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Home , Alternating group, Dihedral group, Group isomorphism

1.6. SYMMETRIC, ALTERNATING, AND DIHEDRAL GROUPS 19

1.6 Symmetric, Alternating, and Dihedral Groups

1.6.1 Symmetric groups Sn

Denote In := {1, 2, ··· , n}. The symmetric Sn is the group of all bijections In → In with functional composition operation.

Def. Fix n. Given distinct elements i1, ··· , ir ∈ {1, ··· , n}, we use (i1i2 ··· ir) to denote the that maps i1 7→ i2, i2 7→ i3, ··· , ir 7→ i1. (i1i2 ··· ir) is called an r-cycle, and r is the or the length of this cycle. A 2-cycle is called a transposition.

Def. The σ1, ··· , σr of Sn are said to be disjoint provided that for each k ∈ In, there is at most one r ∈ 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 ∈ 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 ∈ Z. Then ∼ is an equivalent relationship. So In is partitioned into ∼ equivalence classes. Each class has finitely many elements. Suppose S := {i1, ··· , ir} is a ∼ equivalent class. Then i1, σ(i1), 2 + t t−s σ (i2), ··· , are elements of S. There is a minimal t ∈ Z such that σ (i1) = σ (i1) for k k+s some s ∈ {1, 2, ··· , t}. Since σ is a bijection, we have σ (i1) = σ (i1) for any 2 s−1 k ≥ 0, and S = {i1, σ(i1), σ (i1), ··· , σ (i1)}. 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 ∈ Sn is the least common multiple of the orders of its disjoint cycles.

Two permutations σ1, σ2 ∈ Sn are conjugate in Sn if there exists a permutation g ∈ 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 ∈ 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 ∈ 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 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 . Moreover, det : Wn → {±1} is a . So det ◦f : Sn → {±1} is a group homomorphism. Obviously, det ◦f ((ab)) = −1 for every transposition (ab) ∈ Sn. Hence for σ ∈ 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 of Sn of index 2.

So An is a of Sn with |An| = n!/2. An is called the on n letters.

Lem 1.28. Let r, s be distinct elements of {1, 2, ··· , n}. Then An (n ≥ 3) is generated by the 3-cycles {(rsk) | 1 ≤ k ≤ n, k 6= r, s}.

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 {(rsk) | 1 ≤ k ≤ n, k 6= r, s}.

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) ∈ N, then for any d 6= a, b, c,

(abd) = (ab)(cd)(abc)2(cd)(ab) = [(ab)(cd)](abc)2[(ab)(cd)]−1 ∈ 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 .

Ex. A 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 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 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.