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1.6 Symmetric, Alternating, and Dihedral Groups

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1.6 Symmetric, Alternating, and Dihedral Groups 1.6. SYMMETRIC, ALTERNATING, AND DIHEDRAL GROUPS 19 1.6 Symmetric, Alternating, and Dihedral Groups 1.6.1 Symmetric groups Sn Denote In := f1; 2; ··· ; ng. The symmetric group Sn is the group of all bijections In ! In with functional composition operation. Def. Fix n. Given distinct elements i1; ··· ; ir 2 f1; ··· ; ng, we use (i1i2 ··· ir) to denote the permutation that maps i1 7! i2, i2 7! i3, ··· , ir 7! i1. (i1i2 ··· ir) is called an r-cycle, and r is the order or the length of this cycle. A 2-cycle is called a transposition. Def. The permutations σ1; ··· ; σr of Sn are said to be disjoint provided that for each k 2 In, there is at most one r 2 Ir such that σi(k) 6= k. Thm 1.22. Every nonidentity permutation in Sn is uniquely (up to the order of the factors) a product of disjoint cycles, each of which has length at least 2. 1 2 3 4 5 6 Ex. In S6, 2 4 5 1 3 6 can be expressed as (1 2 4)(3 5). Sketch of Proof of Theorem 1.22. Let σ 2 Sn be the nonidentity permutation. Define a re- k lationship in In, such that i ∼ j in In if and only if σ (i) = j for some k 2 Z. Then ∼ is an equivalent relationship. So In is partitioned into ∼ equivalence classes. Each class has finitely many elements. Suppose S := fi1; ··· ; irg is a ∼ equivalent class. Then i1, σ(i1), 2 + t t−s σ (i2), ··· , are elements of S. There is a minimal t 2 Z such that σ (i1) = σ (i1) for k k+s some s 2 f1; 2; ··· ; tg. Since σ is a bijection, we have σ (i1) = σ (i1) for any integer 2 s−1 k ≥ 0, and S = fi1; σ(i1); σ (i1); ··· ; σ (i1)g. Then σ is the product of disjoint cy- 2 s−1 cles (i1 σ(i1) σ (i1) ··· σ (i1)) produced from these equivalent classes. We may omit the 1-cycles. The uniqueness is easily seen from the uniqueness of the partition. Cor 1.23. The order of a permutation σ 2 Sn is the least common multiple of the orders of its disjoint cycles. Two permutations σ1; σ2 2 Sn are conjugate in Sn if there exists a permutation g 2 Sn −1 such that gσ1g = σ2. Given two cycles of the same length in Sn, say σ1 = (a1a2 ··· ar) and σ2 = (b1b2 ··· br), let g 2 Sn be the permutation that sends ai to bi for i = 1; 2; ··· ; r. Then −1 gσ1g = σ2. Conversely, every permutation that conjugates to σ1 in Sn is a cycle of the same length as σ1. Cor 1.24. Two permutations α; β 2 Sn are conjugate in Sn if and only if they have the same cycle structure. A cycle can be expressed as a product of transpositions by the way that (x1) = (x1x2)(x1x2) and (x1x2 ··· xr) = (x1xr)(x1xr−1) ··· (x1x3)(x1x2). Then we have the following result. Cor 1.25. Every permutation in Sn can be written as a product of (not necessarily disjoint) transpositions. 20 CHAPTER 1. GROUPS 1.6.2 Alternating groups An Def. A permutation is said to be even [resp. odd] if it can be written as a product of even [resp. odd] number of transpositions. Thm 1.26. A permutation in Sn (n ≥ 2) cannot be both even and odd. n Proof. We use matrix theory to prove it. Denote ei := (0; 0; ··· ; 1 ; ··· ; 0) in R . Let i-th term Wn be the set of n × n permutations matrices, i.e. matrices each of whose columns and rows consists of exactly one nonzero entry 1. Define f : Sn ! Wn by σ 7! Pσ, where the i-th row of Pσ is eσ(i). It is easy to check that f is a group isomorphism. Moreover, det : Wn ! {±1g is a group homomorphism. So det ◦f : Sn ! {±1g is a group homomorphism. Obviously, det ◦f ((ab)) = −1 for every transposition (ab) 2 Sn. Hence for σ 2 Sn, it is a product of even [resp. odd] number of transpositions if and only if det ◦f(σ) = 1 [resp. det ◦f(σ) = −1]. Thm 1.27. The set An of all even permutations of Sn forms a subgroup of Sn of index 2. So An is a normal subgroup of Sn with jAnj = n!=2. An is called the alternating group on n letters. Lem 1.28. Let r; s be distinct elements of f1; 2; ··· ; ng. Then An (n ≥ 3) is generated by the 3-cycles f(rsk) j 1 ≤ k ≤ n; k 6= r; sg: Proof. Every element of An is a product of even numbers of transpositions. Since (ab)(cd) = (acb)(acd); (ab)(ac) = (acb); An is generated by all 3-cycles. A 3-cycle is of the form (ras) = (rsa)2; (rab) = (rsb)(rsa)2; (sab) = (rsb)2(rsa); or (abc) = (rsa)2(rsc)(rsb)2(rsa): This shows that An is generated by f(rsk) j 1 ≤ k ≤ n; k 6= r; sg: Lem 1.29. If N is a normal subgroup of An (n > 3) and N contains a 3-cycle, then N = An. Proof. If (abc) 2 N, then for any d 6= a; b; c, (abd) = (ab)(cd)(abc)2(cd)(ab) = [(ab)(cd)](abc)2[(ab)(cd)]−1 2 N: So N contains all 3-cycles. Thus N = An. 1.6. SYMMETRIC, ALTERNATING, AND DIHEDRAL GROUPS 21 Def. A group G is simple if G has no proper normal subgroups. Ex. A cyclic group is simple if and only if it is isomorphic to Zp for some prime p. Thm 1.30. The alternating group An is simple when n 6= 4. See textbook (Section 1.6) for a complete proof. The key idea is to show that every non-proper normal subgroup of An contains a 3-cycle. 1.6.3 Dihedral group Dn The subgroup of Sn generated by a = (123 ··· n) and b = (2 n)(3 (n − 1)) ··· (i (n + 2 − i)) ··· is called the dihedral group of degree n, denoted Dn. It is isomorphic to the group of all symmetries of a regular n-gon. Thm 1.31. The dihedral group Dn (n ≥ 3) is a group of order 2n whose generators a and b satisfy: 1. an = b2 = e; ak 6= e if 0 < k < n; 2. ba = a−1b..
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