Permutations

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Note that as f is a bijective function, all n of the elements in X = {1, 2, . . .,n} must occur in the second row. It is for this reason that such functions are termed “permutations”: one can think of them as simply reordering the elements in X. The composite of two permutations f and g is the function obtained by applying f first and then applying g. Since we are writing maps on the right, we denote this by fg. It is easy to calculate the permutation obtained by composing two permutations written in the above two-row notation: ( ) 1 2 3 4 2 4 1 3 ( ) 1 2 3 4 3 2 4 1 ◦ Here 1 ↦→ 2 by the first permutation and then 2 ↦→ 2 by the second. Thus the composite does the first then the second, so 1 ↦→ 2. Equally the composite has the following effects: Hence ( ) 1 2 3 4 ◦ 2 4 1 3 Similarly = 2 ↦→ 4 ↦→ 1, 3 ↦→ 1 ↦→ 3, 4 ↦→ 3 ↦→ 4 ( ) 1 2 3 4 ◦ 3 2 4 1 ( ) 1 2 3 4 = 3 2 4 1 ( ) 1 2 3 4 = 2 4 1 3 ( ) 1 2 3 4 2 1 3 4 ( ) 1 2 3 4 1 4 3 2 (Since 1 ↦→ 3 ↦→ 1, 2 ↦→ 2 ↦→ 4, 3 ↦→ 4 ↦→ 3 and 4 ↦→ 1 ↦→ 2.) Note this already illustrates one phenomenon: in general, for two permutations f and g. Cycle Notation fg ≠ gf The above two-row notation is quite inefficient and also difficult to understand the permutations in great detail. For example, the permutation ( ) 1 2 3 4 1 4 3 2 fixes both 1 and 3, while swaps round 2 and 4. It would be nice to have a more efficient way to describe this element (ideally one which misses out 1 and 3 since they are not moved by the permutation). 69
Definition 11.2 Let x 1 , x 2 , . . ., x r be r distinct elements of {1, 2, . . .,n} (so 1 r n). The r-cycle (x 1 x 2 . . .x r ) is the permutation in S n which maps x 1 ↦→ x 2 , x 2 ↦→ x 3 , . . ., x r−1 ↦→ x r , x r ↦→ x 1 and fixes all other points in {1, 2, . . .,n}. Such a cycle may be described by drawing the points x i in a circular picture. Thus the cycle could also be written as (x 2 x 3 . . .x r x 1 ), or (x 3 x 4 . . .x r x 1 x 2 ), etc. For example, the above permutation ( ) 1 2 3 4 1 4 3 2 could be written more simply as (2 4) or (4 2). This tells us that this permutation fixes both 1 and 3. What about the identity permutation This is the permutation ( ) 1 2 3 . . . n . 1 2 3 . . . n This is often written as the cycle (1). (Such a cycle fixes all elements except 1 and moves 1 to 1: so it really is the identity.) Of course, it could also be written (x) for any x ∈ {1, 2, . . .,n}, but to avoid confusion its probably best to stick to x = 1. Definition 11.3 Two cycles (x 1 x 2 . . .x r ) and (y 1 y 2 . . .y s ) in S n are disjoint if no element in {1, 2, . . .,n} is moved by both cycles. If these cycles are non-identity (i.e., if r 2 and s 2) then this condition can be expressed as {x 1 , x 2 , . . .,x r } ∩ {y 1 , y 2 , . . .,y s } = ∅. The crucial observation that will enable us to make use of cycles is the following: Theorem 11.4 Every permutation (of n points) can be written as a product of disjoint cycles. The proof is omitted. A proof is not too hard, but it is more helpful to give an example to illustrate and this should be fairly convincing of the truth of the theorem. 70
Page 1: Section 11 Permutations Definition
Page 5 and 6: Definition 11.10 The order of a per

Permutations

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