Sylow Andrew Clarey Sylow Theorems Definitions/ Looking at the Structure of Arbitrary Groups Theorems Groups, Lagrange’s, Normality Class Equation, Cauchy’s Andrew Clarey First Sylow Theorem Occidental College Theorem Examples Proof Mentor: Professor Nalsey Tinberg Additional Proofs

Second Sylow Theorem December 3, 2015

Third Sylow Theorem All material comes from Saracino, Abstract Algebra unless Results Cyclic subgroups otherwise stated. Simple Groups Additional Examples

References 1 / 26 Overview

Sylow 1 Definitions/ Theorems Theorems Groups, Subgroups Andrew Clarey Lagrange’s, Normality Definitions/ Class Equation, Cauchy’s Theorem Theorems Groups, Subgroups 2 First Sylow Theorem Lagrange’s, Normality Theorem Class Equation, Cauchy’s Theorem Examples First Sylow Theorem Proof Theorem Additional Proofs Examples Proof 3 Second Sylow Theorem Additional Proofs 4 Third Sylow Theorem Second Sylow Theorem 5 Results Third Sylow Cyclic subgroups Theorem

Results Simple Groups Cyclic subgroups Additional Examples Simple Groups Additional Examples 6 References References 2 / 26 Definitions/Theorems

Sylow Theorems Andrew Clarey A set G is called a [denoted (G, ∗)] if:

Definitions/ i) G has a binary operator ∗. We write a ∗ b as ab. Theorems Groups, Subgroups ii) ∗ is associative Lagrange’s, Normality iii) there is an element e ∈ G such that Class Equation, Cauchy’s Theorem x ∗ e = e ∗ x = x, ∀x ∈ G

First Sylow iv) for each x ∈ G, ∃ y ∈ G such that x ∗ y = y ∗ x = e. We Theorem write y = x −1. Theorem Examples Proof A group G is called cyclic if ∃ x ∈ G such that Additional Proofs n G = {x |n ∈ Z} = hxi. Then x is called a generator. Second Sylow Theorem Example cyclic groups are Z, Zn. Third Sylow The of a group G, denoted |G|, is the number of Theorem

Results elements in the group. Cyclic subgroups Simple Groups Additional Examples

References 3 / 26 Definitions/Theorems

Sylow Theorems A subset H of a group (G, ∗) is called a of G if Andrew Clarey all h ∈ H form a group under ∗.

Definitions/ Theorems Theorem: Let H be a nonempty subset of a group G. Groups, Subgroups Lagrange’s, Then H is a subgroup iff: Normality Class Equation, Cauchy’s Theorem i) ∀a, b ∈ H, ab ∈ H −1 First Sylow ii) ∀a ∈ H, a ∈ H Theorem Theorem We write H ≤ G. Examples If H G, then a Left/Right of H in G is a subset Proof 6 Additional Proofs of the form aH/Ha where a ∈ G and Second Sylow Theorem aH/Ha = {ah/ha|h ∈ H}.

Third Sylow Two elements x, y ∈ G are conjugate if ∃g ∈ G such that Theorem y = g −1xg. Results −1 Cyclic subgroups If H 6 G, then gHg 6 G is a conjugate subgroup of Simple Groups Additional Examples G, ∀g ∈ G.

References 4 / 26 Definitions/Theorems

Sylow Theorems Andrew Clarey Lagrange’s Theorem: Let G be a finite group and let

Definitions/ H 6 G. Then |H| | |G|, as |G| = |H|[G : H] where [G : H] Theorems is the number of Left/Right . Groups, Subgroups Lagrange’s, Normality Let H 6 G. Then the number of Left/Right Cosets of H Class Equation, Cauchy’s Theorem in G is [G : H], called the index. First Sylow Theorem Let H 6 G. Then we say H is a if Theorem ∀h ∈ H, g ∈ G, ghg −1 ∈ H. We write H G. Examples E Proof Additional Proofs Theorem: Let H 6 G. Then the following are equivalent: Second Sylow Theorem i) H E G Third Sylow −1 Theorem ii) gHg = H, ∀g ∈ G

Results iii) gH = Hg, ∀g ∈ G Cyclic subgroups Simple Groups Additional Examples

References 5 / 26 Definitions/Theorems

Sylow Theorems If H is the only subgroup in G of order |H| then H E G. Andrew Clarey If H E G then G/H is a group called the Definitions/ whose elements are of the form gH, ∀g ∈ G, and whose Theorems Groups, Subgroups operation is ∗ such that aH ∗ bH = (a ∗ b)H. Lagrange’s, Normality Class Equation, If G, H are groups, then we can define a function φ: Cauchy’s Theorem G → H as a homomorphism if φ(g1g2) = φ(g1)φ(). First Sylow Theorem Define a surjection φ from G → G/H where g → gH. Theorem Examples Proof The of φ is given by Ker(φ) = {g ∈ G|φ(g) = eH }, Additional Proofs where eH is the identity in H and it is a normal subgroup. Second Sylow Theorem The Normalizer of H 6 G is the subset Third Sylow N(H) = {g ∈ G|gHg −1 = H}. Theorem Results The Center of a group G is the set of elements Cyclic subgroups Simple Groups Z(G) = {a ∈ G|ag = ga, ∀g ∈ G}. Additional Examples

References 6 / 26 Definitions/Theorems

Sylow Theorems

Andrew Clarey The Centralizer of a g ∈ G is the set of elements Definitions/ Theorems Z(g) = {a ∈ G|ag = ga} Groups, Subgroups Lagrange’s, Normality Theorem: The Class Equation of a group G states: Class Equation, Cauchy’s Theorem |G| = |Z(G)| + [G : Z(g1)] + ··· + [G : Z(gk )], First Sylow g1,..., gk ∈/ Z(G), where each gi is a representative of a Theorem Theorem which contains at least 2 elements. Examples Proof Cauchy’s Theorem: Let G be an , and let p Additional Proofs be a prime such that p | |G|. Then G contains an element Second Sylow Theorem of order p. That is, ∃x ∈ G so that p is the lowest Third Sylow non-zero number such that x p = e. Theorem

Results Cyclic subgroups Simple Groups Additional Examples

References 7 / 26 First Sylow Theorem

Sylow Theorems

Andrew Clarey

Definitions/ A subgroup of a group G is called a p-Sylow subgroup if n + n Theorems its order is p , p a prime and n ∈ Z , such that p | |G| Groups, Subgroups n+1 Lagrange’s, and p |G|. Normality - Class Equation, Cauchy’s Theorem First Sylow Theorem: Let G be a finite group, p a First Sylow prime, k ∈ +. Theorem Z Theorem i) If pk | |G|, then G has a subgroup of order pk . In Examples Proof particular, G has a p-Sylow subgroup. Additional Proofs k ii) Let H be any p-Sylow subgroup of G. If K 6 G, |K| = p , Second Sylow −1 Theorem then for some g ∈ G we have K ⊆ gHg . In particular, K

Third Sylow is contained in some p-Sylow subgroup of G. Theorem

Results Cyclic subgroups Simple Groups Additional Examples

References 8 / 26 Examples

Sylow Theorems

Andrew Clarey

Definitions/ 2 4 2 2 Theorems Say |G| = 2 · 3 · 5 · 7 . Then we know there will be at least Groups, Subgroups Lagrange’s, one of each: Normality Class Equation, 2-Sylow subgroup of order 4, Cauchy’s Theorem 3-Sylow subgroup of order 81, First Sylow Theorem 5-Sylow subgroup of order , Theorem Examples 7-Sylow subgroup of order . Proof Additional Proofs We also know there will be subgroups of order 2, 3, 9, 27, 5, Second Sylow and 7. Theorem

Third Sylow Theorem

Results Cyclic subgroups Simple Groups Additional Examples

References 9 / 26 Examples

Sylow Theorems 2 Andrew Clarey Let G = A4, a group of order 12 = 2 · 3

Definitions/ Theorems A = {e, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3), (1, 2, 3), (1, 2, 4), Groups, Subgroups 4 Lagrange’s, Normality (1, 3, 4), (1, 3, 2), (1, 4, 3), (1, 4, 2), (2, 3, 4), (2, 4, 3)} Class Equation, Cauchy’s Theorem First Sylow So, a 2-Sylow subgroup of G would be a subgroup of order 4, Theorem Theorem an example is: Examples Proof Additional Proofs H = {e, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)} Second Sylow Theorem Third Sylow In fact this is the only one and therefore is normal, and all Theorem subgroups of order 2 and 4 are contained within it. Results Cyclic subgroups Simple Groups Additional Examples

References 10 / 26 Examples

Sylow Theorems 1 3 Andrew Clarey Say G = SL2(Z3). Then |G| = 24 = 2 · 3 and −1 ≡ 2mod3. The only 2-Sylow subgroup is: Definitions/ Theorems Groups, Subgroups " # " # " # " # Lagrange’s, 1 0 0 −1 1 1 −1 1 Normality , , , , Class Equation, 0 1 1 0 1 −1 1 1 Cauchy’s Theorem " # " # " # " # First Sylow −1 0 0 1 −1 −1 1 −1 Theorem , , , Theorem 0 −1 −1 0 −1 1 −1 −1 Examples Proof Additional Proofs and there are 4 3-Sylow subgroups: Second Sylow Theorem " # " # " # " # Third Sylow  1 1   1 0   0 1   0 −1  Theorem , , , Results 0 1 1 1 −1 −1 1 −1 Cyclic subgroups Simple Groups Additional Examples

References 11 / 26 Proof

Sylow Theorems k Andrew Clarey We now prove that if p | |G|, then G has a subgroup of order pk . In particular, G has a p-Sylow subgroup, part i of the First Definitions/ Theorems Sylow Theorem. Groups, Subgroups Lagrange’s, Normality + k Class Equation, Let G be a group, p a prime, k ∈ Z such that p | |G|. We Cauchy’s Theorem will proceed with induction on |G|. If |G| = 2 the result is First Sylow Theorem trivial, and we are done. So, let’s assume the theorem is true Theorem Examples for all groups of order less than |G| and show it is true for |G|. Proof Additional Proofs Second Sylow Case 1: Assume ∃H < G such that p [G : H]. Theorem - |G| = [G : H]|H| so pk must divide |H|. Third Sylow Theorem By the inductive hypothesis, Since |H| < |G|, H has a Results subgroup of order pk , therefore G does as well. Cyclic subgroups Simple Groups Additional Examples

References 12 / 26 Proof

Sylow Theorems

Andrew Clarey Case 2: Assume @H < G such that p - [G : H]. Definitions/ Theorems So, ∀H < G, p | [G : H]. Groups, Subgroups Lagrange’s, P Normality By the Class Equation: |G| = |Z(G)| + [G : Z(gi )]. Class Equation, Cauchy’s Theorem Since p | |G| and p | [G : Z(gi )] ∀i, then p | |Z(G)|.

First Sylow Theorem ⇒ By Cauchy’s Theorem Z(G) has a subgroup of order p, say Theorem A. Then A G. Examples E Proof k−1 Additional Proofs So, |G/A| = |G|/p ⇒ p | |G/A|. But |G/A| < |G|. Second Sylow Theorem ⇒ The inductive hypothesis applies to G/A. So, G/A has a Third Sylow subgroup of order pk−1, say J. Theorem

Results Cyclic subgroups Simple Groups Additional Examples

References 13 / 26 Proof

Sylow Theorems Andrew Clarey Define φ : G → G/A

Definitions/ Theorems Let H = {g ∈ G|φ(gA) ∈ J}. H 6= ∅, eH = A. Then H ≤ G. Groups, Subgroups Lagrange’s, −1 −1 Normality Show that g1, g2 ∈ H ⇒ g1g2 ∈ H, i.e. Show g1g2 A ∈ J Class Equation, Cauchy’s Theorem −1 −1 But, g1g2 A = g1A(g2A) ∈ J as J < G/A, and A < H as First Sylow Theorem A ∈ J. Theorem Examples So, map φ : H → J by h → hA which is onto by definition. Proof Additional Proofs Then Ker(φ): H ∩ A = A, and therefore H/A ∼ J. Second Sylow = Theorem So, J has the form H/A for some H < G, where Third Sylow Theorem pk−1 = |H/A| = |H|/|A| = |H|/p. So, |H| = pk as required. Results Cyclic subgroups Simple Groups Additional Examples

References 14 / 26 additional proof 1

Sylow Theorems

Andrew Clarey

Definitions/ Theorems Groups, Subgroups We proceed by induction on |G|. If|G| = 2, the result is trivially Lagrange’s, Normality true. Now assume the statement is true for all groups of order Class Equation, Cauchy’s Theorem less than |G|. First Sylow Theorem Case 1: If G has a proper subgroup H such that pk divides |H|, Theorem Examples then, by our inductive assumption, H has a subgroup of order Proof k Additional Proofs p and we are done. Second Sylow Theorem

Third Sylow Theorem

Results Cyclic subgroups Simple Groups Additional Examples

References 15 / 26 additional proof 1

Sylow k Theorems Case 2: We may assume that p does not divide the order of Andrew Clarey any proper subgroup of G. Next, consider the class equation for G: Definitions/ P Theorems |G| = |Z(G)| + [G : Z(gi )] Groups, Subgroups Lagrange’s, where we sum over a representative of each conjugacy class. Normality k k Class Equation, Since p divides |G| = [G : Z(g )]|Z(g )| and p does not Cauchy’s Theorem i i

First Sylow divide |Z(gi )|, we know p must divide [G : Z(gi )], ∀gi ∈/ Z(G). Theorem Thus, from Cauchy’s Theorem, we see that Z(G) contains an Theorem Examples element of order p, say x. Since x is in the center of G, hxi is Proof Additional Proofs a normal subgroup of G and we may form the factor group Second Sylow G/hxi. Now observe that pk−1 divides |G/hxi|. Thus, by the Theorem k−1 Third Sylow inductive hypothesis, |G/hxi| has a subgroup of order p and Theorem this subgroup has the form H/hxi where H is a subgroup of G. Results Finally, note that |H/hxi| = pk−1 and |hxi| = p imply that Cyclic subgroups k 2 Simple Groups |H| = p and this completes the proof. Additional Examples

References 16 / 26 additional proof 2

Sylow We divide the proof into two cases. Theorems Andrew Clarey Case 1: p divides the order of the center Z(G) of G. By

Definitions/ Cauchy’s Theorem, Z(G) must have an element of order p, say Theorems x. By induction, the quotient group G/hxi must have a Groups, Subgroups Lagrange’s, k−1 Normality subgroup P of order p . Then the pre- of P in Z(G) is Class Equation, k Cauchy’s Theorem the desired subgroup of order p .

First Sylow Theorem Case 2: assume that p does not divide the order of the center Theorem Examples of G. Again: Proof |G| = |Z(G)| + P[G : Z(g )], Additional Proofs i

Second Sylow where the sum is over all the distinct conjugacy classes of G; Theorem that is, conjugacy class with more than one element. Since p Third Sylow Theorem fails to divide the order of the center, ∃i such that k Results p - [G : Z(gi )]. Then p must divide the order of the subgroup Cyclic subgroups Simple Groups Z(gi ) as |G| = [G : Z(gi )]|Z(gi )|. Again, by induction, G will Additional Examples have a p-Sylow subgroup.3 References 17 / 26 Second Sylow Theorem

Sylow Theorems Second Sylow Theorem: All p-Sylow subgroups of G are Andrew Clarey conjugate to each other. Consequently, a p-Sylow subgroup is Definitions/ Theorems normal iff it is the only p-Sylow subgroup. Groups, Subgroups Lagrange’s, Proof : Normality Class Equation, Cauchy’s Theorem Let K, H be p-Sylow subgroups of G. Then by the second part −1 First Sylow of the First Sylow Theorem K ⊆ gHg . But K and H have Theorem −1 Theorem the same order, so K = gHg . Examples Proof Next, if there is a p-Sylow subgroup H E G and K is any Additional Proofs −1 Second Sylow p-Sylow subgroup, then K = gHg , so K = H. Therefore H Theorem is the only p-Sylow subgroup. Third Sylow Theorem Finally, if H is the only p-Sylow subgroup, then Results |gHg −1| = |H| ⇒ H = gHg −1 and H is normal. Cyclic subgroups Simple Groups Additional Examples

References 18 / 26 Third Sylow Theorem

Sylow Theorems

Andrew Clarey We give the Third Sylow Theorem without proof: Definitions/ Theorems Groups, Subgroups Third Sylow Theorem: Let H be any p-Sylow subgroup of G. Lagrange’s, Normality Then the number of p-Sylow subgroups in G is [G : N(H)]. Class Equation, Cauchy’s Theorem This number divides |G| and has the form 1 + jp for some First Sylow Theorem j ≥ 0 and this number divides [G : H]. Theorem Examples Proof Note: [G : H] = [G : N(H)][N(H): H]. Additional Proofs

Second Sylow Theorem These theorems allow us to look at the structure of arbitrary Third Sylow Theorem groups in order to try and classify them.

Results Cyclic subgroups Simple Groups Additional Examples

References 19 / 26 Cyclic Theorem

Sylow Cyclic Theorem: Let G be a group of order pq, where p and Theorems Andrew Clarey q are primes and p < q. Then if p - q − 1, G is cyclic.

Definitions/ Theorems Examples: Groups, Subgroups Lagrange’s, Every group of order 15 is cyclic. 15 = 3 · 5 and Normality Class Equation, 3 5 − 1 = 4. So by the theorem, all groups of order 15 Cauchy’s Theorem -

First Sylow are cyclic. Theorem Every group of order 35 is cyclic. 35 = 5 · 7 and Theorem Examples 5 7 − 1 = 6. So by the theorem, all groups of order 35 Proof - Additional Proofs are cyclic. Second Sylow Theorem Every group of order 119 is cyclic. 119 = 7 · 17 and

Third Sylow 7 - 17 − 1 = 16. So by the theorem, all groups of order Theorem 119 are cyclic. Results Cyclic subgroups (It can be proven that when p | q − 1, ∃ a non-abelian group of Simple Groups Additional Examples order pq. Moreover, all non-abelian groups of order pq are References isomorphic to each other.) 20 / 26 Simple Groups

Sylow Theorems Andrew Clarey A group G is called simple if its only normal subgroups are {e} Definitions/ and G. Theorems Groups, Subgroups Lagrange’s, Normality Examples: Class Equation, Cauchy’s Theorem 3 2 First Sylow No group of order 200 is simple. 200 = 2 · 5 . Theorem Theorem Consider the 5-Sylow subgroup H of 25 elements. The Examples Proof number of 5-Sylow subgroups is [G : N(H)] = 1 + 5j | 8. Additional Proofs The only possibility is 1, so H is the only 5-Sylow Second Sylow Theorem subgroup and is normal by the Second Sylow Theorem, Third Sylow and therefore G cannot be simple. Theorem

Results Cyclic subgroups Simple Groups Additional Examples

References 21 / 26 Simple Groups

Sylow Theorems

Andrew Clarey 3 Definitions/ No group of order 56 is simple. 56 = 2 · 7. Theorems Groups, Subgroups The number of 2-Sylow subgroups is 1 + 2k and divides 7, Lagrange’s, Normality Class Equation, therefore is 1 or 7. The number of 7-Sylow subgroups is Cauchy’s Theorem 1 + 7j and divides 8, and therefore is 1 or 8. If either is 1, First Sylow Theorem then we are done. So, let’s say there are 8 7-Sylow Theorem Examples subgroups. They have trivial intersection, which gives Proof Additional Proofs 8 · 6 = 48 elements. But, 56 − 48 = 8 elements, which

Second Sylow only allows for one 2-Sylow subgroup, and therefore this Theorem 2-Sylow subgroup is normal and the group is not simple.4 Third Sylow Theorem

Results Cyclic subgroups Simple Groups Additional Examples

References 22 / 26 groups of order 30

Sylow Theorems Let’s find all the groups of order 30 = 2 · 3 · 5 Andrew Clarey

Definitions/ Theorems So, we know there are Sylow subgroups A, B, and C of order 2, Groups, Subgroups 3, and 5 respectively. The number of 5-Sylow subgroups is Lagrange’s, Normality [G : N(C)], so must divide 6. But it is also of the form 1 + 5j, Class Equation, Cauchy’s Theorem so either 1 or 6. Similarly, the 3-Sylow subgroups are First Sylow Theorem 1 + 3k | 10, either 1 or 10. Theorem Examples Suppose there are six 5-Sylow subgroups and 10 3-Sylow Proof Additional Proofs subgroups. Any two 5-Sylow subgroups must have trivial Second Sylow Theorem intersection since they are both order 5. All six 5-Sylow groups

Third Sylow would give 6 · 4 = 24 elements of order 5 in G. Similarly, the Theorem 3-Sylow subgroups give 20 elements of order 3 in G. By our Results Cyclic subgroups assumption, this would imply |G| ≥ 44, which is impossible. Simple Groups Additional Examples

References 23 / 26 groups of order 30

Sylow Theorems

Andrew Clarey So, we know the group is not simple, but let’s continue to Definitions/ Theorems explore its structure. Groups, Subgroups Lagrange’s, Normality So, either there is only one 5-Sylow or one 3-Sylow subgroup. Class Equation, Cauchy’s Theorem Therefore, either B or C is normal, and BC is a subgroup of G, First Sylow Theorem of order |B| · |C|/|B ∩ C| = 15. So BC is cyclic, say BC = hxi. Theorem Examples Since hxi has index 2 (as |G|/|BC| = 30/15 = 2) it is normal Proof Additional Proofs in G. If we let A = hyi, then G = hxihyi since hxihyi has order Second Sylow 30. We must have yxy −1 = x t for some t. If we knew Theorem the value of t, we could determine the structure of G. Third Sylow Theorem

Results Cyclic subgroups Simple Groups Additional Examples

References 24 / 26 groups of order 30

Sylow Theorems

Andrew Clarey −1 Definitions/ So yxy must have order 15, because x does, and therefore Theorems (t, 15) = 1, so t = 1, 2, 4, 7, 8, 11, 13, 14. We also have Groups, Subgroups Lagrange’s, −1 −1 t −1 −1 t t t t2 Normality y(yxy )y = yx y = (yxy ) = (x ) = x . Class Equation, Cauchy’s Theorem t2 t2−1 2 First Sylow So, x = x ⇒ x = e, and thus 15 | (t − 1). This rules Theorem out t = 2, 7, 8, 13, so there are at most four possibilities for t, Theorem Examples so at most four nonisomorphic groups of order 30. Proof Additional Proofs In fact, there are four: Second Sylow Theorem 30, S3 × 5, 3 × D5, and D15. Third Sylow Z Z Z Theorem

Results Cyclic subgroups Simple Groups Additional Examples

References 25 / 26 References

Sylow Theorems

Andrew Clarey

Definitions/ Theorems 1 www.math.uconn.edu/∼kconrad/blurbs/grouptheory/ Groups, Subgroups Lagrange’s, sylowpf.pdf Normality Class Equation, 2 Cauchy’s Theorem Contemporary Abstract Algebra, Gallian First Sylow 3 www.math.drexel.edu/∼rboyer/courses/math533 03/ Theorem Theorem sylow thm.pdf Examples Proof 4 http://turnbull.mcs.st-and.ac.uk/∼colva/topics/ch3.pdf Additional Proofs Second Sylow 5 Abstract Algebra: A First Course Second Edition, Saracino Theorem

Third Sylow Theorem

Results Cyclic subgroups Simple Groups Additional Examples

References 26 / 26