α = Permutation groups

Transcription

1 Permutation groups A permutation of a set A is another name for a bijection from A to A, that is, any function ϕ: A A which is both one-to-one and onto. When A is a finite set with n objects, it is customary to take those objects to be the numbers 1, 2, 3,, n. We can then conveniently represent permutations of A with the notation 1 2 n ϕ =. ϕ(1) ϕ(2) ϕ(n) As any pair of permutations can be composed to produce another permutation, we see that the set of permutations of A forms a group under composition, known as the symmetric group of degree n and denoted S n. These groups were the first groups to be studied by nineteenth century algebraists. For instance, given α,β S 3 with we find that α = 1 2 3, β = 1 2 3,

2 αβ = βα = 1 2 3, showing that S 3 is non-abelian. If we think of the vertices of an equilateral triangle as labeled with the numbers in A = {1, 2, 3}, then each of the symmetries of the triangle can be identified with a particular element of S 3. So we find that D 3 is a subgroup of S 3. (And, more generally, D n is a subgroup of S n.) Another way to represent permutations is to use cycle notation. Using this convention, the permutation ϕ = S is rewritten in the form ϕ = (1 2)(3 5 6), that is, as a product of two disjoint cycles, the leftmost of length 2 and the rightmost of length 3. If also ψ = (1 4 6) S 6, then ϕψ = ( ) is a 6-cycle. Proposition Every permutation of a finite set can be represented as a cycle or product of disjoint cycles. Proof Let α be a permutation of the set A = {1,2,,n}. If a 1 is any element of A, then put

3 a 2 =α(a 1 ),a 3 =α(a 2 ) =α 2 (a 1 ), and so on until we arrive at the first repeated element, i.e., α j (a 1 ) =α i (a 1 ) with j > i. But then α j i (a 1 ) = a 1, so if we put p = j i, we have that α(a p ) =α p (a 1 ) = a 1, and so α produces a cycle out of elements a 1,a 2,,a p. If p = n, or if α fixes all the other elements of A, then α = ( a 1 a 2 a m ) is a single cycle and we are done. If not, let b 1 be an element of A different from all these and generate elements b 2 =α(b 1 ),b 3 =α(b 2 ) =α 2 (b 1 ), as before. None of the b s could be identical with any of the a s, for then α k (b 1 ) =α l (a 1 ) b 1 =α l k (a 1 ), meaning that b 1 is one of the a s, which is impossible. Continuing in this way until we exhaust the elements of A, we get that α = ( a 1 a 2 a p )( b 1 b 2 b q )L( c 1 c 2 c r ). This completes the proof. // Proposition Disjoint cycles commute. ( ) and β = ( b 1 b 2 b q ) be Proof Let α = a 1 a 2 a p disjoint cycles and permutations of the set S = { a 1,a 2,,a p,b 1,b 2,,b q,,c 1,c 2,,c r } where we understand that the c s are the elements of S fixed by both α and β. Then, viewing α and β as functions, we see that not only does

4 (αβ)(a i ) =α(β(a i )) =α(a i ) = a i+1 = β(a i+1 ) = β(α(a i )) = (βα)(a i ) for any of the a s, and (αβ)(b j ) =α(β(b j )) =α(b j+1 ) = b j+1 = β(b j ) = β(α(b j )) = (βα )(b j ) for any of the b s, but (αβ)(c k ) =α(β(c k )) =α(c k ) = c k = β(c k ) = β(α(c k )) = (βα)(c k ) for any of the c s. Since this takes care of all the elements of S, αβ = βα. // Theorem The order of a permutation of a finite set written in disjoint cycle form is the least common multiple of the lengths of the cycles. Proof A cycle of length n has order n. (Why?) So if α and β are cycles of lengths m and n, respectively, and k = lcm(m,n), then (αβ) k = (α k )(β k ) = (e)(e) = e. Thus, the order of αβ, call it l, must divide k. But e = (αβ) l =α l β l α l = β l. However, α l must fix all the elements that α does (and possible more), and β l must similarly fix all the elements that β does; since they are the same element and α and β are

5 disjoint cycles, it follows that α l = β l must fix every element of the set. In other words, α l = β l = e. This implies that m, the order of α, and n, the order of β, must divide l. So k must divide l, too. Since we showed above that l must divide k, it follows that l, the order of αβ, equals k = lcm(m,n). The argument for the case of more than two disjoint cycles is a simple extension of this argument because of the fact that lcm(m,n,p) = lcm( lcm(m,n), p). // Cycles of length 2 are called transpositions since they simply transpose some pair of elements in the underlying set. Theorem Every permutation in S n is a product of transpositions. Proof The identity is a product of transpositions since e = (1 2)(1 2). Also, any permutation which is itself a cycle can be written in the form ( a 1 a 2 a k ) = (a 1 a k )(a 1 a k 1 )L(a 1 a 2 ), so is the product of transpositions. Finally, every permutation is the product of disjoint cycles and the decomposition of these cycles into transpositions will use disjoint transpositions, so we re done. //

6 Lemma The identity can only be represented as a product of an even number of transpositions. Proof Suppose e = τ 1 τ 2 Lτ k where each τ i is a transposition. Since k = 1 is impossible and k = 2 is even, let s suppose that k > 2. Consider the following three possibilities for the product τ k 1 τ k : (1) These two transpositions involve a total of only two elements of the underlying set, that is τ k 1 = τ k = (a b). Here, τ k 1 τ k = e so e is the product of k 2 transpositions, which is even by (strong) induction. Thus k is even. (2) The two transpositions involve a total of three elements of the underlying set, that is τ k = (a b) = (b a) and τ k 1 = (a c) = (c a). We can then rewrite τ k 1 τ k in the form τ k 1 τ k = (a b)(b c) since the permutations on both sides of this last equation send a to b, b to c, and c to a. Notice however that a is fixed by the rightmost transposition (b c) in this last equation. (3) The two transpositions involve a total of four elements of the underlying set, that is τ k = (a b) = (b a) and τ k 1 = (c d) = (d c). Now we can write τ k 1 τ k = (a b)(c d) and once again, a is fixed by the rightmost transposition. What we have shown is that e = τ 1 τ 2 Lτ k is either a product of an even number of transpositions, or is a product of transpositions, the rightmost of which

7 fixes the element a. In the latter case, we can repeat the argument above with the product τ k 2 τ k 1, which will ensure that either k is even or or we can write e as a product of k transpositions in such a way that the two rightmost ones fix a. Repeating this process, we either show k to be even or that e is the product of k transpositions in such a way that only the first transposition does not fix a. This last situation is impossible however, since the resulting product of k transpositions will not fix a whereas e does. So it must be that k is even. // Theorem If the permutation α can be written as a product of transpositions in two ways: α =σ 1 σ 2 Lσ k = τ 1 τ 2 Lτ l (where all the σ s and τ s are transpositions), then k and l have the same parity, that is, they are either both odd or both even. Proof e =σ 1 σ 2 Lσ k τ 1 1 τ 2 1 Lτ l 1 =σ 1 σ 2 Lσ k τ 1 τ 2 Lτ l, so by the lemma, k + l is even. The result follows. //

8 Corollary The set of even permutations in S n (those which are the product of an even number of transpositions) forms a subgroup of S n denoted A n and called the alternating group of degree n. Proof An exercise. // Proposition The order of S n is n!, and the order of A n is n!/2. Proof Easy. // The names of the groups S n and A n come from the study of polynomials in n variables x 1,x 2,,x n, a key ingredient in the resolution of the problem of the quintic formula by Galois and Abel in the 1820s. A symmetric polynomial in these n variables is one which is unchanged by any transposition of variables. For instance, when n = 4, the polynomial x x x x 4 2 is a symmetric polynomial, but (x 1 x 2 )(x 1 x 3 )(x 1 x 4 )(x 2 x 3 )(x 2 x 4 )(x 3 x 4 ) is not instead, it changes sign after a

9 transposition of any pair of varibles. Such a polynomials is called alternating polynomials. Since every permutation is the product of transpositions, the symmetric polynomials are those left invariant under the action of the group elements in S n on the variables in the polynomial. Similarly, since a single transposition of variables changes the sign of an alternating polynomial, the action of the elements of A n on the variables are preciselyt those that leave the polynomial invariant. In the same way that the dihedral groups D n arise as the rigid motions of the plane that leave invariant the regular n-gon, we can also study those rigid motions of 3-space that leave invariant regular solids. For instance, the group of symmetries of the tetrahedron is A 4. (See pp )