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Home , Dihedral group, Subgroup, Symmetric group

Notes on the symmetric

1 Computations in the

Recall that, given a X, the set SX of all from X to itself (or, more briefly, of X) is group under composition. In particular, for each n ∈ N, the symmetric group Sn is the group of per- mutations of the set {1, . . . , n}, with the group operation equal to . 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 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 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 of Sn. For example, the Hn defined by Hn = {σ ∈ Sn : σ(n) = n} is easily checked to be a of Sn isomorphic to Sn−1 (see the home- work for a generalization of this). If n = n1 + n2 for two positive n1, n2 then the subset

H = {σ ∈ Sn : σ({1, . . . , n1}) = {1, . . . , n1}} is also a subgroup of Sn. Note that, if σ ∈ H, then automatically

σ({n1 + 1, . . . , n2}) = {n1 + 1, . . . , n2},

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 Dn, the group of of a regular n-gon in the plane, is (isomorphic to) a subgroup of Sn by looking at the 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 : {1, . . . , n} → {1, . . . , n}, 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 :  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 of Sn can be factored into a product of cycles, in an way if we are careful.

Definition 1.1. Let {a1, . . . , ak} be a subset of {1, . . . , n} 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 {a1, . . . , ak}. Note that, again for k > 1, the k-cycle (a1, . . . , ak) moves i ⇐⇒ i ∈ Supp(a1, . . . , ak). Two cycles (a1, . . . , ak) and (b1, . . . , b`) are disjoint if

Supp(a1, . . . , ak) ∩ Supp(b1, . . . , b`) = ∅, i.e. the sets {a1, . . . , ak} and {b1, . . . , b`} 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 {1, . . . , n} 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 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 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 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 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 ∈ {1, . . . , n}, 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 ∈ {1, . . . , n}, 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 ∈ 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 ∈ {1, . . . , n}. 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. On the other hand, if j 6= σ(ai) −1 for any i, then (σ(a1), . . . , σ(ak))(j) = j. But j 6= σ(ai) ⇐⇒ σ (j) 6= ai, so

−1 −1 σ · (a1, . . . , ak) · σ (j) = σ · (a1, . . . , ak)(σ (j)) −1 −1 = σ((a1, . . . , ak)(σ (j))) = σ(σ (j)) = j.

−1 Hence σ·(a1, . . . , ak)·σ (j) = (σ(a1), . . . , σ(ak))(j) for every j ∈ {1, . . . , n}, proving the formula.

In the discussion above, we have seen numerous computations with cy- cles, and in particular examples where the product of two cycles either is or is not again a cycle. More generally, we have the following factorization result for elements of Sn:

Theorem 1.3. Let σ ∈ Sn, σ 6= 1. Then σ is a product of disjoint cycles of lengths at least 2. Moreover, the terms in the product are unique up to order.

Remark 1.4. 1) Just as with factoring natural numbers into primes, a single cycle is a “product” of cycles (it is a product of one cycle). 2) We could also allow 1 in this framework, with the convention that 1 is the empty product, i.e. the “product” of no cycles. 3) Since disjoint cycles commute, we can always change the order in a prod- uct of disjoint cycles and the answer will be the same. However, the theorem says that is the only ambiguity possible.

Example 1.5. For the group S4, the only possibilities for products of dis- joint cycles are: 1) the identity 1; 2) a transposition, i.e. a 2-cycle; 3) a 3-cycle; 4) a 4-cycle; 5) a product of two disjoint 2-cycles. There is just 4 one identity 1. The number of transpositions is = 6. For a 3-cycle 2 (a1, a2, a3), there are 4 choice for a1, then 3 choices for a2, then 2 choices for a3, giving a total of 4 · 3 · 2 = 24 choices for the ordered triple (a1, a2, a3). However, as an element of S4,(a1, a2, a3) = (a2, a3, a1) = (a3, a1, a2), so the total number of different elements of S4 which are 3-cycles is 24/3 = 8. Like- wise, the total number of 4-cycles, viewed as elements of S4, is 4·3·2·1/4 = 6. Finally, to count the number of products (a, b)(c, d) of two disjoint 2-cycles, note that, as above, there are 6 choices for (a, b). The choice of (a, b) then determines c and d since {c, d} = {1, 2, 3, 4} − {a, b}. But since (a, b)(c, d) = (c, d)(a, b), we should divide the total number by 2, to get 6/2 = 3 elements

5 of S4 which can be written as a product of two disjoint 2-cycles. As a check, adding up the various possibilities gives 1 + 6 + 8 + 6 + 3 = 24, as expected. In this notation, D4 is the subgroup of S4 given by

D4 = {1, (1, 2, 3, 4), (1, 3)(2, 4), (1, 4, 3, 2), (2, 4), (1, 3), (1, 2)(3, 4), (1, 4)(2, 3)}. The proof of the theorem gives a procedure which is far easier to under- stand and implement in practice than it is to explain in the abstract. For example, suppose we are given a concrete permutation corresponding to a 2 × n matrix, for example 1 2 3 4 5 6 7 8 9 σ = , 3 6 5 7 1 4 9 8 2 here is how to write σ as a product of disjoint cycles of lengths at least 2: beginning with 1, we see that σ(1) = 3, σ(3) = 5, σ(5) = 1, so we have returned to where we started. Write down the cycle (1, 3, 5). Searching for elements not in the support of (1, 3, 5), we see that the first such is 2. Then σ(2) = 6, σ(6) = 4, σ(4) = 7, σ(7) = 9, σ(9) = 2 so that we are again where we started. Write down the cycle (2, 6, 4, 7, 9). Search again for elements not in the support of (1, 3, 5) or (2, 6, 4, 7, 9). The only remaining element is 8, and σ(8) = 8. As we omit 1-cycles, we are left with the factorization σ = (1, 3, 5)(2, 6, 4, 7, 9) = (2, 6, 4, 7, 9)(1, 3, 5). To describe this procedure in general, we introduce a new idea.

Definition 1.6. Let σ ∈ Sn. We define a relation ∼σ on {1, . . . , n} as follows: for i, j ∈ {1, . . . , n}, i ∼σ j if there exists an r ∈ Z such that σr(i) = j. The orbit of i under σ is the set

r Oσ(i) = {σ (i): r ∈ Z}. Before giving examples, we note the following:

Proposition 1.7. The relation ∼σ is an and the equiv- alence class [i] of i for ∼σ is the orbit Oσ(i). 0 Proof. Clearly i = σ (i), so that i ∼σ i. Hence ∼σ is reflexive. If i ∼σ j, then r −r by definition there exists an r ∈ Z such that σ (i) = j. Thus i = σ (j), so that j ∼σ i and ∼σ is symmetric. To see that it is transitive, suppose r that i ∼σ j and j ∼σ k. Thus there exist r, s ∈ Z such that σ (i) = j and s r+s s r s σ (j) = k. Then σ (i) = σ (σ (i)) = σ (j) = k. Thus i ∼σ k and so ∼σ is transitive, hence an equivalence relation. By definition, the equivalence class [i] is Oσ(i).

6 By general facts about equivalence relations, the orbits Oσ(i) partition {1, . . . , n} into disjoint , in the sense that every integer j ∈ {1, . . . , n} is in exactly one orbit Oσ(i). Note that Oσ(i) = {i} ⇐⇒ σ fixes i. For example, the identity permutation has the n orbits {1}, {2},..., {n}.A k- cycle (a1, . . . , ak) with k ≥ 2 has one orbit {a1, . . . , ak} with k elements, and the remaining orbits are one element orbits {i} for the i ∈ {1, . . . , n} such that i 6= ar for any r. There are then n − k one element orbits and one orbit with k elements, for a total of n − k + 1 orbits. For another example, given the σ ∈ S9 described above, σ = (1, 3, 5)(2, 6, 4, 7, 9), the orbits of σ are {1, 3, 5}, {2, 4, 6, 7, 9}, and {8}. More generally, if σ is a product of disjoint cycles ρ1 ··· ρ` of lengths at least 2, then the orbits Oσ(i) are the supports of the cycles ρi in addition to one element orbits for each i ∈ {1, . . . , n} which is not in the support of any ρi. Note that, while σ determines the orbits, the orbits do not completely determine σ. For example, σ0 = (1, 5, 3)(2, 9, 7, 6, 4) has the same set of orbits as σ. On the other hand, the orbits do determine the “shape” of σ, in other words they tell us in this case that σ is a product of a disjoint 3-cycle and a 5-cycle, and they tell us that the support of the 3-cycle is {1, 3, 5} and the support of the 5-cycle is {2, 4, 6, 7, 9}. Let us make two remarks about the equivalence relation ∼σ above. First, d since Sn is finite, every element σ has finite order, say σ = 1 with d > 0. Then σd−1 = σ−1, so we can write every negative power of σ as a positive power as well. Thus r Oσ(i) = {σ (i): r ∈ N}. Second, given i ∈ {1, . . . , n}, the orbit of i is the set {i, σ(i), σ2(i),... } by the above discussion. Inductively, set a1 = i and define ak+1 = σ(ak), so that k−1 ak = σ (i). Thus the orbit Oσ(i) is just {a1, a2,... }, with σ(ak) = ak+1. Note that the set {a1, a2,... } is finite, so there have to exist r, s ≥ 1 with r 6= s and ar = as. Let k ∈ N be the smallest integer ≥ 1 such that ak+1 = a` for some ` with 1 ≤ ` ≤ k, i.e. ak+1 is the first term in the sequence which is equal to a previous term. Equivalently, k is the largest positive integer such that a1, . . . , ak are all distinct. We claim that ak+1 = a1 = i, in other words that the sequence starts to repeat only when we come back to the starting point a1 = i. For suppose that ak+1 = a` with 1 ≤ ` ≤ k. Then k `−1 k−`+1 σ (i) = σ (i), so that σ (i) = i. Hence ak−`+2 = a1, in other words there is a repetition at stage k − ` + 2. Since k + 1 was the smallest positive integer greater than 1 for which some repetition occurs, k − ` + 2 ≥ k + 1, so that ` ≤ 1. But also ` ≥ 1, so that ` = 1 and ak+1 = a1, as claimed. Once k we have ak+1 = a1, i.e. σ (i) = i, then ak+2 = σ(i) = a2, ak+3 = a3, . . . . In

7 general, given r ∈ Z, writing r = kq + ` with 0 ≤ ` ≤ k − 1, it follows that r ` kq ` k q ` σ (i) = σ (σ (i)) = σ ((σ ) (i)) = σ (i) = a`+1, where 1 ≤ ` + 1 ≤ k, and hence that

2 k−1 Oσ(i) = {i, σ(i), σ (i), . . . , σ (i)} = {a1, . . . , ak}.

Using the above, let us show that an arbitrary σ ∈ Sn can be factored into a product of disjoint cycles as in the statement of Theorem 1.3. First, given i ∈ {1, . . . , n}, either σ(i) = i, which happens ⇐⇒ Oσ(i) = {i}, or Oσ(i) has at least 2 elements. In this second case, write Oσ(i) = {a1, . . . , ak} as above, where k = #(Oσ(i)) and σ(aj) = aj+1 for 1 ≤ j ≤ k − 1 and σ(ak) = a1. Thus, if ρ is the k-cycle (a1, . . . , ak), then ρ(aj) = aj+1 = σ(aj) for all aj, and ρ(r) = r if r∈ / {a1, . . . , ak} = Oσ(i) = Supp ρ. Now list the orbits of σ as O1,...,ON , say, where O1,...,OM are the orbits with at least two elements and OM+1,...,ON are the one-element orbits. For each orbit Or with r ≤ M, we have found a cycle ρr of length at least 2, with Supp ρr = Or, ρr(i) = σ(i) if i ∈ Or, and ρr(i) = i if i∈ / Or. Then the cycles ρ1, . . . , ρM are disjoint, and by inspection σ = ρ1 ··· ρM . This proves that σ is a product of disjoint cycles, and the uniqueness up to order is easy to check from the construction.  The proof shows the following:

Corollary 1.8. Let σ ∈ Sn, and suppose that the orbits of σ are O1,...,ON , where #(Oi) = ki ≥ 2 if i ≤ M and #(Oi) = 1 if i > M. Then

σ = ρ1 ··· ρM , where each ρi is a ki-cycle, Supp ρi = Oi, and hence, if i 6= j, ρi and ρj are disjoint.

2 The sign of a permutation

We now look for further ways to factor elements of Sn, not in general uniquely. First, we note that the k-cycle (1, . . . , k) is a product of transpo- sitions: (1, . . . , k) = (1, k)(1, k − 1) ··· (1, 3)(1, 2), since the right hand side sends 1 to 2, 2 to 1 and then to 3, 3 to 1 and then to 4, and so on, and finally k to 1. There is nothing special about choosing the cycle (1, . . . , k): if a1, . . . , ak are k distinct elements of {1, . . . , n}, then

(a1, . . . , ak) = (a1, ak)(a1, ak−1) ··· (a1, a3)(a1, a2).

8 Hence every k-cycle is a product of k − 1 transpositions. Since every per- mutation σ ∈ Sn is a product of k-cycles and every k-cycle is a product of transpositions, we conclude:

Theorem 2.1. Every element σ ∈ Sn is a product of transpositions. Intuitively, to permute a set with n elements, it is enough to successively n switch two at a time. This says that the transpositions (i, j) with i < j 2 generate the group Sn. In fact, as in the homework, Sn can be generated by 2 elements, although in many ways the most natural generating set is given by the n − 1 elements (1, 2), (2, 3),..., (n − 1, n). It is easy to see that there is in general no way to write a permutation uniquely as a product of transpositions. For example, we can always insert the identity which is a product (i, j)(i, j) of a transposition with itself. For another example, corresponding to the fact that (1, 2, 3) = (2, 3, 1) = (3, 1, 2) and the above recipe for writing a 3-cycle as a product of transpositions, we have (1, 3)(1, 2) = (1, 2)(2, 3) = (2, 3)(1, 3).

If this product is taking place in Sn, n ≥ 5, then we have many more ways of writing (1, 2, 3) as a product of transpositions, for example (1, 2, 3) = (4, 5)(1, 3)(2, 4)(2, 4)(1, 2)(4, 5). Despite the lack of uniqueness, there is one feature that all of the ways of writing a permutation as a product of transpositions have in common:

Theorem 2.2. Let σ ∈ Sn, and suppose that σ = τ1 ··· τk = ρ1 ··· ρ`, where the τi and ρj are all transpositions. Then k ≡ ` (mod 2). In other words, a given element σ of Sn can be written as an even number of transpositions or as an odd number of transpositions, but not both.

Thus every element of Sn has a well-defined parity, i.e. is either even or odd, depending on whether it can be written as an even number of trans- positions or as an odd number of transpositions. We will describe three different proofs of the theorem; each one is instructive. First Proof. Consider the positive integer Y Nn = (j − i), 1≤i

9 of Nn is not important (exercise: show that Nn = (n − 1)!(n − 2)! ··· 2!1!); clearly, Nn is just a large positive integer, and the main point is that it is nonzero since no factor is zero. Given σ ∈ Sn, consider what happens when Y we consider instead the product (σ(j)−σ(i)). This is again a product 1≤i

Note that ε(σ) just depends on σ, not on how σ is expressed as a product of transpositions. The two main facts we need about ε(σ) are

1. For all σ1, σ2 ∈ Sn, ε(σ1σ2) = ε(σ1)ε(σ2) (i.e. ε is multiplicative); and 2. If τ = (a, b) is a transposition, then ε(τ) = −1.

Assuming (1) and (2), let us finish the proof. If σ = τ1 ··· τk is a product of transpositions, then using (1) repeatedly and then (2), we see that

k ε(σ) = ε(τ1 ··· τk) = ε(τ1) ··· (τk) = (−1) .

Hence ε(σ) = 1 if σ is a product of an even number of transpositions and ε(σ) = −1 if σ is a product of an even number of transpositions. In particular, if also σ = ρ1 ··· ρ` where the ρi are transpositions, then ε(σ) = (−1)` = (−1)k, and hence k ≡ ` (mod 2). We will not write down the proof of (1); it follows from carefully looking at how ε(σ) is defined. However, we will prove (2). Suppose that τ = (a, b) is a transposition, where we may assume that a < b. We examine for which pairs i < j, 1 ≤ i < j ≤ n, the difference τ(j) − τ(i) is negative. Note that, if neither i nor j is a or b, then τ(j)−τ(i) = j −i > 0. Thus we may assume that at least one of i, j is either a or b. For the moment, we assume that (i, j) 6= (a, b), i.e. either j = a or i = b, but not both. If j = a, then i < a, hence i 6= b, and τ(a) − τ(i) = b − i > a − i > 0. Thus the difference of the pairs a − i don’t change sign. Similarly, if i = b and j > b, then τ(j) − τ(b) = j − a > j − b > 0. So the only possible sign changes come from differences j −a, with j > a, or b−i with i < b. Consider first the case of j − a. If j > b, then τ(j) − τ(a) = j − b > 0. If b > j > a,

10 then τ(j) − τ(a) = j − b < 0, and there are b − a − 1 such j, so these contribute a factor of (−1)b−a−1. Likewise, when we look at the difference b − i, then τ(b) − τ(i) = a − i. The sign is unchanged if i < a but becomes negative if a < i < b. Again, there are b − a − 1 such i, so these contribute another factor of (−1)b−a−1 which cancels out the first factor of (−1)b−a−1. The only remaining pair of integers that we have not yet considered is the pair where i = a and j = b. Here τ(b) − τ(a) = a − b = −(b − a), so there is one remaining factor of −1 which is not canceled out. Thus, the total number of sign changes is ε(τ) = (−1)b−a−1(−1)b−a−1(−1) = −1. Finally, we note a fact which is useful in Modern Algebra II: it didn’t matter that we chose the integers between 1 and n and looked at their differences. We could have started with any sequence t1, . . . , tn of, say, real numbers such that the ti are all distinct and looked at the product Y (ti−tj). Comparing this product with the product where we permute 1≤i

P (σ1σ2)(ei) = eσ1σ2(i);

P (σ1)P (σ2)(ei) = P (σ1)(eσ2(i)) = eσ1σ2(i).

11 Hence P (σ1σ2) and P (σ1)P (σ2) have the same value on each basis vector, hence on all vectors, and thus P (σ1σ2) = P (σ1)P (σ2). Now, to convert the matrix P (σ) to a number, we have the det. Define ε(σ) = det P (σ). As it stands, ε(σ) is just a real number. However, using the multiplicative property of the determinant, for all σ1, σ2 ∈ Sn, we have

ε(σ1σ2) = det(P (σ1σ2)) = det(P (σ1)P (σ2)) = det P (σ1) det P (σ2) = ε(σ1)ε(σ2).

Clearly ε(1) = det I = 1. Moreover, if τ = (i, j) is a transposition, then P (τ) is obtained from I by switching the ith and jth columns of I. Another well-known property of the determinant says that, in this case, det P (τ) = − det I = −1. Putting this together, we see that, if σ = τ1 ··· τk is a product of transpositions, then, as in the first proof,

k ε(σ) = ε(τ1 ··· τk) = ε(τ1) ··· ε(τk) = (−1) , and hence k is either always even or always odd. Third Proof. In this proof, we don’t define the function ε directly, but just use information about the orbits of σ to see that σ cannot simultaneously be the product of an even and an odd number of transpositions. For any σ ∈ Sn, define N(σ) to be the number of orbits of σ, including the one- element orbits. For example, as we saw earlier, if σ is an r-cycle, then N(σ) = n − r + 1. The main point of the third proof is then:

Claim 2.3. If σ ∈ Sn is a product of k transpositions, i.e. σ = τ1 ··· τk where each τi is a transposition, then

N(σ) ≡ n − k (mod 2).

To see that Claim 2.3 implies the theorem, suppose that σ = τ1 ··· τk = ρ1 ··· ρ`, where the τi and ρj are transpositions. Then we have both that N(σ) ≡ n−k (mod 2) and that N(σ) ≡ n−` (mod 2), hence k ≡ ` (mod 2).

Proof of Claim 2.3. We prove Claim 2.3 by induction on k. If k = 1, then σ is a transposition, hence a 2-cycle, and hence N(σ) = n−2+1 = n−1 by the discussion before the statement of the claim, so the congruence in the claim is in fact an equality. By induction, suppose that the claim is true for all products of k transpositions, and consider a product of k + 1 transpositions, say σ = τ1 ··· τkτk+1. If we set ρ = τ1 ··· τk, then by the inductive hypothesis N(ρ) ≡ n−k (mod 2), and we must show that N(ρτk+1) ≡ n−k−1 (mod 2). Clearly, it is enough (since 1 ≡ −1 (mod 2)) to show the following:

12 Claim 2.4. If ρ ∈ Sn and τ ∈ Sn is a transposition, then N(ρτ) ≡ N(ρ) + 1 (mod 2).

In fact, N(ρτ) = N(ρ) ± 1.

Proof of Claim 2.4. Label the orbits of ρ (including the one-element orbits) as O1,...,ON , where N = N(ρ). As usual, we can assume that O1,...,OM have at least two elements and that OM+1,...,ON are the one-element orbits. We must show that ρτ has N ± 1 orbits. Since τ is a transposition, we can write τ = (a, b) for some a, b ∈ {1, . . . , n}, a 6= b.

Case I. There is some orbit Oi of ρ, necessarily with at least two elements, such that a, b ∈ Oi. In this case, we claim that N(ρτ) = N(ρ) + 1. For each j ≤ M, let Oj correspond to the kj-cycle µj, so that ρ = µ1 ··· µM , where the µj are disjoint and hence commute. For the cycle µi, we know that a, b ∈ Supp µi = Oi, and a, b are not in the support of any µj, j 6= i. Then (a, b) commutes with all of the µj, j 6= i, and hence

ρτ = µ1 ··· µM τ = µ1 ··· (µiτ) ··· µM .

If we write µi = (i1, . . . , iki ), then one of the ir, say it, is a, and another, say is, is b, and after switching a and b we can assume that t < s. Thus we can write µi = (i1, . . . , it−1, a, it+1, . . . , is−1, b, is+1, . . . , iki ). Of course, it is possible that t = 1 and s = 2, say. Now consider the product

µiτ = (i1, . . . , it−1, a, it+1, . . . , is−1, b, is+1, . . . , iki )(a, b).

We see that a is sent to is+1, is+1 is sent to is+2,..., iki−1 to iki , iki to i1,. . . , and finally it−1 is sent back to a. As for b, it is sent to it+1, it+1 is sent to it+2, . . . , and finally is−1 is sent back to b. In other words,

µiτ = (i1, . . . , it−1, a, it+1, . . . , is−1, b, is+1, . . . , iki )(a, b) 0 00 = (a, is+1, . . . , iki , i1, . . . , it−1)(b, it+1, . . . , is−1) = µ µ , say, where µ0 and µ00 are two cycles, disjoint from each other and from all of the other µj and the remaining one-element orbits. (Note: it is possible that one or both of µ0, µ00 is a one-cycle. How could this happen?) It follows that the orbits of ρτ are the Oj for j 6= i together with the two new orbits O0,O00 coming from µ0 and µ00. Counting them, we see that there are N + 1 orbits in all, hence N(ρτ) = N(ρ) + 1. Case II. In this case, we assume that a and b are in different orbits, say a ∈ Oi and b ∈ Oj with i 6= j. We let µi, µj be the corresponding cycles.

13 Note that it is possible that either Oi or Oj is a one-element orbit, in which case we will simply define µi or µj to be the corresponding one-cycle (in particular, as an element of Sn, it would then be the identity). Then

ρτ = µ1 ··· µM τ = µ1 ··· (µiµjτ) ··· µM , where we have simply moved the µi and µj next to each other since all of the µi commute. In this case, we claim that N(ρτ) = N(ρ) − 1. Write as before

µi = (i1, . . . , it−1, a, it+1, . . . , iki ) and µj = (`1, . . . , `s−1, b, `s+1, . . . , `kj ). A calculation similar to that in Case I shows that

µiµjτ = (i1, . . . , it−1, a, it+1, . . . , iki )(`1, . . . , `s−1, b, `s+1, . . . , `kj )(a, b)

= (a, `s+1, . . . , `kj , `1, . . . , `s−1, b, it+1, . . . , iki , i1, . . . , it−1) = µ, say. Thus the orbits of ρτ are the Ok for k 6= i, j together with O = Supp µ. Counting these up, we see that there are N − 1 orbits in all, hence N(ρτ) = N(ρ) − 1.

Definition 2.5. If σ ∈ Sn, we define ε(σ) = 1, if σ is a product of an even number of transpositions, and ε(σ) = −1, if σ is a product of an odd number of transpositions. Likewise, we define σ to be even if σ is a product of an even number of transpositions (i.e. ε(σ) = 1) and odd if σ is a product of an odd number of transpositions (i.e. ε(σ) = −1). The content of the previous theorem is then that ε is well-defined. Finally we define An, the , to be the subset of Sn consisting of even permutations.

Lemma 2.6. Let σ1 and σ2 be elements of Sn. If σ1 and σ2 are both even or both odd, then the product σ1σ2 is even. If one of σ1, σ2 is even and the other is odd, then the product σ1σ2 is odd. Equivalently, ε(σ1σ2) = ε(σ1)ε(σ2).

Proof. Suppose that σ1 can be written as a product τ1 ··· τk of k transposi- tions and that σ2 can be written as a product ρ1 ··· ρ` of ` transpositions. Then σ1σ2 = τ1 ··· τkρ1 ··· ρ` is a product of k + ` transpositions. (Confus- ingly, for the product σ1σ2 we take the sum of the numbers of transposi- tions.) From this, the proof of the lemma is clear: if k and ` are both even or both odd, then k + ` is even, and if one of k, ` is even and the other is odd, then k + ` is odd.

Corollary 2.7. An is a subgroup of Sn.

Proof. If σ1, σ2 ∈ An, then σ1 and σ2 are both even, hence σ1σ2 is even, hence σ1σ2 ∈ An and An is closed under taking products. The element 1 is

14 even, i.e. 1 ∈ An, either formally (1 is the product of 0 transpositions and 0 is even) or by writing 1 = (a, b)(a, b), which is possible as long as n ≥ 2. Finally, if σ ∈ An, then σ = τ1 ··· τk is the product of k transpositions, where k is even. Then

−1 −1 −1 −1 σ = (τ1 ··· τk) = τk ··· τ1 = τk ··· τ1,

−1 i.e. σ is the product of the k transpositions τ1, . . . , τk in the opposite order. In particular, σ−1 is the product of an even number of transpositions, −1 hence σ ∈ An. Thus An is a subgroup of Sn. Example 2.8. First note that, as we have already seen, a k-cycle is a product of k − 1 transpositions if k ≥ 1 (and a 1-cycle is a product of two transpositions). Thus a k-cycle is is even if k is odd and odd if k is even. Hence we can determine the parity of any product of cycles (whether or not they are disjoint). This gives the following description of the subgroups An for small n: A1 = S1 = {1}. A2 = {1}. A3 consists of 1 together with the two different 3-cycles in S3: A3 = {1, (1, 2, 3), (1, 3, 2)}. A4 consists of 1, the 8 3-cycles, and the 3 products of two disjoint 2-cycles. Hence #(A4) = 12 = 1 1 2 #(S4), and in hindsight we also see that #(A2) = 1 = 2 #(S2) and that 1 #(A3) = 3 = 2 #(S3). This is not a coincidence: 1 Proposition 2.9. For n ≥ 2, #(An) = 2 #(Sn) = n!/2.

Proof. Since every element of Sn is either even or odd, Sn is the disjoint union of An, the set of even permutations, and Sn − An, the set of odd permutations. We will find a from An to Sn − An. This will imply that #(An) = #(Sn − An), hence

#(Sn) = #(An) + #(Sn − An) = 2#(An),

1 hence #(An) = 2 #(Sn). To find a bijection from An to Sn − An, choose an odd permutation ρ. For example, we can take ρ = (1, 2). Note that this is only possible for n ≥ 2, since for n = 1 S1 = {1} and every permutation is trivially even. Then define a function f : An → Sn − An by the rule:

f(σ) = ρ · σ, the product of σ with the fixed odd permutation ρ. If σ ∈ An, then σ is even and hence ρσ is odd, so that f is really a function from An to Sn − An. To show that f is a bijection, it is enough to find an , i.e. a

15 function g : Sn − An → An such that f ◦ g and g ◦ f are both the identity (on their respective domains). Define g : Sn − An → An by the rule:

g(σ) = ρ−1σ.

If ρ is odd, then ρ−1 is odd, hence, if σ is also odd, then ρ−1σ is even. Thus g is really a function from Sn − An to An. Finally, if σ ∈ An, then −1 −1 g ◦ f(σ) = ρ ρσ = σ, and if σ ∈ Sn − An, then f ◦ g(σ) = ρρ σ = σ. Thus −1 f ◦ g = IdSn−An and g ◦ f = IdAn , so that g = f and f is a bijection.

The group An is a group of fundamental importance, for reasons which will gradually emerge this semester, and will play a major role in Modern Algebra II as well.

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