21 Symmetric and alternating groups Recall. The symmetric group on n letters is the group Sn = Perm(f1; : : : ; ng) 21.1 Theorem (Cayley). If G is a group of order n then G is isomorphic to a subgroup of Sn. Proof. Let S be the set of all elements of G. Consider the action of G on S G × S ! S; a · b := ab This action defines a homomorphism %: G ! Perm(S). Check: this homomor- phism is 1-1. It follows that G is isomorphic to a subgroup of Perm(S). Finally, ∼ since jSj = n we have Perm(S) = Sn. 21.2 Notation. Denote [n] := f1; : : : ; ng If σ 2 Sn, σ :[n] ! [n] then we write 1 2 3 : : : n ! σ = σ(1) σ(2) σ(3) : : : σ(n) 21.3 Definition. A permutation σ 2 Sn is a cycle of length r (or r-cycle) if there are distinct integers i1; : : : ; ir 2 [n] such that σ(i1) = i2; σ(i2) = i3; : : : ; σ(ir) = i1 and σ(j) = j for j 6= i1; : : : ; ir. A cycle of length 2 is called a transposition. 83 Note. The only cycle of length 1 is the identity element in Sn. 21.4 Notation. If σ is a cycle as above then we write σ = (i1 i2 : : : ir) 21.5 Example. In S5 we have 1 2 3 4 5! = (2 4 5 3) 1 4 2 5 3 Note: (2 4 5 3) = (4 5 3 2) = (5 3 2 4) = (3 2 4 5). 21.6 Definition. Permutations σ; τ 2 Sn are disjoint if fi 2 [n] j σ(i) 6= ig \ fj 2 [n] j τ(j) 6= jg = ? 21.7 Proposition. If σ; τ are disjoint permutations then στ = τσ. Proof. Exercise. 21.8 Proposition. Every non-identity permutation σ 2 Sn is a product of dis- joint cycles of length ≥ 2. Moreover, this decomposition into cycles is unique up to the order of factors. 21.9 Example. Let σ 2 S9 1 2 3 4 5 6 7 8 9! σ = 4 7 1 3 2 6 5 9 8 Then σ = (1 4 3)(2 7 5)(8 9). 84 Proof of proposition 21.8. Consider the action of Z on the set [n] given by k · i = σk(i) for k 2 Z, i 2 [n]. Notice that k Orb(i) = fσ (i) j k 2 Zg Define σi :[n] ! [n] ( σ(j) if j 2 Orb(i) σi(j) = j otherwise Notice that σi is a bijection since σ(Orb(i)) = Orb(i). Thus σi 2 Sn. Check: 1) σi is a cycle of length jOrb(i)j. 2) if Orb(i1);:::; Orb(ir) are all distinct orbits of [n] containing more than one element then σi1 ; : : : ; σir are non-trivial, disjoint cycles and σ = σi1 · ::: · σir Uniqueness of decomposition - easy. 21.10 Proposition. Every permutation σ 2 Sn is a product of (not necessarily disjoint) transpositions. Proof. By Proposition 21.8 it is enough to show that every cycle is a product of transpositions. We have: (i1 i2 i3 : : : ir) = (i1 ir)(i1 ir−1) · ::: · (i1 i3)(i1 i2) Note. For σ 2 Sn we have a bijection σ × σ :[n] × [n] ! [n] × [n] 85 given by σ × σ(i; j) = (σ(i); σ(j)). Define Sσ := f(i; j) 2 [n] × [n] j i > j and σ(i) < σ(j)g 21.11 Definition. A permutation σ 2 Sn is even (resp. odd) if the number of elements of Sσ is even (resp. odd). 21.12 Theorem. 1) The map sgn: Sn ! Z=2Z defined by ( 0 if σ is even sgn(σ) = 1 if σ is odd is a homomorphism. 2) If σ is a transposition then sgn(σ) = 1, so this homomorphism is non-trivial. Proof. 1) Let σ; τ 2 Sn. Denote sσ = jSσj. We want to show sτσ ≡ sτ + sσ (mod 2) + + Let [n] := f(i; j) 2 [n] × [n] j i > jg. Define subsets Pσ;Rσ;Pτ ;Rτ ⊆ [n] as follows: −1 −1 Pσ :=f(i; j) j σ (i) > σ (j)g −1 −1 Rσ :=f(i; j) j σ (i) < σ (j)g Pτ :=f(i; j) j τ(i) > τ(j)g Rτ :=f(i; j) j τ(i) < τ(j)g Notice that sσ = jRσj and sτ = jRτ j. Notice also that (i; j) 2 Sτσ iff either (τ(i); τ(j)) 2 Pσ \ Rτ or (τ(j); τ(i)) 2 Rσ \ Pτ . This gives sτσ = jPσ \ Rτ j + jRσ \ Pτ j 86 On the other hand we have: sσ = jRσj = jRσ \ Pτ j + jRσ \ Rτ j sτ = jRτ j = jPσ \ Rτ j + jRσ \ Rτ j Therefore sσ + sτ = jRσ \ Pτ j + jPσ \ Rτ j + 2jRσ \ Rτ j = sτσ + 2jRσ \ Rτ j and so sτ + sσ ≡ sτσ (mod 2). 2) Exercise. 21.13 Definition/Proposition. The set An = fσ 2 Sn j σ is eveng is a normal subgroup of Sn. It is called the alternating group on n letters. Proof. It is enough to notice that An = Ker(sgn). Note. We have ∼ Sn=An = Z=2Z n! Since jSnj = n! thus jAnj = 2 . 21.14 Proposition. If σ 2 Sn then σ is even (resp. odd) iff σ is a product of an even (resp. odd) number of transpositions. 87 Proof. If σ = τ1: : .τm where τ1; : : : ; τm are transpositions then m m X X sgn(σ) = sgn(τ1: : .τm) = sgn(τi) = 1 i=1 i=1 Thus sgn(σ) = 0 iff m is even and sgn(σ) = 1 iff m is odd. Note. If follows that if a permutation σ 2 Sn is a product of an even number of transpositions then it cannot be written as a product of an odd number of transpositions (and vice versa). 21.15 Corollary. A permutation σ 2 Sn is even iff σ = σ1σ2: : .σr Pr where σi is a cycle of length mi and i=1(mr + 1) is even. Proof. It is enough to notice that by the proof of Proposition 21.10 a cycle of length m is a product of m + 1 transpositions. Note. The usual notation for the sign of a permutation is ( 1 if σ is even sgn(σ) = −1 if σ is odd ∼ where {−1; 1g = Z=2Z is the multiplicative group of units in Z. 88 22 Simplicity of alternating groups 22.1 Theorem. The alternating group An is simple for n ≥ 5. 22.2 Lemma. For n ≥ 3 every element of An is a product of 3-cycles. Proof. It is enough to show that if n ≥ 3 and τ, σ are transpositions in Sn then τσ is a product of 3-cycles. Case 1) τ, σ are disjoint transpositions: τ = (i j), σ = (k l) for distinct elements i; j; k; l 2 [n]. Then we have τσ = (i j k)(j k l) Case 2) τ, σ are not disjoined: τ = (i j), σ = (j k). Then τσ = (i j k) 0 22.3 Lemma. If n ≥ 5 and σ, σ are 3-cycles in Sn then σ0 = τστ −1 for some τ 2 An Proof. Check: if (i1 i2 : : : ir) is a cycle in Sn then for any ! 2 Sn we have −1 !(i1 i2 : : : ir)! = (!(i1) !(i2) :::!( ir)) 0 If σ = (i1 i2 i3), σ = (j1 j2 j3) then take ! 2 Sn such that !(ik) = jk for k = 1; 2; 3. We have σ0 = !σ!−1 89 If ! 2 An we can then take τ := !. Assume then that ! 62 An. Since n ≥ 5 there are r; s 2 [n] such that (r s) and σ = (i1 i2 i3) are disjoint cycles. Take τ = !(r s). Then τ 2 An. Moreover, since (r s) commutes with σ we have τστ −1 = !(r s)σ(r s)−1!−1 = !σ!−1 = σ0 22.4 Corollary. If n ≥ 5 and H is a normal subgroup of An such that H contains some 3-cycle then H = An. Proof. By Lemma 22.3 H contains all 3-cycles, and so by Lemma 22.2 it contains all elements of An. Proof of Theorem 22.1. Let n ≥ 5, H C An and H 6= f(1)g. We need to show that H = An. By Corollary 22.4 it will suffice to show that H contains some 3-cycle. Let (1) 6= σ be an element of H with the maximal number of fixed points in [n]. We will show that σ is 3-cycle. Take the decompositon of σ into dosjoint cycles: σ = σ1σ2 · ::: · σm Case 1) σ1; : : : ; σm are transpositions. Since σ 2 An we must then have m ≥ 2. Say, σ1 = (i j), σ2 = (k l). Take s 6= i; j; k; l and let τ = (k l s) 2 An. Since H is normal in An we have τστ −1σ−1 2 H Check: 1) τστ −1σ−1 6= (1) since τστ −1σ−1(k) 6= k 90 2) τστ −1σ−1 fixes every element of [n] fixed by σ 3) τστ −1σ−1 fixes i; j. Thus τστ −1σ−1 has more fixed points than σ which is impossible by the definition of σ. Case 2) σr is a cycle of length ≥ 3 for some 1 ≤ r ≤ m. We can assume r = 1: σ1 = (i j k : : : ). If σ = σ1 and σ1 is a 3-cycle we are done. Otherwise σ must move at least two more elements, say p; q. In such case take τ = (k p q). We have τστ −1σ−1 2 H Check: 1) τστ −1σ−1 6= (1) since τστ −1σ−1(k) 6= k 2) τστ −1σ−1 fixes every element of [n] fixed by σ 3) τστ −1σ−1 fixes j.