21 Symmetric and alternating groups
Recall. The symmetric group on n letters is the group
Sn = Perm({1, . . . , n})
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 |S| = n we have Perm(S) = Sn.
21.2 Notation. Denote [n] := {1, . . . , n}
If σ ∈ 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 ∈ Sn is a cycle of length r (or r-cycle) if there are distinct integers i1, . . . , ir ∈ [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 ∈ Sn are disjoint if
{i ∈ [n] | σ(i) 6= i} ∩ {j ∈ [n] | τ(j) 6= j} = ∅
21.7 Proposition. If σ, τ are disjoint permutations then στ = τσ.
Proof. Exercise.
21.8 Proposition. Every non-identity permutation σ ∈ 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 ∈ 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 ∈ Z, i ∈ [n]. Notice that
k Orb(i) = {σ (i) | k ∈ Z}
Define σi :[n] → [n] ( σ(j) if j ∈ Orb(i) σi(j) = j otherwise
Notice that σi is a bijection since σ(Orb(i)) = Orb(i). Thus σi ∈ Sn. Check:
1) σi is a cycle of length |Orb(i)|.
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 ∈ 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 ∈ Sn we have a bijection σ × σ :[n] × [n] → [n] × [n]
85 given by σ × σ(i, j) = (σ(i), σ(j)). Define
Sσ := {(i, j) ∈ [n] × [n] | i > j and σ(i) < σ(j)}
21.11 Definition. A permutation σ ∈ 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 ∈ Sn. Denote sσ = |Sσ|. We want to show
sτσ ≡ sτ + sσ (mod 2)
+ + Let [n] := {(i, j) ∈ [n] × [n] | i > j}. Define subsets Pσ,Rσ,Pτ ,Rτ ⊆ [n] as follows:
−1 −1 Pσ :={(i, j) | σ (i) > σ (j)} −1 −1 Rσ :={(i, j) | σ (i) < σ (j)}
Pτ :={(i, j) | τ(i) > τ(j)}
Rτ :={(i, j) | τ(i) < τ(j)}
Notice that sσ = |Rσ| and sτ = |Rτ |. Notice also that (i, j) ∈ Sτσ iff either (τ(i), τ(j)) ∈ Pσ ∩ Rτ or (τ(j), τ(i)) ∈ Rσ ∩ Pτ . This gives
sτσ = |Pσ ∩ Rτ | + |Rσ ∩ Pτ |
86 On the other hand we have:
sσ = |Rσ| = |Rσ ∩ Pτ | + |Rσ ∩ Rτ |
sτ = |Rτ | = |Pσ ∩ Rτ | + |Rσ ∩ Rτ | Therefore
sσ + sτ = |Rσ ∩ Pτ | + |Pσ ∩ Rτ | + 2|Rσ ∩ Rτ | = sτσ + 2|Rσ ∩ Rτ | and so sτ + sσ ≡ sτσ (mod 2).
2) Exercise.
21.13 Definition/Proposition. The set
An = {σ ∈ Sn | σ is even} 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 |Sn| = n! thus |An| = 2 .
21.14 Proposition. If σ ∈ 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 ∈ 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 ∈ 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, 1} = 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 ∈ [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 ∈ An
Proof. Check: if (i1 i2 . . . ir) is a cycle in Sn then for any ω ∈ Sn we have
−1 ω(i1 i2 . . . ir)ω = (ω(i1) ω(i2) . . . ω( ir))
0 If σ = (i1 i2 i3), σ = (j1 j2 j3) then take ω ∈ Sn such that ω(ik) = jk for k = 1, 2, 3. We have σ0 = ωσω−1
89 If ω ∈ An we can then take τ := ω.
Assume then that ω 6∈ An. Since n ≥ 5 there are r, s ∈ [n] such that (r s) and σ = (i1 i2 i3) are disjoint cycles. Take τ = ω(r s). Then τ ∈ An. Moreover, since (r s) commutes with σ we have
τστ −1 = ω(r s)σ(r s)−1ω−1 = ωσω−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= {(1)}. 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 ∈ 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) ∈ An. Since H is normal in An we have
τστ −1σ−1 ∈ 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 ∈ 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.
Thus τστ −1σ−1 has more fixed points than σ which is again impossible by the definition of σ.
As a consequence σ must be a 3-cycle.
22.5. Classification of simple finite groups.
1) cyclic groups Z/pZ, p – prime
2) alternating groups An, n ≥ 5 3) finite simple groups of Lie type, e.g. projective special linear groups
PSLn(F) := SLn(F)/Z(SLn(F))
F -finite field, n ≥ 2 (and n > 2 if F = F2 or F = F3).
91 4) 26 sporadic groups (the smallest: Mathieu group M11, |M11| = 7920, the biggest: the Monster M, |M| ≈ 8 · 1053).
92 23 Solvable groups
Recall. Every finite group G has a composition series:
{e} = G0 ⊆ ... ⊆ Gk = G where Gi−1 C Gi and Gi/Gi−1 is a simple group.
23.1 Definition. A group G is solvable if it has a composition series
{e} = G0 ⊆ ... ⊆ Gk = G ∼ such that for every i the group Gi/Gi−1 is a simple abelian group (i.e. Gi/Gi−1 = Z/piZ for some prime pi).
23.2 Example.
1) Every finite abelian group is solvable.
2) For n ≥ 5 the symmetric group Sn has a composition series
{(1)} ⊆ An ⊆ Sn
and so Sn is not solvable.
23.3 Proposition. A finite group G is solvable iff it has a normal series
{e} = H0 ⊆ ... ⊆ Hl = G such that Hj/Hj−1 is an abelian group for all j.
Proof. Exercise.
93 Recall. 1) If G is a group the [G, G] is the commutator subgroup of G [G, G] = h{aba−1b−1 | a, b ∈ G}i
2) [G, G] is the smallest normal subgroup of G such that G/[G, G] is abelian: if G/H for some H C G then [G, G] ⊆ H.
23.4 Definition. For a group G the derived series of G is the normal series · · · ⊆ G(2) ⊆ G(1) ⊆ G(0) = G where Gi+1 = [G(i),G(i)] for i ≥ 1. The group G(i) is called the i-th derived group of G.
23.5 Theorem. A group G is solvable iff G(n) = {e} for some n ≥ 0.
Proof. Exercise.
23.6 Theorem. 1) Every subgroup of a solvable group is solvable. 2) Ever quotient group of a solvable group is solvable. 3) If H CG, and both H and G/H are solvable groups then G is also solvable.
Proof.
1) If H ⊆ G then H(i) ⊆ G(i). Thus if G(n) = {e} then H(n) = {e}.
2) For H C G take the canonical epimorphism f : G → G/H. We have f(G(i)) = (G/H)(i)
94 Therefore if G(n) = {e} then (G/H)(n) = {e}.
(m) (n) 3) Assume that H C G, and that H , (G/H) are trivial groups. Consider the canonical epimorphism f : G → G/H. We have
f(G(n)) = (G/H)(n) = {e}
Therefore G(n) ⊆ Ker(f) = H. As a consequence we obtain
(m) G(n+m) = G(n) ⊆ H(m) = {e}
23.7 Theorem (Feit-Thompson). Every finite group of odd order is solvable.
Proof. See: W. Feit, J.G. Thompson, Solvability of groups of odd order, Pacific Journal of Mathematics 13(3) (1963), 775-1029.
23.8 Corollary. There are no non-abelian finite simple groups of odd order.
Proof. Let G 6= {e} be a simple group of odd order. By Theorem 23.7 G is solvable so [G, G] 6= G. Since [G, G] C G, by simplicity of G we must have [G, G] = {e}, and so G is an abelian group.
95