Math 3121 Abstract Algebra I Lecture 16 Sections 36-37

Final Tues, Dec 9 Can have review sheet – both sides 8.5 by 11 inches. Comprehensive Review Thurs

HW for Section 34 Do Hand in (Due Dec 2): Don’t hand in: Pages 310-311: 2, 4, 7 Don’t hand in: Pages 310-311: 1, 3

HW on Section 35 Don’t hand in: Do hand in: Pages 319-321: 1, 3, 5, 7

Section 36: Sylow Theorems Primary subgroups Group actions again Counting mod p Sylow p-group of a group First Sylow Theorem Second Sylow Theorem Third Sylow Theorem

Counting Orbits Let G be a finite group, and let X be a finite G-group. Pick xi, one in each orbit of G on X. Let XG = {x in G| g x = x, for all g in G}, and let s = | XG |. Then

Counting Primary Groups Actions Theorem: Let G be a group of order pn and let X finite G-set. Then Proof: |Gxi| divides |G| and |Gxi|>1, for i > s. Thus p divides |Gxi|, for i > s. Since p also divides |G|, p divides |G| - |XG|. Thus |X| ≡ |XG| mod p.

P-Group Definition: Let p be a prime. A group G is a p-group if every element of p has order that is a power of a p. A subgroup of a group G is a p-subgroup of G if the subgroup is a itself a p-group.

Cauchy’s Theorem Theorem (Cauchy). Let p be a prime. Let G be a finite group and let p divide |G|. Then G has an element of order p, and thus a subgroup of order p. Proof: Let X = {(g1, …, gp) | gi in G and g1 g2 … gp = e} Then |X| = |G|p-1 since Gp-1 = {(g1, …, gp-1) | gi in G } has |G|p-1 elements, and is in one-to-one correspondence with X via: (g1, …, gp-1) in Gp-1 ↦ (g1, …, gp-1, gp) with gp = (g1 g2 … gp-1)-1 Since p divides |G|, p divides |X|. Let σ be the cycle (1, 2, …, p) in Sp. Let σ act on X by σ ((g1, …, gp)) = (gσ(1), gσ(2), …, gσ(p))= (g2, …, gp, g1) The group < σ > acts on X, making X a < σ >-group. By the previous theorem, |X< σ >|≡ |X| mod p. Thus p divides |X< σ >|. Since |X< σ >| contains the p-tuple (e, …, e), it must contain some other element (a1, …, ap) different than (e, …, e). Since (a1, …, ap) is fixed by σ, ai = a1 ,for all i = 1, …, p. Then a = ai = a1 is of order p.

Action on the set of Subgroups of Group Let G be a group. And let S be the collection of all subgroups of G. S is a G-set under the action of conjugation. That is if H is a subgroup, and g is in G, then the action of g on H gives g Hg-1. For any subgroup H of G, GH is the subgroup of G and GH = {g in G | g Hg-1 = H} = N[H] called the normalizer of H.

Fact If H is a finite subgroup of a group G, then N[H] = {g in G | g h g-1 in H, for all h in H}

Lemma Lemma: Let H be a p-subgroup of a finite group G. Then (N[H] : H) ≣ (G : H) mod p Proof: Let X be the set of all left cosets of H in G, and let H act on X by left translation. Then X is a H-set. Note that |X| = (G : H). XH = { x H | h x H = x H, for all h in H} These are the cosets of H that sit within N[H]. Thus |XH |= ( N[H] : H). Counting mod p: (N[H] : H) ≣ (G : H) mod p.

Corollary Corollary: Let H be a p-subgroup of a finite group G. If p divides (G: H) then N[H] is not equal to H. Proof: If p divides (G: H), then p divides (N[H]: H) (by the lemma). Thus (N[H]: H) > 1. Thus N[H] is not equal to H.

First Sylow Theorem Theorem (First Sylow Theorem): Let G be a finite group of order pn m where n ≥ 1 and where p does not divide m. Then G contains a subgroup of order pi for all integers i such that 1 ≤ i ≤ n. Every subgroup H of G of order pi is a normal subgroup of order pi+1, for 1 ≤ i ≤ n.

Proof of Sylow’s First Theorem Proof: 1) By Cauchy’s theorem, G contains a subgroup of order p. Use induction: Suppose G contains a subgroup of order pi for i < n. Let H be a subgroup of order pi. p divides (G : H). By the lemma, p divides (N[H] : H). Since H is normal in N[H], N[H]/H is a group. p divides |N[H]/H|. By Cauchy’s theorem, N[H]/H contains a subgroup of order p. Call it K. Let  be the canonical map from N[H] to N[H]/H. Then -1[K] is of order pi+1. 2) Note that in the above construction H < -1[K] ≤ N[H]. Since H is normal in N[H], it is normal in -1[K] which is of order pi+1.

Definition of Sylow p-subgroup Definition: A Sylow p-subgroup P of a group G is a maximal p-subgroup of G.

Second Sylow Theorem Theorem: Let P1 and P2 be Sylow p-subgroups of a finite group G. Then P1 and P2 are conjugate. Proof: Let X be the left cosets of P1, and let P2 act on X by y(x H) = (y x)H. Then X is a P2 set. By the first theorem in this section, |XP2| ≡ |X| mod p. But |X|= (G : P1) is not divisible by p. Thus |XP2| is not zero. Let x P1, be in XP2. Then y x P1 = x P1 for all y in P2. Thus x-1 y x in P1 for all y in P2. Thus x-1 P2 x ≤ P1. Since P1| = |P2|, x-1 P2 x = P1

Third Sylow Theorem Theorem: If G is a finite group and p divides |G| then the number of Sylow p-subgroups is congruent to 1 module p and divides |G|. Proof: Suppose P is a Sylow p-subgroup of G. Let X be the set of all Sylow p-subgroups of G and let P act on X by conjugation. XP = { T in X | xTx-1 = T, for all x in P}. This means that P is contained in N[T], for all T in XP. Now we show that XP = {P}: Note that P is in XP. Let T be in XP. Now T and P are both Sylow p-subgroups of N[T]. By the Sylow second theorem, T is conjugate to P in N[T]. Thus there is a y in N[T] such that P = yTy-1. Since T is normal in N[T], P = T. By the first theorem of this section, |X| ≡ |XP| mod p. Thus |X| ≡ 1 mod p. Finally, consider the action of G on X. Since all T in X are conjugate, there is only one orbit in X under G. By theorem 16.16, |X| divides |G|.

Examples of Sylow p-groups Find p-groups and Sylow p-groups of Z16. Solution: |Z16| = 216 . Thus the possible p-groups have order: 2, 4, 8, 16. They are: {0, 8}, {0, 4, 8, 12}, {0, 2, 4, 6, 8, 10, 12, 14}, and Z16 itself. Z16 is a Sylow p-subgroup of itself. Note that there is only one of each primary order. Find p-groups and Sylow p-groups of Z48. Solution: |Z48| = 248 . Thus the possible p-groups have order: 2, 4, 8, 16, and 3. They are: {0, 24}, {0, 6, 12, 18}, {0, 3, 6, 12, 15, 18, 21}, and {0, 16, 32}. {0, 3, 6, 12, 15, 18, 21}, and {0, 16, 32} are Sylow p-subgroups with p = 2 and p=3.

S3 These are both Sylow p-groups. Find p-groups of Sylow p-groups of S3 Solution: (First Sylow) For p = 2: {(), (12)}, {(), (23)}, {(), (13)} For p = 3: {(), (123), (132)} These are both Sylow p-groups. (Second Sylow) Note that each of the Sylow 2-groups are conjugate via conjugation by (123): (1 2 3) {(), (12)} (1 2 3) -1 = {(), (23)} (1 2 3) {(), (23)} (1 2 3) -1 = {(), (13)} (1 2 3) {(), (13)} (1 2 3) -1 = {(), (12)} Note that (12) acts on the set of Sylow 2-groups: (1 2 ) {(), (12)} (1 2 ) -1 = {(), (12)} (1 2 ) {(), (23)} (1 2 ) -1 = {(), (13)} (1 2 ) {(), (13)} (1 2 ) -1 = {(), (23)} Check out the third Sylow theorem: 3 = 1 mod 2 and 3 divides 6.

First Sylow Theorem: S4 Find p-groups and Sylow p-groups of S4. Solution: |S4.| = 24 = 23 3 (First Sylow) For p = 2, there are 6+3 = 9: {(), (1 2)}, {(), (1 3)}, {(), {(), (1 4)}, {(), (2 3)}, {(), (2 4)}, {(), (3 4)}, { (), (1 2) (3 4)}, { (), (1 3) (2 4)}, { (), (1 4) (2 3)} For p = 4, there are 3+1 = 4: {(), (1 2 3 4), (1 3) (2 4), (1 4 3 2)}, (), (1 2 4 3), (1 4) (2 3), (1 3 4 2)}, (), (1 3 2 4), (1 2) (3 4), (1 4 2 3)}, (), (1 4) (2 3), ( 1 2) (3 4), (1 3) (2 4) }, For p = 8, there are 3 dihedral subgroups: {(), (1 2 3 4), (1 3) (2 4), (1 4 3 2), (1 3), (2 4), (1 2) (3 4), (1 4) (2 3)}, (), (1 2 4 3), (1 4) (2 3), (1 3 4 2), (1 4), (2 3), ( 1 2) (3 4), (1 3) (2 4) }, (), (1 3 2 4), (1 2) (3 4), (1 4 2 3), (1 3) (2 4), (1 4)(2 3), (12), ( 3 4)},

First Sylow Theorem: S4 For p = 3: {(), (1 2 3), (1 3 2)}, {(), (1 2 4), (1 4 2)}, {(), (1 3 4), (1 4 3)}, {(), (2 3 4), (2 4 3)},

No group of order 15 is Simple Example: Suppose G is of order 15. Then G must have a subgroup of order 5. This is a Sylow 5-subgroup. The number of such subgroups must be 1 mod 5 and divides 15 Thus 1, 6, 11. However, only 1 divides 15. It must be normal. Thus G is not simple.

The Alternating group on five letters There must be a subgroup of order 5. Try: 1, 6, 11, 16, 21. Only 1 and 6 divide 60. Here is one Sylow 5-subgroup: {(), (1 2 3 4 5), (1 3 5 2 4), (1 4 2 5 3), (1 5 4 3 2)} Permute the numbers 1, …, 5 to get the other Sylow 5-subgroups: Try conjugating by (4 5) {(), (1 2 3 5 4), (1 3 4 2 5), (1 5 2 4 3), (1 4 5 3 2)} Note that there are 24 5-cycles, thus 6 different Sylow 5-groups.