Ch6 Addition Proofs of Theorems on permutations.

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1 Ch6 Addition Proofs of Theorems on permutations. Definition Two permutations ρ and π of a set A are disjoint if every element moved by ρ is fixed by π and every element moved by π is fixed by ρ. (In other words, the sets of objects moved by each permutation are disjoint sets.) Theorem Disjoint permutations on a set commute. Proof Let ρ and π be disjoint permutations on A. Let a A. There are three cases, (i) a is moved only by ρ, (ii) a is moved only by π and (iii) a is not moved by either. (Since ρ and π are disjoint there is no fourth case.) Case (i), so assume a is fixed by π. Let b = ρ (a), so b a since ρ moves a. Note that if ρ were to fix b then ρ (b) = b = ρ (a) But ρ is injective, so b = a, contradicting b a. Hence ρ moves b. But ρ and π are disjoint so π fixes b. We can now justify every step in ρ π (a) = ρ (π (a)) by definition of, = ρ (a), since π fixes a, = b, by definition of b, = π (b), since π fixes b, = π (ρ (a)) by definition of b, = π ρ (a). by definition of. Case (ii), so assume a is fixed by ρ. Just interchange ρ π in the proof of case (i) to get a proof in this case. Case (iii) Simply ρ π (a) = ρ (π (a)) = ρ (a), since π fixes a, = a since ρ fixes a, = π (a) = π (ρ (a)) = π ρ (a). In all cases we get ρ π (a) = π ρ (a), an equality of elements in A. True for all a A means ρ π = π ρ, an equality of functions, and hence π and ρ commute. 1

2 Cancellation Law Assume that α, β and γ are permutations on a set A. If γ α = γ β then α = β. If α γ = β γ then α = β. Proof Assume γ α = γ β. Then α = 1 A α = ( γ 1 γ ) α = γ 1 (γ α) composition is associative = γ 1 (γ β) by assumption = ( γ 1 γ ) β = 1 A β = β. Leave the other case to students. Definition The positive powers ρ n of a permutation are defined inductively by setting ρ 1 = ρ and ρ k+1 = ρ ρ k for all k N. The negative powers of a permutation are defined by ρ n = (ρ 1 ) n for all n N. Finally, we set ρ 0 = 1 A. Lemma If β and γ are cycles on A that both move an element a A, and β r (a) = γ r (a) for all r 1 then β = γ. (This result allows us to go from knowing what β and γ do to one element to knowing that they act identically on all elements, that is, they are the same permutation.) Proof A cycle can start at any point so we can write β = (a, a 1, a 2,..., a m ) where a i = β i (a) for all i m and β (a m ) = a. (1) Similarly, γ = (a, b 1, b 2,..., b n ) where b j = γ j (a) for all j n and γ (b n ) = a. (2) Without loss of generality assume m n, (if not the case, relabel β as γ and γ as β.) Then from (1), for all i m we have a i = β i (a) = γ i (a) by assumption in Theorem (with r = i), = b i, by (2), Thus we could write γ as (a, a 1, a 2,..., a m, b m+1,...b n ). But then b m+1 = γ m+1 (a) by (2), = β m+1 (a) by assumption in Theorem (r = m + 1), = β (β m (a)) = β (a m ) = a, by (1). 2

3 So the cycle in γ goes back to a. Thus n, the cycle length of γ, equals m. So γ and β contain the same elements and are the same length hence γ = β. Theorem A permutation on a finite set A can be expressed as a product of disjoint cycles uniquely apart from the order of the cycles. Proof Let π = α 1 α 2... α s = β 1 β 2... β t be two factorizations into disjoint cycles. Proof is by induction on n = max (s, t). If n = 1 then π = α 1 = β 1 and the two factorizations are identical. Assume result true for n = k. Assume we have a permutation π that has two factorizations as above with max (s, t) = k + 1. Let a A be an element moved by β t. By disjointedness a is unmoved by all β i, 1 i t 1. Thus for r 1 we have π r (a) = (β 1 β 2... β t ) r (a) = (β 1 β 2... β t ) (β 1 β 2... β t ) (β 1 β 2... β t ) (a) = β r t β r t 1... β r 2 β r 1 (a), disjoint permutations commute, = β r t β r t 1... β r 2 (a), since a is fixed by β 1, and continue,. = β r t (a), since a is fixed by β t 1. Thus π r (a) = β r t (a). (We cannot yet use the Lemma above since π need not be a cycle.) Since a is moved by π it must be moved by some α j, 1 j s and unmoved by all the other α i, i j. By relabelling, we can assume that a is moved by α s alone. As above, π r (a) = α r s (a) for all r 1. Combine to get α r s (a) = β r t (a) for all r 1. Since both α s and β t are cycles we can now apply the Lemma above and deduce α s = β t. By the Cancellation Law we get α 1 α 2... α s 1 = β 1 β 2... β t 1. Now max (s 1, t 1) = max (s, t) 1 = k and we can use the inductive hypothesis to conclude that s = t and, on relabelling the β i, α 1 = β 1,..., α t 1 = β t 1. 3

4 Theorem The order of a cycle is equal to its length. Proof Let σ be a cycle on A. be the length of σ. Let d be the order of σ, so σ d = 1 A and let l We will show that d = l by showing that d l and d l. Let a A be moved by σ. Recall that we can write the cycle starting with any element moved by it. Hence From this we see that (a) Proof that d l. σ = (a, a 1, a 2,..., a l 1 ) = ( a, σ (a), σ 2 (a),..., σ l 1 (a) ). σ (a) a, σ (a) a, σ l 1 (a) a, σ l (a) = a. Starting from σ d = 1 A we have σ d (a) = a. From the list we see that σ d (a) = a σ j (a) for any 0 j l 1. In other words, d j for any 0 j l 1. Hence d l. (b) Proof that d l. Let b A be given. There are two cases. Case (i) If b is moved by σ, then b is in ( a, σ (a), σ 2 (a),..., σ l 1 (a) ), i.e. b = σ j (a) for some 0 j l 1. Consider σ l (b) = σ l ( σ j (a) ) = σ l+j (a) = σ j+l (a) Thus σ l (b) = b.. = σ j ( σ l (a) ) = σ j (a), see last line in above list, = b since b = σ j (a). Case (ii) The second case is b fixed by σ. But then trivially b is fixed by σ l, i.e. σ l (b) = b. In both cases we get σ l (b) = b. True for all b A means that σ l = 1 A. But d is the order of σ so, by Lemma, d l. In particular, d l. 4

5 Combine d l and d l to deduce d = l. Theorem Suppose that π = π 1 π 2 is a decomposition into a product of two disjoint permutations then the order of π is the least common multiple of the orders of π 1 and π 2. Proof Let d be the order of π. Let d 1 be the order of π 1 and d 2 the order of π 2. Set f = lcm (d 1, d 2 ). We will show d = f by showing that d f and d f. (a) Proof that d f. From the definition of f = lcm (d 1, d 2 ) we have d 1 f and d 2 f which in turn mean there exist integers a 1 and a 2 such that f = a 1 d 1 and f = a 2 d 2. Then π f = (π 1 π 2 ) f = (π 1 π 2 ) (π 1 π 2 )... (π 1 π 2 ) = π f 1 π f 2 reordering allowed since π 1 and π 2 disjoint, = π a 1d 1 1 π a 2d 2 2 = ( ) π d a1 ( 1 1 π d 2 2 ) a2 = (1 A ) a 1 (1 A ) a 2, = 1 A. That is π f = 1 A. But the order of π is d so, by the Lemma, d f. In particular, d f. (b) Proof that d f. Let a A be given. There are two cases. (i) Suppose a is moved by π 1. But then a is fixed by π 2 since the two cycles are disjoint. Recalling that d is the order of π we start from a = π d (a) = (π 1 π 2 ) d (a) = ( π d 1 π d 2) (a), reordering allowed since π1 and π 2 are disjoint, = π d 1 ( π d 2 (a) ), by definition of composition, = π d 1 (a) since a is fixed by π 2 and thus by π d 2. Thus π d 1 (a) = a. (ii) In the second case a is fixed by π 1. Trivially it is then fixed by π d 1, i.e. π d 1 (a) = a. 5

6 So in all cases we have π d 1 (a) = a. True for all a A means that π d 1 = 1 A. But d 1 is the order of π 1 and so, by the Lemma, d 1 d. Repeat the argument, replacing π 1 by π 2 and vice-versa to get π d 2 = 1 A and thus d 2 d. (Student must do this.) Thus d 1 d and d 2 d, in which case, d is a common multiple of d 1 and d 2. Yet f is the least of all such common multiples, hence f d. Combine d f and f d to get d = f. Theorem Suppose that π = π 1 π 2... π m is a decomposition into a product of disjoint permutations, then the order of π is the least common multiple of the orders of the permutations π 1, π 2,..., π m. Proof by induction on m. If m = 1 then the result holds by a previous Theorem. Assume result holds for m = k. Let π have a decomposition into k + 1 disjoint permutations, π 1 π 2... π k+1, and let o i =order(π i ). By the induction hypothesis the order of π 1 π 2... π k equals lcm (o 1, o 2,..., o k ). The earlier Theorem on the order of the composition of two disjoint permutations means that the order of (π 1 π 2... π k ) π k+1 equals lcm (lcm (o 1, o 2,..., o k ), o k+1 ) = lcm (o 1, o 2,..., o k, o k+1 ) by question 4, Sheet 2. 6

6 Permutations Very little of this section comes from PJE.

6 Permutations Very little of this section comes from PJE Definition A permutation (p147 of a set A is a bijection ρ : A A Notation If A = {a b c } and ρ is a permutation on A we can express the action