Notes on the symmetric group 1 Computations in the symmetric group Recall that, given a set X, the set SX of all bijections from X to itself (or, more briefly, permutations of X) is group under function composition. In particular, for each n 2 N, the symmetric group Sn is the group of per- mutations of the set f1; : : : ; ng, with the group operation equal to function composition. Thus Sn is a group with n! elements, and it is not abelian if n ≥ 3. If X is a finite set with #(X) = n, then any labeling of the elements of X as x1; : : : ; xn defines an isomorphism from SX to Sn (see the homework for a more precise statement). We will write elements of Sn using Greek let- ters σ; τ; ρ, : : : , we denote the identity function by 1, and we just use the multiplication symbol · or juxtaposition to denote composition (instead of the symbol ◦). Recall however that functions work from right to left: thus στ(i) = σ(τ(i)), in other words evaluate first τ at i, and then evaluate σ on this. We say that σ moves i if σ(i) 6= i, and that σ fixes i if σ(i) = i. There are many interesting subgroups of Sn. For example, the subset Hn defined by Hn = fσ 2 Sn : σ(n) = ng is easily checked to be a subgroup of Sn isomorphic to Sn−1 (see the home- work for a generalization of this). If n = n1 + n2 for two positive integers n1; n2 then the subset H = fσ 2 Sn : σ(f1; : : : ; n1g) = f1; : : : ; n1gg is also a subgroup of Sn. Note that, if σ 2 H, then automatically σ(fn1 + 1; : : : ; n2g) = fn1 + 1; : : : ; n2g; and in fact it is easy to check that H is isomorphic to Sn1 × Sn2 . There are many other subgroups of Sn. For example, the dihedral group Dn, the group of symmetries of a regular n-gon in the plane, is (isomorphic to) a subgroup of Sn by looking at the permutation of the vertices of the n-gon. 1 Note that #(Dn) = 2n, and hence that Dn is a proper subgroup of Sn if n ≥ 4. We shall see (Cayley's theorem) that, if G is a finite group, then there exists an n such that G is isomorphic to a subgroup of Sn (and in fact one can take n = #(G)). Thus the groups Sn are as complicated as all possible finite groups. To describe a function σ : f1; : : : ; ng ! f1; : : : ; ng, not necessarily a per- mutation, we can give a table of its values, recording i and then σ(i), as follows: i 1 2 ... n σ(i) σ(1) σ(2) ... σ(n) Of course, we could describe the same information by a 2 × n matrix: 1 2 : : : n : σ(1) σ(2) : : : σ(n) The condition that σ is a permutation is then the statement that the integers 1; : : : ; n each occur exactly once in the second row of the matrix. For example, if σ is given by the matrix 1 2 3 4 5 6 7 8 ; 2 4 8 1 5 7 3 6 then σ(1) = 2 and σ(5) = 5. This notation is however cumbersome and not well suited to calculation. We describe a much more efficient way to write down elements of Sn by first writing down some special ones, called cycles, and then showing that very element of Sn can be factored into a product of cycles, in an essentially unique way if we are careful. Definition 1.1. Let fa1; : : : ; akg be a subset of f1; : : : ; ng with exactly k elements; equivalently, a1; : : : ; ak are distinct. We denote by (a1; : : : ; ak) the following element of Sn: (a1; : : : ; ak)(ai) = ai+1; if i < k; (a1; : : : ; ak)(ak) = a1; (a1; : : : ; ak)(j) = j; if j 6= ai for any i. We call (a1; : : : ; ak) a k-cycle or cycle of length k. For k > 1, we define the support Supp(a1; : : : ; ak) to be the set fa1; : : : ; akg. Note that, again for k > 1, the k-cycle (a1; : : : ; ak) moves i () i 2 Supp(a1; : : : ; ak). Two cycles (a1; : : : ; ak) and (b1; : : : ; b`) are disjoint if Supp(a1; : : : ; ak) \ Supp(b1; : : : ; b`) = ;; i.e. the sets fa1; : : : ; akg and fb1; : : : ; b`g are disjoint. 2 Remark 1.2. 1) A 1-cycle (a1) is the identity function 1, no matter what a1 is. For this reason, we will generally only consider cycles of length at least 2. If σ = (a1; : : : ; ak) with k ≥ 2, then σ is never the identity, since σ(a1) = a2 6= a1. 2) A 2-cycle (a1; a2) is also called a transposition. It is the unique permu- tation of f1; : : : ; ng which switches a1 and a2 and leaves all other elements alone. 3) From the description in 2), it is clear that (a1; a2) = (a2; a1). But for k ≥ 3, the order of the elements a1; : : : ; ak is important: for example σ1 = (1; 3; 2) 6= σ2 = (1; 2; 3), because σ1(1) = 3 but σ2(1) = 2. 4) However, there are a few ways to change the order without changing the element of Sn: clearly (a1; a2; : : : ; ak) = (a2; a3; : : : ; ak; a1) = (a3; a4; : : : ; ak; a1; a2) = ··· = (ak; a1; : : : ; ak−2; ak−1): In other words, you can start the cycle anywhere, at ai, say, but then you have to list the elements in order: the next one must be ai+1, and so on, with the understanding that once you reach ak, the next one has to be a1, then a2, and then so on up to ai−1. Clearly, this the only way you can change the order. By convention, we often start with the smallest ai. Of course, after that, there is no constraint on the sizes of the consecutive members of the cycle. −1 5) It is easy to see that the inverse (a1; a2; : : : ; ak) = (ak; ak−1; : : : ; a1). In other words, the inverse of the k-cycle (a1; a2; : : : ; ak) is the k-cycle where the elements are written in the opposite order. In particular, for a transposition −1 (a1; a2), (a1; a2) = (a2; a1) = (a1; a2), i.e. a transposition has order 2. 6) Generalizing the last line of 5), it is easy to see that the order of a k-cycle σ = (a1; a2; : : : ; ak) is exactly k. In fact, σ(ai) = ai+1, so that r σ (ai) = ai+r, with the understanding that the addition i + r is to be taken mod k, but using the representatives 1; : : : ; k for addition mod k instead of the more usual ones (in this course) of 0; : : : ; k − 1. In particular, we see r r that σ (ai) = ai for all i () r is a multiple of k, and since σ (j) = j for r j 6= ai, we see that k is the smallest positive integer r such that σ = 1. Note however that, if σ is a k-cycle, its powers σr need not be k-cycles. For example, (1; 2; 3; 4)2 = (1; 3)(2; 4); and (1; 3)(2; 4) is not a k-cycle for any k. 3 7) Suppose that σ1 = (a1; : : : ; ak) and σ2 = (b1; : : : ; b`) are two disjoint cy- cles. Then it is easy to see that σ1σ2 = σ2σ1, i.e. \disjoint cycles commute." To check this we have to check that, for all j 2 f1; : : : ; ng, σ1σ2(j) = σ2σ1(j). First, if j = ai for some i, then σ1(ai) = ai+1 (with our usual conventions on adding mod k) but σ2(ai) = ai since ai 6= br for any r. For the same reason, σ2(ai+1) = ai+1. Thus, for all i with 1 ≤ i ≤ k, σ1σ2(ai) = σ1(σ2(ai)) = σ1(ai) = ai+1; whereas σ2σ1(ai) = σ2(σ1(ai)) = σ2(ai+1) = ai+1 = σ1σ2(ai): Similarly, σ1σ2(br) = br+1 = σ2σ1(br) for all r, 1 ≤ r ≤ `. Finally, if j is not an ai or a br for any i; r, then σ1σ2(j) = σ1(σ2(j)) = σ1(j) = j; and similarly σ2σ1(j) = j. Thus, for all possible j 2 f1; : : : ; ng, σ1σ2(j) = σ2σ1(j), hence σ1σ2 = σ2σ1. Note that non-disjoint cycles might or might not commute. For example, (1; 2)(1; 3) = (1; 3; 2) 6= (1; 2; 3) = (1; 3)(1; 2); whereas (1; 2; 3; 4; 5)(1; 3; 5; 2; 4) = (1; 4; 2; 5; 3) = (1; 3; 5; 2; 4)(1; 2; 3; 4; 5): 8) Finally, we have the beautiful formula: if σ 2 Sn is an arbitrary element (not necessarily a k-cycle), and (a1; : : : ; ak) is a k-cycle, then −1 σ · (a1; : : : ; ak) · σ = (σ(a1); : : : ; σ(ak)): −1 In other words, σ · (a1; : : : ; ak) · σ is again a k-cycle, but it is the k-cycle where the elements a1; : : : ; ak have been \renamed" by σ. To prove this formula, it suffices to check that −1 σ · (a1; : : : ; ak) · σ (j) = (σ(a1); : : : ; σ(ak))(j) for every j 2 f1; : : : ; ng. First, if j is of the form σ(ai) for some i, then by definition (σ(a1); : : : ; σ(ak))(σ(ai)) = σ(ai+1), with the usual remark that if i = k then we interpret k + 1 as 1. On the other hand, −1 −1 σ · (a1; : : : ; ak) · σ (σ(ai)) = σ · (a1; : : : ; ak)(σ (σ(ai)) = σ · (a1; : : : ; ak)(ai) = σ((a1; : : : ; ak)(ai)) = σ(ai+1): 4 Thus both sides agree if j = σ(ai) for some i.