# THE SYLOW THEOREMS AND THEIR APPLICATIONS

Size: px
Start display at page:

Transcription

1 THE SYLOW THEOREMS AND THEIR APPLICATIONS AMIN IDELHAJ Abstract. This paper begins with an introduction into the concept of group actions, along with the associated notions of orbits and stabilizers, culminating in the proofs of Cayley s theorem and the orbit-stabilizer theorem. Then, we build up to the three Sylow theorems, and subsequently give some of their applications. Contents 1. Introduction 1 2. Group Actions, Orbits, and Stabilizers 2 3. The Sylow Theorems 3 4. Applications of the Sylow Theorems 5 Acknowledgements 8 References 8 1. Introduction One of the important results in the theory of finite groups is Lagrange s theorem, which states that the order of any subgroup of a group must divide the order of the group. One might suggest a possible converse to this theorem, that for any number dividing the order of a group, there exists a subgroup of that order. This is not true. For example, the group A 4 of even permutations on the set {1, 2, 3, 4} has order 12, yet there does not exist a subgroup of order 6. The Sylow theorems do provide us with a sort of partial converse to Lagrange s theorem, by asserting the existence of certain subgroups (called Sylow p-subgroups) of any group with a given order, and gives some information about their properties. In this paper, we wish to build up to the proofs of the Sylow theorems by use of group actions and the orbit-stabilizer theorem. Definition 1.1. Let G be a group and let X be a set. An action of G on X is a group homomorphism φ : G Sym(X), where Sym(X) is the group of permutations of X. Notation 1.2. When the action of G on X is unambiguous, we implicitly denote it by G X. By abuse of notation, we will use g to denote both the group element and the permutation φ(g) it induces on X. If x X is any point, we denote the action of g on it by gx. While this notation is also used for group multiplication, it is clear from context which one is intended. Date: July

2 2 AMIN IDELHAJ 2. Group Actions, Orbits, and Stabilizers In this section, we discuss two important concepts regarding group actions: orbits and stabilizers. Using these concepts, we prove Cayley s theorem and the orbitstabilizer theorem. Throughout, we let the general action be that of G on X. Definition 2.1. : The orbit of a point x X is the set of points to which x is transported by the action of G: Orb(x) = {gx : g G}. The cardinality of the orbit is called its length. Definition 2.2. The stabilizer of a point x X, denoted Stab(x), is the set {g G : gx = x} G. Definition 2.3. The transporter from x to y is the set of elements that send x to y: Trans(x, y) = {g G : gx = y} G. Theorem 2.4. The stabilizer of any element in X is a subgroup of G. Proof. If g and h are in Stab(x), (gh)x = g(hx) = gx = x, implying that Stab(x) is closed under multiplication. In addition, x = (g 1 g)x = g 1 (gx) = g 1 x, implying that the inverse of every element in Stab(x) also lies in Stab(x). The identity is in Stab(x) as it is in the kernel of the associated homomorphism, and fixes every point in X. Thus, Stab(x) G. Theorem 2.5. The orbits of X form a partition. Proof. It is clear that every point in X is in its own orbit. Now, we wish to prove that if two orbits overlap, they coincide. Consider x Orb(y) Orb(z). Then x = gy = hz for some g, h G. Thus, if w = mz for some m G, then w = mh 1 gy. Thus, Orb(z) Orb(y). By an analogous argument, Orb(y) Orb(z), implying that they coincide. Definition 2.6. An action is: (1) Faithful if the kernel of the associated homomorphism φ : G Sym(X) is trivial. (2) Transitive if there is only one orbit. (3) Regular if it is both faithful and transitive. With this, we are now able to prove Cayley s theorem and the orbit-stabilizer theorem. Theorem 2.7 (Cayley). Every group of order n is isomorphic to some subgroup of S n. Proof. We consider the action of G on itself by left multiplication. Because this action is faithful, G embeds as a subgroup of Sym(G), and because Sym(G) = S n, G is isomorphic to a subgroup of S n. Theorem 2.8 (Orbit-Stabilizer). When a group G acts on a set X, the length of the orbit of any point is equal to the index of its stabilizer in G: Orb(x) = [G : Stab(x)] Proof. The first thing we wish to prove is that for any two group elements g and g, gx = g x if and only if g and g are in the same left coset of Stab(x). We know

3 THE SYLOW THEOREMS AND THEIR APPLICATIONS 3 this because if gx = g x, then g 1 g fixes x. Thus, g gstab(x), and since g also lies in its own left coset of Stab(x), g Stab(x) = g Stab(x). Now define a mapping φ : G/Stab(x) Orb(x) by φ(gstab(x)) = gx. This map is surjective because for y Orb(x), we can choose a g Trans(x, y), for which g Stab(x) maps to y under φ. Our previous result proves that this map is injective, and thus, we have a bijection. Orb(x) = G/Stab(x) = [G : Stab(x)]. 3. The Sylow Theorems Definition 3.1. Let G be a finite group of order p n m where p m. A Sylow p- subgroup of G is one of order p n. By Lagrange s theorem, such a subgroup would be a maximal p-subgroup of G. Example 3.2. If we consider the group Z 100 under addition, it has order 100 = Thus, a Sylow 2-subgroup is a subgroup of order 4, while a Sylow 5-subgroup is a subgroup of order 25. We now state the three Sylow theorems, and dedicate the rest of this section to their proofs. Theorem 3.3 (Sylow s first theorem). If p is a prime number and p G, then there exists a Sylow p-subgroup of G. Theorem 3.4 (Sylow s second theorem). For a given prime p, all Sylow p-subgroups of G are conjugate to each other. Theorem 3.5 (Sylow s third theorem). We denote the number of Sylow p-subgroups of G by n p. Then the following results hold: (1) n p p 1. (2) If G = p n m so that p m, then n p m. (3) If P is any Sylow p-subgroup of G, then n p = [G : N(P )], where N(P ) is the normalizer of P in G. We now wish to prove Sylow s first theorem, beginning with a combinatorial lemma that will be employed in this proof: ( p r ) ( ) m n Lemma 3.6. If p m, then p p r, where are the binomial coefficients. k Proof. If we expand the binomial coefficient, we get: ( p r ) m = p r p r m 1 k=0 p r 1 (p r m k) (p r k) k=0 Now, we wish to prove that if p divides a term p n m k in the numerator, then it divides the corresponding term p n k in the denominator. If we take k = p l q where either l < r and p q or l = r, we have that pr m k p r k = pr l m q, which is not p r l q divisible by p. Thus, the whole product is not divisible by p. Proof of Sylow s first theorem. We denote the set of all subsets of G with order p n by Ω. This consists of taking the set of combinations ( of p n elements out of the p n m p n ) m elements of G, and thus we have Ω =. Now, we let G act on Ω by left p n

4 4 AMIN IDELHAJ multiplication. Since Ω is partitioned by its orbits, the sum of the lengths of its orbits is equal to Ω. By Lemma 3.6, we find that p Ω, and as such, there must exist at least one orbit whose length is coprime to p. Choose ω Ω to be a set for which p Orb(ω). Now, define H := Stab(ω). We can consider its action on ω by left multiplication, since H permutes the elements of ω. The orbits of this action are, by definition, right cosets of H, and thus, the length of each orbit is [G : H]. Thus, H ω = p n, and thus, H is a p-group. By the orbit-stabilizer theorem, Orb(ω) H = G = p n m. Since H is a p-group and p Orb(ω), we have that H = p n, implying that H is a Sylow p-subgroup of G. Example 3.7. Consider a group with order 72, and look at its prime factorization: 72 = By Sylow I, we know that there exists a subgroup of order 8 as well as a subgroup of order 9, without further information on the group structure. In order to prove Sylow II and III, we will need the following lemma, regarding the actions of p-groups on finite sets. Lemma 3.8. If G is a p-group and X is finite, then X p X G, where X G is the set of points x X that are fixed by every g G. Proof. Consider a set of representatives x 1,..., x n X for the G-orbits of X. We have that X = n i=1 Orb(x i) = n i=1 [G : Stab(x i)]. If x i is not a fixed point, then [G : Stab(x i )] = p r for some r > 0. If x i is a fixed point, then its orbit is the singleton {x i }, and thus, Orb(x i ) = 1. Thus in modulo p, each fixed point contributes 1 to the above sum, and every other term contributes 0, thereby giving the desired result. Proof of Sylow s second theorem. Consider two Sylow p-subgroups of G, call them R and S. We shall have R act on G/S by left multiplication. By Lemma 3.6, we have that the number of fixed points of G/S under the action of R is congruent to [G : S] mod p. Suppose G = p n m where p m. Since S is a Sylow p-subgroup, [G : S] = m 0 mod p, and thus, there must be some non-zero number of fixed points in the action. Denote one of the fixed points by gs. We have rgs = gs for any r R. Thus, g 1 rgs = S which implies g 1 Rg S. Since g 1 Rg and S are finite sets of the same size, the inclusion is an equality and we have that the two subgroups are conjugate. The following fact, which is used extensively in the next section, is an immediate consequence of Sylow s second theorem. Corollary 3.9. A Sylow p-subgroup of G is unique if and only if it is normal in G. In particular, it is unique if the group is abelian. Proof. If a Sylow p-subgroup is unique, then it is equal to all its conjugations and thus normal. If there are multiple Sylow p-subgroups, they must be conjugate to each other, so none of them can be closed under conjugation, prohibiting normality. Since any subgroup of an abelian group is normal, a Sylow p-subgroup must be unique. Finally, we turn our attention the third Sylow theorem.

5 THE SYLOW THEOREMS AND THEIR APPLICATIONS 5 Proof of part 1 of Sylow s third theorem. We consider S := Syl p (G) to be the set of all Sylow p-subgroups of G. Fix a particular subgroup P, and have it act on S by conjugation. Note that S is finite since G has only finitely many subsets. Thus, Lemma 3.8 applies and we find that S = n p (G) p S P. Because P is closed under conjugation by one of its elements, it is a fixed point, so now we consider any other fixed point, call it Q. Because pq = Qp for every p P, we know that P N(Q). In addition, Q N(Q), and thus, P and Q are subgroups of N(Q). In particular, by Langrange s theorem, p n N(Q), and since p n+1 G, it follows that p n+1 N(Q). Therefore, P and Q are Sylow p-subgroups of N(Q), and by Sylow s second theorem, they are conjugate in N(Q). But since Q N(Q), this can only be achieved if P = Q. This shows that there is only one fixed point in S, whereby we now have that n p = S p S P = 1 Proof of part 2 of Sylow s third theorem. We now consider the action of the entire group G, on Syl p (G). Because Sylow p-subgroups of G are conjugate, there is only one orbit, whose length is n p. Using the orbit-stabilizer theorem, it follows that n p G = p n m. Since n p 1 mod p, p and n p must be relatively prime. Thus, n p p n m can occur only if n p m. Proof of part 3 of Sylow s third theorem. We again look at the action of G on Syl p (G). By orbit-stabilizer, n p = [G : Stab(P )]. Under the conjugation action, Stab(P ) = N(P ), giving us our desired result. We conclude this section with a theorem whose proof is similar in form to that of the third Sylow theorem. Theorem Every p-subgroup of G is contained in some Sylow p-subgroup of G Proof. We consider H to be a p-subgroup of G, and let it act on Syl p (G) by conjugation. There must be at least one fixed point under the action since Syl p (G) p 1. We denote a fixed point by Q and note that H is a subgroup of N(Q). H Q is a subgroup of N(Q) since Q is normal in N(Q). H Q is a p-subgroup of G since its order is H Q = H Q H Q. Because Q is not contained properly in any other p-subgroup of G, we now have that Q = H Q, implying that H Q. 4. Applications of the Sylow Theorems In this section, we wish to explore some applications of the Sylow theorems to combinatorics, arithmetic, and finite group theory, in order to demonstrate their significance. We begin with a lemma that does not invoke any of the Sylow theorems, but will be useful throughout this section. Lemma 4.1. Every group of prime order is cyclic. Proof. By Lagrange s theorem, we know that for any element of G, we have g G. However, if G has prime order and g is not the identity, then this can only be satisfied if g = G, making G cyclic. Now, we begin the applications. Theorem 4.2 (Wilson). A natural number p is prime if and only if (p 1)! p 1

6 6 AMIN IDELHAJ Proof. We will only prove the only if part of this statement, since it uses the Sylow theorems. Consider the symmetric group S p for p prime. Since p p! and p 2 p!, we have that the Sylow p-subgroups are cyclic groups of order p. There are (p 1)! many p-cycles in S p, and every Sylow p-subgroup contains precisely (p 1) such cycles, while sharing none. Thus it follows that there are precisely (p 2)! distinct Sylow p-subgroups. By the third Sylow theorem, we have that n p = (p 2)! p 1, thus implying that (p 1)! p p 1 p 1. Theorem 4.3. Every group of order 15 is cyclic. Proof. A group of order 15 has a subgroup of order 3 and a subgroup of order 5. By Sylow III, we have that n and n 3 5, meaning there is only 1 Sylow 3-subgroup a. By an analogous argument, there is only 1 Sylow 5-subgroup b, which is also cyclic. By Sylow II, these two subgroups must be normal. We know that their intersection must be trivial, since any element in their intersection has an element whose order divides both 3 and 5. We know that HK = H K H K = 15 = G. Thus, every element in G is of the form a i b j. It follows that G = H K and therefore cyclic by the Chinese Remainder Theorem. Theorem 4.4. Every finite p-group is isomorphic to some subgroup of the upper unitriangular group. n 1 Proof. The order of GL(n, p) is n 1 k=0 (pn p k ) = p n(n 1) 2 k=0 (pn k 1). Since p p n k 1, a Sylow p-subgroup of GL(n, p) has order p n(n 1) 2. We know that the unitriangular group is a subgroup of the general linear group, since it is closed under multiplication and all of its matrices have determinant 1. We note that a unitriangular matrix can take any value in its superdiagonal entries, thus it has n(n 1) 2 entries that are not fixed to be 0 or 1. Since they are allowed to take any value in F p, there are p n(n 1) 2 such matrices. Thus, over a prime field, the unitriangular group is a Sylow p-subgroup of the general linear group. If G = p k = n, we can embed G in S n, which is in turn embeded GL(n, p) via the permutation matrices. Thus, G is a p-subgroup of GL(n, p). By Theorem 3.10, G is contained in some Sylow p-subgroup H of GL(n, p). By Sylow s second theorem, H is conjugate to the unitriangular group, which implies that the latter has a subgroup isomorphic to G. We now move to one of the more substantial theorems in this section. There exists a theorem which says that except for a particular set of 26 groups, called the sporadic groups, all finite simple groups are isomorphic to a cyclic group of prime order, an alternating group of degree at least 5, or a simple group of the Lie type (the definition of which is a bit too complicated to go into here). While the proof of this theorem is thousands of pages long, we will prove a simpler result that demonstrates the power of the Sylow theorems, which are very important in the full proof of the classification theorem. To begin, we prove the following lemma. Lemma 4.5. The order of a non-abelian simple group divides the factorial of the index of every proper subgroup. Proof. G acts on G/H by left multiplication. This action is clearly transitive, and thus, the homomorphism φ : G Sym(G/H) has a kernel that is not all of G. However, since the kernel must be a normal subgroup, it must be trivial. Thus, the

7 THE SYLOW THEOREMS AND THEIR APPLICATIONS 7 homomorphism is injective, and G is isomorphic to a subgroup of Sym(G/H), and thus, G Sym(G/H) = [G : H]!. Theorem 4.6. The smallest non-abelian simple group is A 5, and is unique up to isomorphism. Proof. In order to prove this theorem, we first eliminate all possibilities of simple non-abelian groups with order less than 60. Lemma 4.1 eliminates groups of prime order, for they are cyclic and thus abelian. Any group of order p k m where m < p and k 0 will have a single Sylow p- subgroup, since n p p 1 and n p m is only satisfied by n p = 1. Uniqueness of a Sylow p-subgroup implies normality by the second Sylow theorem, eliminating groups of order 6, 10, 14, 15, 18, 20, 21, 22, 26, 28, 33, 34, 35, 38, 39, 42, 44, 46, 50, 51, 52, 54, 55, 57, and 58. If p is a prime such that q = 2 p 1 is a Mersenne prime, we claim that a group of order 2 p q also cannot be simple and non-abelian. We know that n q q 1 and n q 2 p. Either n q = 1, in which case the q-sylow subgroup is normal, or n q = 2 p. In the latter case, assume H 1,..., H 2 p are the Sylow q-subgroups. Since they are all of prime order, the intersection of any two of them is trivial. Thus, the total number of non-identity elements of order q is 2 p (q 1) = G 2 p. The Sylow 2-subgroups of G must be contained entirely within the remaining 2 p elements, which means that this subgroup is unique and normal. This fact disqualifies groups of order 12 and 56. We know that 45 = 5 3 2, so n 5 9 and n 5 5 1, only satisfied by n = 5. Similarly, 40 = 5 2 3, so n 5 8 and n 5 5 1, implying that n 5 = 1. Thus, groups of those orders have normal subgroups and are thus disqualified. If G is simple and non-abelian, its order must divide the factorial of any Sylow number n p. Let P be a Sylow p-subgroup of G. By Sylow III, [G : N(P )] = n p. Note that P G because it is a p-group and has a non-trivial center. Since P is not the trivial group, it cannot be normal in G. In particular N(P ) is a proper subgroup of G. By lemma 4.5, G [G : N(P )]! = n p!. Groups of order 24, 36, and 48 fail to meet this condition, and are thus disqualified. If G = 30, we have that n 3 = 1 or 10, and n 5 = 1 or 6. If G is simple, then it has 10 subgroups of order 3 and 6 subgroups of order 5. However, since these groups are all cyclic of prime order, any non-trivial element of G is contained in at most one of these groups. However, that would require that G 10(3 1) + 6(5 1) = 44, which is false. Thus, G is not simple. We now wish to prove that A 5 is simple, and then that any simple group of order 60 is isomorphic to it. We note that the conjugacy classes of A 5 have sizes 1, 12, 12, 15, and 20. A non-trivial normal subgroup would contain the identity and at least one other conjugacy class, with its order being some sum of one and at least one of the other above numbers. However, the only such sum which divides 60 (required by Lagrange) is to sum all of them up. Thus, the two normal subgroups have order 1 (identity) and 60 (whole group), making A 5 simple. If a simple group of order 60 has a subgroup H of index 5, then we could consider the 5 distinct left cosets of H. Now G acts on them transitively by left multiplication, we see that there is a non-trivial homomorphism φ : G S 5. Clearly,

8 8 AMIN IDELHAJ φ(g) S 5 since the latter is not simple, and if φ(g) < 60, the homomorphism s kernel would be non-trivial, violating simplicity of G. Thus, φ(g) = 60. Now, we apply the signature homomorphism sgn φ(g) : φ(g) {1, 1}. Since φ(g) is simple and the homomorphism is non-injective, it is trivial and φ(g) A 5. We now know that φ(g) = A 5 since both have order 60. Now, it remains to be shown that every simple group of order 60 has a subgroup of index 5. The four possibilities for n 2 are 1, 3, 5, and 15. Clearly, n 2 1 by simplicity, and we also know that n 2 3 since 60 3!. We know that n 5 = 6, since the only other possibility by Sylow III would be n 5 = 1, violating simplicity of G. Similarly, n 3 = 10, because n 3 1 by simplicity again, and n 3 4 because 60 4!. Thus, there are 20 elements of order 3 and 24 elements of order 5 in G. Two distinct Sylow 2-subgroups of G would intersect in up to 2 elements, meaning that if n 2 = 15, the group would need to have at least 75 elements, which it does not. Thus, n 2 = 5, implying that the normalizer of a Sylow 2-subgroup would have index 5. By the previous discussion, G is isomorphic to A 5. Acknowledgements I would like to thank my mentors, Nir Gadish and Oishee Banerjee, for their guidance throughout the research process. It is with their support that I was able to write a paper on group theory, which I previously had no background in. I would also like to thank Peter May for organizing this program and giving me the opportunity to explore this topic. Finally, I would like to thank my professor, Laci Babai, for his guidance and inspiration. References [1] I. N. Herstein. Topics in Algebra. John Wiley & Sons [2] Keith Conrad. Group Actions. kconrad/blurbs/grouptheory/gpaction.pdf [3] Keith Conrad. Consequences of the Sylow Theorems. kconrad/blurbs/grouptheory/sylowapp.pdf [4] Gabe Cunningham. Sylow Theorems and The General Linear Group. dav/sylow.pdf

### CONSEQUENCES OF THE SYLOW THEOREMS

CONSEQUENCES OF THE SYLOW THEOREMS KEITH CONRAD For a group theorist, Sylow s Theorem is such a basic tool, and so fundamental, that it is used almost without thinking, like breathing. Geoff Robinson 1.

### 3. G. Groups, as men, will be known by their actions. - Guillermo Moreno

3.1. The denition. 3. G Groups, as men, will be known by their actions. - Guillermo Moreno D 3.1. An action of a group G on a set X is a function from : G X! X such that the following hold for all g, h

### Math 250A, Fall 2004 Problems due October 5, 2004 The problems this week were from Lang s Algebra, Chapter I.

Math 250A, Fall 2004 Problems due October 5, 2004 The problems this week were from Lang s Algebra, Chapter I. 24. We basically know already that groups of order p 2 are abelian. Indeed, p-groups have non-trivial

### DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY - PART 3/3. Contents

DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY - PART 3/3 T.K.SUBRAHMONIAN MOOTHATHU Contents 1. Cayley s Theorem 1 2. The permutation group S n 2 3. Center of a group, and centralizers 4 4. Group actions

### Definitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations

Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of

### GROUP ACTIONS RYAN C. SPIELER

GROUP ACTIONS RYAN C. SPIELER Abstract. In this paper, we examine group actions. Groups, the simplest objects in Algebra, are sets with a single operation. We will begin by defining them more carefully

### Selected exercises from Abstract Algebra by Dummit and Foote (3rd edition).

Selected exercises from Abstract Algebra by Dummit Foote (3rd edition). Bryan Félix Abril 12, 2017 Section 4.1 Exercise 1. Let G act on the set A. Prove that if a, b A b = ga for some g G, then G b = gg

### LECTURES 11-13: CAUCHY S THEOREM AND THE SYLOW THEOREMS

LECTURES 11-13: CAUCHY S THEOREM AND THE SYLOW THEOREMS Recall Lagrange s theorem says that for any finite group G, if H G, then H divides G. In these lectures we will be interested in establishing certain

### Math 429/581 (Advanced) Group Theory. Summary of Definitions, Examples, and Theorems by Stefan Gille

Math 429/581 (Advanced) Group Theory Summary of Definitions, Examples, and Theorems by Stefan Gille 1 2 0. Group Operations 0.1. Definition. Let G be a group and X a set. A (left) operation of G on X is

### School of Mathematics and Statistics. MT5824 Topics in Groups. Problem Sheet I: Revision and Re-Activation

MRQ 2009 School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet I: Revision and Re-Activation 1. Let H and K be subgroups of a group G. Define HK = {hk h H, k K }. (a) Show that HK

### Stab(t) = {h G h t = t} = {h G h (g s) = g s} = {h G (g 1 hg) s = s} = g{k G k s = s} g 1 = g Stab(s)g 1.

1. Group Theory II In this section we consider groups operating on sets. This is not particularly new. For example, the permutation group S n acts on the subset N n = {1, 2,...,n} of N. Also the group

### GRE Subject test preparation Spring 2016 Topic: Abstract Algebra, Linear Algebra, Number Theory.

GRE Subject test preparation Spring 2016 Topic: Abstract Algebra, Linear Algebra, Number Theory. Linear Algebra Standard matrix manipulation to compute the kernel, intersection of subspaces, column spaces,

### Exercises on chapter 1

Exercises on chapter 1 1. Let G be a group and H and K be subgroups. Let HK = {hk h H, k K}. (i) Prove that HK is a subgroup of G if and only if HK = KH. (ii) If either H or K is a normal subgroup of G

### MA441: Algebraic Structures I. Lecture 26

MA441: Algebraic Structures I Lecture 26 10 December 2003 1 (page 179) Example 13: A 4 has no subgroup of order 6. BWOC, suppose H < A 4 has order 6. Then H A 4, since it has index 2. Thus A 4 /H has order

### GROUP ACTIONS EMMANUEL KOWALSKI

GROUP ACTIONS EMMANUEL KOWALSKI Definition 1. Let G be a group and T a set. An action of G on T is a map a: G T T, that we denote a(g, t) = g t, such that (1) For all t T, we have e G t = t. (2) For all

### Chapter 25 Finite Simple Groups. Chapter 25 Finite Simple Groups

Historical Background Definition A group is simple if it has no nontrivial proper normal subgroup. The definition was proposed by Galois; he showed that A n is simple for n 5 in 1831. It is an important

### Groups and Symmetries

Groups and Symmetries Definition: Symmetry A symmetry of a shape is a rigid motion that takes vertices to vertices, edges to edges. Note: A rigid motion preserves angles and distances. Definition: Group

### its image and kernel. A subgroup of a group G is a non-empty subset K of G such that k 1 k 1

10 Chapter 1 Groups 1.1 Isomorphism theorems Throughout the chapter, we ll be studying the category of groups. Let G, H be groups. Recall that a homomorphism f : G H means a function such that f(g 1 g

### SF2729 GROUPS AND RINGS LECTURE NOTES

SF2729 GROUPS AND RINGS LECTURE NOTES 2011-03-01 MATS BOIJ 6. THE SIXTH LECTURE - GROUP ACTIONS In the sixth lecture we study what happens when groups acts on sets. 1 Recall that we have already when looking

### Math 210A: Algebra, Homework 5

Math 210A: Algebra, Homework 5 Ian Coley November 5, 2013 Problem 1. Prove that two elements σ and τ in S n are conjugate if and only if type σ = type τ. Suppose first that σ and τ are cycles. Suppose

### D-MATH Algebra I HS 2013 Prof. Brent Doran. Solution 3. Modular arithmetic, quotients, product groups

D-MATH Algebra I HS 2013 Prof. Brent Doran Solution 3 Modular arithmetic, quotients, product groups 1. Show that the functions f = 1/x, g = (x 1)/x generate a group of functions, the law of composition

### Solutions to odd-numbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 3

Solutions to odd-numbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 3 3. (a) Yes; (b) No; (c) No; (d) No; (e) Yes; (f) Yes; (g) Yes; (h) No; (i) Yes. Comments: (a) is the additive group

### FINITE GROUP THEORY: SOLUTIONS FALL MORNING 5. Stab G (l) =.

FINITE GROUP THEORY: SOLUTIONS TONY FENG These are hints/solutions/commentary on the problems. They are not a model for what to actually write on the quals. 1. 2010 FALL MORNING 5 (i) Note that G acts

### Lecture 5.6: The Sylow theorems

Lecture 5.6: The Sylow theorems Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley (Clemson) Lecture 5.6:

### Algebra Exercises in group theory

Algebra 3 2010 Exercises in group theory February 2010 Exercise 1*: Discuss the Exercises in the sections 1.1-1.3 in Chapter I of the notes. Exercise 2: Show that an infinite group G has to contain a non-trivial

### Group Theory

Group Theory 2014 2015 Solutions to the exam of 4 November 2014 13 November 2014 Question 1 (a) For every number n in the set {1, 2,..., 2013} there is exactly one transposition (n n + 1) in σ, so σ is

### Groups Subgroups Normal subgroups Quotient groups Homomorphisms Cyclic groups Permutation groups Cayley s theorem Class equations Sylow theorems

Group Theory Groups Subgroups Normal subgroups Quotient groups Homomorphisms Cyclic groups Permutation groups Cayley s theorem Class equations Sylow theorems Groups Definition : A non-empty set ( G,*)

### φ(xy) = (xy) n = x n y n = φ(x)φ(y)

Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =

### Algebra Exam Fall Alexander J. Wertheim Last Updated: October 26, Groups Problem Problem Problem 3...

Algebra Exam Fall 2006 Alexander J. Wertheim Last Updated: October 26, 2017 Contents 1 Groups 2 1.1 Problem 1..................................... 2 1.2 Problem 2..................................... 2

### Before you begin read these instructions carefully.

MATHEMATICAL TRIPOS Part IA Tuesday, 4 June, 2013 1:30 pm to 4:30 pm PAPER 3 Before you begin read these instructions carefully. The examination paper is divided into two sections. Each question in Section

### INTRODUCTION TO THE GROUP THEORY

Lecture Notes on Structure of Algebra INTRODUCTION TO THE GROUP THEORY By : Drs. Antonius Cahya Prihandoko, M.App.Sc e-mail: antoniuscp.fkip@unej.ac.id Mathematics Education Study Program Faculty of Teacher

### MATH 101: ALGEBRA I WORKSHEET, DAY #3. Fill in the blanks as we finish our first pass on prerequisites of group theory.

MATH 101: ALGEBRA I WORKSHEET, DAY #3 Fill in the blanks as we finish our first pass on prerequisites of group theory 1 Subgroups, cosets Let G be a group Recall that a subgroup H G is a subset that is

### Theorems and Definitions in Group Theory

Theorems and Definitions in Group Theory Shunan Zhao Contents 1 Basics of a group 3 1.1 Basic Properties of Groups.......................... 3 1.2 Properties of Inverses............................. 3

### Two subgroups and semi-direct products

Two subgroups and semi-direct products 1 First remarks Throughout, we shall keep the following notation: G is a group, written multiplicatively, and H and K are two subgroups of G. We define the subset

### 9. Finite fields. 1. Uniqueness

9. Finite fields 9.1 Uniqueness 9.2 Frobenius automorphisms 9.3 Counting irreducibles 1. Uniqueness Among other things, the following result justifies speaking of the field with p n elements (for prime

### (1) Let G be a finite group and let P be a normal p-subgroup of G. Show that P is contained in every Sylow p-subgroup of G.

(1) Let G be a finite group and let P be a normal p-subgroup of G. Show that P is contained in every Sylow p-subgroup of G. (2) Determine all groups of order 21 up to isomorphism. (3) Let P be s Sylow

### PROBLEMS FROM GROUP THEORY

PROBLEMS FROM GROUP THEORY Page 1 of 12 In the problems below, G, H, K, and N generally denote groups. We use p to stand for a positive prime integer. Aut( G ) denotes the group of automorphisms of G.

### Course 311: Abstract Algebra Academic year

Course 311: Abstract Algebra Academic year 2007-08 D. R. Wilkins Copyright c David R. Wilkins 1997 2007 Contents 1 Topics in Group Theory 1 1.1 Groups............................... 1 1.2 Examples of Groups.......................

### 5 Group theory. 5.1 Binary operations

5 Group theory This section is an introduction to abstract algebra. This is a very useful and important subject for those of you who will continue to study pure mathematics. 5.1 Binary operations 5.1.1

### Chapter 9: Group actions

Chapter 9: Group actions Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Summer I 2014 M. Macauley (Clemson) Chapter 9: Group actions

### A. (Groups of order 8.) (a) Which of the five groups G (as specified in the question) have the following property: G has a normal subgroup N such that

MATH 402A - Solutions for the suggested problems. A. (Groups of order 8. (a Which of the five groups G (as specified in the question have the following property: G has a normal subgroup N such that N =

### Fix(g). Orb(x) i=1. O i G. i=1. O i. i=1 x O i. = n G

Math 761 Fall 2015 Homework 4 Drew Armstrong Problem 1 Burnside s Lemma Let X be a G-set and for all g G define the set Fix(g : {x X : g(x x} X (a If G and X are finite, prove that Fix(g Stab(x g G x X

### SUPPLEMENT ON THE SYMMETRIC GROUP

SUPPLEMENT ON THE SYMMETRIC GROUP RUSS WOODROOFE I presented a couple of aspects of the theory of the symmetric group S n differently than what is in Herstein. These notes will sketch this material. You

### The Outer Automorphism of S 6

Meena Jagadeesan 1 Karthik Karnik 2 Mentor: Akhil Mathew 1 Phillips Exeter Academy 2 Massachusetts Academy of Math and Science PRIMES Conference, May 2016 What is a Group? A group G is a set of elements

### Examples: The (left or right) cosets of the subgroup H = 11 in U(30) = {1, 7, 11, 13, 17, 19, 23, 29} are

Cosets Let H be a subset of the group G. (Usually, H is chosen to be a subgroup of G.) If a G, then we denote by ah the subset {ah h H}, the left coset of H containing a. Similarly, Ha = {ha h H} is the

### Math 451, 01, Exam #2 Answer Key

Math 451, 01, Exam #2 Answer Key 1. (25 points): If the statement is always true, circle True and prove it. If the statement is never true, circle False and prove that it can never be true. If the statement

### ENTRY GROUP THEORY. [ENTRY GROUP THEORY] Authors: started Mark Lezama: October 2003 Literature: Algebra by Michael Artin, Mathworld.

ENTRY GROUP THEORY [ENTRY GROUP THEORY] Authors: started Mark Lezama: October 2003 Literature: Algebra by Michael Artin, Mathworld Group theory [Group theory] is studies algebraic objects called groups.

### Fall /29/18 Time Limit: 75 Minutes

Math 411: Abstract Algebra Fall 2018 Midterm 10/29/18 Time Limit: 75 Minutes Name (Print): Solutions JHU-ID: This exam contains 8 pages (including this cover page) and 6 problems. Check to see if any pages

### Spectra of Semidirect Products of Cyclic Groups

Spectra of Semidirect Products of Cyclic Groups Nathan Fox 1 University of Minnesota-Twin Cities Abstract The spectrum of a graph is the set of eigenvalues of its adjacency matrix A group, together with

### Quiz 2 Practice Problems

Quiz 2 Practice Problems Math 332, Spring 2010 Isomorphisms and Automorphisms 1. Let C be the group of complex numbers under the operation of addition, and define a function ϕ: C C by ϕ(a + bi) = a bi.

### Simple groups and the classification of finite groups

Simple groups and the classification of finite groups 1 Finite groups of small order How can we describe all finite groups? Before we address this question, let s write down a list of all the finite groups

### Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.

Glossary 1 Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.23 Abelian Group. A group G, (or just G for short) is

GROUP ACTIONS KEITH CONRAD. Introduction The groups S n, A n, and (for n 3) D n behave, by their definitions, as permutations on certain sets. The groups S n and A n both permute the set {, 2,..., n} and

GROUP ACTIONS KEITH CONRAD 1. Introduction The symmetric groups S n, alternating groups A n, and (for n 3) dihedral groups D n behave, by their very definition, as permutations on certain sets. The groups

Answers to Final Exam MA441: Algebraic Structures I 20 December 2003 1) Definitions (20 points) 1. Given a subgroup H G, define the quotient group G/H. (Describe the set and the group operation.) The quotient

### Math 594, HW2 - Solutions

Math 594, HW2 - Solutions Gilad Pagi, Feng Zhu February 8, 2015 1 a). It suffices to check that NA is closed under the group operation, and contains identities and inverses: NA is closed under the group

### Cosets, factor groups, direct products, homomorphisms, isomorphisms

Cosets, factor groups, direct products, homomorphisms, isomorphisms Sergei Silvestrov Spring term 2011, Lecture 11 Contents of the lecture Cosets and the theorem of Lagrange. Direct products and finitely

### 6 Cosets & Factor Groups

6 Cosets & Factor Groups The course becomes markedly more abstract at this point. Our primary goal is to break apart a group into subsets such that the set of subsets inherits a natural group structure.

### 2 Lecture 2: Logical statements and proof by contradiction Lecture 10: More on Permutations, Group Homomorphisms 31

Contents 1 Lecture 1: Introduction 2 2 Lecture 2: Logical statements and proof by contradiction 7 3 Lecture 3: Induction and Well-Ordering Principle 11 4 Lecture 4: Definition of a Group and examples 15

### ABSTRACT ALGEBRA 1, LECTURES NOTES 5: SUBGROUPS, CONJUGACY, NORMALITY, QUOTIENT GROUPS, AND EXTENSIONS.

ABSTRACT ALGEBRA 1, LECTURES NOTES 5: SUBGROUPS, CONJUGACY, NORMALITY, QUOTIENT GROUPS, AND EXTENSIONS. ANDREW SALCH 1. Subgroups, conjugacy, normality. I think you already know what a subgroup is: Definition

### MAT534 Fall 2013 Practice Midterm I The actual midterm will consist of five problems.

MAT534 Fall 2013 Practice Midterm I The actual midterm will consist of five problems. Problem 1 Find all homomorphisms a) Z 6 Z 6 ; b) Z 6 Z 18 ; c) Z 18 Z 6 ; d) Z 12 Z 15 ; e) Z 6 Z 25 Proof. a)ψ(1)

### Modern Algebra I. Circle the correct answer; no explanation is required. Each problem in this section counts 5 points.

1 2 3 style total Math 415 Please print your name: Answer Key 1 True/false Circle the correct answer; no explanation is required. Each problem in this section counts 5 points. 1. Every group of order 6

### RUDIMENTARY GALOIS THEORY

RUDIMENTARY GALOIS THEORY JACK LIANG Abstract. This paper introduces basic Galois Theory, primarily over fields with characteristic 0, beginning with polynomials and fields and ultimately relating the

### Permutation groups/1. 1 Automorphism groups, permutation groups, abstract

Permutation groups Whatever you have to do with a structure-endowed entity Σ try to determine its group of automorphisms... You can expect to gain a deep insight into the constitution of Σ in this way.

### Notes on Group Theory. by Avinash Sathaye, Professor of Mathematics November 5, 2013

Notes on Group Theory by Avinash Sathaye, Professor of Mathematics November 5, 2013 Contents 1 Preparation. 2 2 Group axioms and definitions. 2 Shortcuts................................. 2 2.1 Cyclic groups............................

### Algebra I: Final 2012 June 22, 2012

1 Algebra I: Final 2012 June 22, 2012 Quote the following when necessary. A. Subgroup H of a group G: H G = H G, xy H and x 1 H for all x, y H. B. Order of an Element: Let g be an element of a group G.

### ALGEBRA QUALIFYING EXAM PROBLEMS

ALGEBRA QUALIFYING EXAM PROBLEMS Kent State University Department of Mathematical Sciences Compiled and Maintained by Donald L. White Version: August 29, 2017 CONTENTS LINEAR ALGEBRA AND MODULES General

### Pseudo Sylow numbers

Pseudo Sylow numbers Benjamin Sambale May 16, 2018 Abstract One part of Sylow s famous theorem in group theory states that the number of Sylow p- subgroups of a finite group is always congruent to 1 modulo

### Connectivity of Intersection Graphs of Finite Groups

Connectivity of Intersection Graphs of Finite Groups Selçuk Kayacan arxiv:1512.00361v2 [math.gr] 2 Aug 2016 Department of Mathematics, Istanbul Technical University, 34469 Maslak, Istanbul, Turkey. e-mail:

### Definition List Modern Algebra, Fall 2011 Anders O.F. Hendrickson

Definition List Modern Algebra, Fall 2011 Anders O.F. Hendrickson On almost every Friday of the semester, we will have a brief quiz to make sure you have memorized the definitions encountered in our studies.

### Exercises MAT2200 spring 2013 Ark 4 Homomorphisms and factor groups

Exercises MAT2200 spring 2013 Ark 4 Homomorphisms and factor groups This Ark concerns the weeks No. (Mar ) and No. (Mar ). Plans until Eastern vacations: In the book the group theory included in the curriculum

### A SIMPLE PROOF OF BURNSIDE S CRITERION FOR ALL GROUPS OF ORDER n TO BE CYCLIC

A SIMPLE PROOF OF BURNSIDE S CRITERION FOR ALL GROUPS OF ORDER n TO BE CYCLIC SIDDHI PATHAK Abstract. This note gives a simple proof of a famous theorem of Burnside, namely, all groups of order n are cyclic

### HOMEWORK Graduate Abstract Algebra I May 2, 2004

Math 5331 Sec 121 Spring 2004, UT Arlington HOMEWORK Graduate Abstract Algebra I May 2, 2004 The required text is Algebra, by Thomas W. Hungerford, Graduate Texts in Mathematics, Vol 73, Springer. (it

### Math 120: Homework 6 Solutions

Math 120: Homewor 6 Solutions November 18, 2018 Problem 4.4 # 2. Prove that if G is an abelian group of order pq, where p and q are distinct primes then G is cyclic. Solution. By Cauchy s theorem, G has

### Abstract Algebra Study Sheet

Abstract Algebra Study Sheet This study sheet should serve as a guide to which sections of Artin will be most relevant to the final exam. When you study, you may find it productive to prioritize the definitions,

### Math 2070BC Term 2 Weeks 1 13 Lecture Notes

Math 2070BC 2017 18 Term 2 Weeks 1 13 Lecture Notes Keywords: group operation multiplication associative identity element inverse commutative abelian group Special Linear Group order infinite order cyclic

### CONJUGATION IN A GROUP

CONJUGATION IN A GROUP KEITH CONRAD 1. Introduction A reflection across one line in the plane is, geometrically, just like a reflection across any other line. That is, while reflections across two different

### School of Mathematics and Statistics. MT5824 Topics in Groups. Handout 0: Course Information

MRQ 2009 School of Mathematics and Statistics MT5824 Topics in Groups Handout 0: Course Information Lecturer: Martyn Quick, Room 326. Prerequisite: MT4003 Groups. Lectures: Mon (odd), Wed & Fri 10am, Lecture

### ALGEBRA PH.D. QUALIFYING EXAM September 27, 2008

ALGEBRA PH.D. QUALIFYING EXAM September 27, 2008 A passing paper consists of four problems solved completely plus significant progress on two other problems; moreover, the set of problems solved completely

### Johns Hopkins University, Department of Mathematics Abstract Algebra - Spring 2009 Midterm

Johns Hopkins University, Department of Mathematics 110.401 Abstract Algebra - Spring 2009 Midterm Instructions: This exam has 8 pages. No calculators, books or notes allowed. You must answer the first

### 120A LECTURE OUTLINES

120A LECTURE OUTLINES RUI WANG CONTENTS 1. Lecture 1. Introduction 1 2 1.1. An algebraic object to study 2 1.2. Group 2 1.3. Isomorphic binary operations 2 2. Lecture 2. Introduction 2 3 2.1. The multiplication

### MATH HL OPTION - REVISION SETS, RELATIONS AND GROUPS Compiled by: Christos Nikolaidis

MATH HL OPTION - REVISION SETS, RELATIONS AND GROUPS Compiled by: Christos Nikolaidis PART B: GROUPS GROUPS 1. ab The binary operation a * b is defined by a * b = a+ b +. (a) Prove that * is associative.

### Group Theory (Math 113), Summer 2014

Group Theory (Math 113), Summer 2014 George Melvin University of California, Berkeley (July 8, 2014 corrected version) Abstract These are notes for the first half of the upper division course Abstract

### Finite groups determined by an inequality of the orders of their elements

Publ. Math. Debrecen 80/3-4 (2012), 457 463 DOI: 10.5486/PMD.2012.5168 Finite groups determined by an inequality of the orders of their elements By MARIUS TĂRNĂUCEANU (Iaşi) Abstract. In this note we introduce

### MATH 1530 ABSTRACT ALGEBRA Selected solutions to problems. a + b = a + b,

MATH 1530 ABSTRACT ALGEBRA Selected solutions to problems Problem Set 2 2. Define a relation on R given by a b if a b Z. (a) Prove that is an equivalence relation. (b) Let R/Z denote the set of equivalence

### Definitions, Theorems and Exercises. Abstract Algebra Math 332. Ethan D. Bloch

Definitions, Theorems and Exercises Abstract Algebra Math 332 Ethan D. Bloch December 26, 2013 ii Contents 1 Binary Operations 3 1.1 Binary Operations............................... 4 1.2 Isomorphic Binary

### for all i = 1,..., k and we are done.

FINITE PERMUTATION GROUPS AND FINITE CLASSICAL GROUPS 23 5. Primitivity and related notions In this section we study some key properties of group actions. We will use this material in the next chapter

### 0 Sets and Induction. Sets

0 Sets and Induction Sets A set is an unordered collection of objects, called elements or members of the set. A set is said to contain its elements. We write a A to denote that a is an element of the set

### ABSTRACT ALGEBRA: REVIEW PROBLEMS ON GROUPS AND GALOIS THEORY

ABSTRACT ALGEBRA: REVIEW PROBLEMS ON GROUPS AND GALOIS THEORY John A. Beachy Northern Illinois University 2000 ii J.A.Beachy This is a supplement to Abstract Algebra, Second Edition by John A. Beachy and

### Tutorial on Groups of finite Morley rank

Tutorial on Groups of finite Morley rank adeloro@math.rutgers.edu Manchester, 14-18 July 2008 First tutorial Groups of finite Morley rank first arose in a very model-theoretic context, the study of ℵ 1

### Math 121 Homework 5: Notes on Selected Problems

Math 121 Homework 5: Notes on Selected Problems 12.1.2. Let M be a module over the integral domain R. (a) Assume that M has rank n and that x 1,..., x n is any maximal set of linearly independent elements

### MA441: Algebraic Structures I. Lecture 18

MA441: Algebraic Structures I Lecture 18 5 November 2003 1 Review from Lecture 17: Theorem 6.5: Aut(Z/nZ) U(n) For every positive integer n, Aut(Z/nZ) is isomorphic to U(n). The proof used the map T :

### Homework 2 /Solutions

MTH 912 Group Theory 1 F18 Homework 2 /Solutions #1. Let G be a Frobenius group with complement H and kernel K. Then K is a subgroup of G if and only if each coset of H in G contains at most one element

### AUTOMORPHISM GROUPS AND SPECTRA OF CIRCULANT GRAPHS

AUTOMORPHISM GROUPS AND SPECTRA OF CIRCULANT GRAPHS MAX GOLDBERG Abstract. We explore ways to concisely describe circulant graphs, highly symmetric graphs with properties that are easier to generalize

### Algebra-I, Fall Solutions to Midterm #1

Algebra-I, Fall 2018. Solutions to Midterm #1 1. Let G be a group, H, K subgroups of G and a, b G. (a) (6 pts) Suppose that ah = bk. Prove that H = K. Solution: (a) Multiplying both sides by b 1 on the

### INVERSE LIMITS AND PROFINITE GROUPS

INVERSE LIMITS AND PROFINITE GROUPS BRIAN OSSERMAN We discuss the inverse limit construction, and consider the special case of inverse limits of finite groups, which should best be considered as topological

### M3P10: GROUP THEORY LECTURES BY DR. JOHN BRITNELL; NOTES BY ALEKSANDER HORAWA

M3P10: GROUP THEORY LECTURES BY DR. JOHN BRITNELL; NOTES BY ALEKSANDER HORAWA These are notes from the course M3P10: Group Theory taught by Dr. John Britnell, in Fall 2015 at Imperial College London. They

### NOTES ON FINITE FIELDS

NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining