ∈ Sn . Consider α(1), α2 (1), . . At some point this sequence will begin to repeat itself. Suppose that αt (1) = αs (1) where s < t. Then αt−s (1) = 1. Pick the smallest r1 > 0 such that αr1 (1) = 1. Let α1 be the r1 -cycle given by the sequence 1, α(1), α2 (1), . . , αr1 −1 (1) Now pick the smallest number i2 ̸= αi (1) for any i.

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