Two subgroups and semi-direct products
|
|
- Erick Farmer
- 6 years ago
- Views:
Transcription
1 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 HK of G as follows: HK = {hk : h H, k K}. It is not in general a subgroup. Note the following: Proposition 1.1. Suppose that H K = {1}. Then the function F : H K G defined by F (h, k) = hk is injective and its image is HK. In particular, if H and K are both finite and H K = {1}, then #(HK) = #(H) #(K). Proof. Clearly, the image of F is HK by definition. To see that F is injective, suppose that F (h 1, k 1 ) = F (h 2, k 2 ). Then by definition h 1 k 1 = h 2 k 2. Thus h 1 2 h 1 = k 2 k1 1. Since H is a subgroup, h 1 2 h 1 H, and since K is a subgroup, k 2 k1 1 K. Thus h 1 2 h 1 = k 2 k1 1 H K = {1}, and so h 1 2 h 1 = k 2 k1 1 = 1. It follows that h 1 2 h 1 = 1, so that h 1 = h 2, and similarly k 2 k1 1 = 1 so that k 2 = k 1. Thus (h 1, k 1 ) = (h 2, k 2 ) and F is injective. Corollary 1.2. Suppose that G is finite, that H and K are two subgroups of G with H K = {1} and that #(H) #(K) = #(G). Then the function F : H K G defined by F (h, k) = hk is a bijection from H K to G. 1
2 Proof. Proposition 1.1 says that F : H K G is injective, but both H K and G have the same number of elements, so F is surjective as well, hence a bijection. The next proposition provides some examples where the hypothesis H K = {1} is satisfied. Proposition 1.3. Suppose that H and K are finite subgroups of a group G. (i) If the orders #(H) and #(K) are relatively prime, then H K = {1}. (ii) If #(H) = p is a prime number and H is not contained in K, then H K = {1}. For example, if #(H) = #(K) = p and H K, then H K = {1}. Proof. (i) was a homework problem (by Lagrange s theorem, the order of H K must divide both #(H) and #(K) and hence must be 1). To see (ii), note that, since #(H) is prime, it is isomorphic to Z/pZ, and so the only subgroups of H are either H or {1}. But H K is a subgroup of H, not equal to H since otherwise H K. So H K = {1}. We now consider the case where one of the subgroups, say H, is a normal subgroup of G. In this case, we have the following: Proposition 1.4. Suppose that H G and that K G. Then HK is a subgroup of G, not necessarily normal. Moreover, H HK and K HK. Proof. HK is closed under multiplication since (h 1 k 1 )(h 2 k 2 ) = h 1 (k 1 h 2 k 1 1 )k 1k 2, and this is in HK since, as H is normal, k 1 h 2 k 1 1 H. Identity: since 1 H and 1 K, 1 = 1 1 HK. Inverses: (hk) 1 = k 1 h 1 = (k 1 h 1 k)k 1 and this is in HK since, again by normality, k 1 h 1 k H. Clearly H HK and K HK: for example, 1 K and, for all h H, h = h 1 HK. Thus H HK, hence H HK, and the proof that K HK is similar. Since H G, H HK as well. The question as to what the quotient group HK/H looks like is answered by the Second Isomorphism Theorem: Theorem 1.5. Let H and K be two subgroups of G, with H G. Then HK is a subgroup of G, H HK, H K is a normal subgroup of K, and HK/H = K/(H K). 2
3 In particular, if H and K are both finite, then #(HK) = #(H) #(K) #(H K). Proof. We have seen that HK G and that H HK. Also, H K K, and, if x H K, then, for all k K, kxk 1 H since x H and H is normal, and kxk 1 K since x K and K is a subgroup. Thus H K is a normal subgroup of K. To prove the theorem, we shall find a surjective homomorphism f from K to HK/H whose kernel is H K. The First Isomorphism Theorem then says that K/(H K) = HK/H. Define f(k) = kh, the H-coset containing k. Thus f is the composition π i, where i: K HK is the inclusion and π : HK HK/H is the quotient homomorphism. Hence f is a homomorphism since it is a composition of two homomorphisms, and clearly Ker f = Ker π K = H K. We will be done if we show that f is surjective, in other words that every coset xh, with x HK, is equal to some coset of the form kh with k K. But since x HK, x = hk for some h H and k K. Since H is normal, hk = kh for some h H, and clearly kh H = kh. Thus xh = hkh = kh H = kh = f(k), so that f is surjective. The final equality holds since then #(HK)/#(H) = #(K)/#(H K). In case neither of H, K is normal, a somewhat more involved argument along the lines of the proof of Proposition 1.1 shows the following counting formula: Proposition 1.6. Let H and K be two subgroups of G, and define F : H K G as before by F (h, k) = hk. Then the image of F is HK. Moreover, for all x HK, there is a bijection from the inverse image F 1 (x) to the subgroup H K. In particular, if H and K are both finite, then #(HK) = #(H) #(K) #(H K). 2 Direct products and semi-direct products We turn now to a construction which generalizes the direct product of two groups. To set the background, we start with the following: 3
4 Theorem 2.1. Suppose that H and K are two subgroups of a group G such that (i) H K = {1}; (ii) HK = G; (iii) For all h H and k K, hk = kh. Then G = H K. Proof. Define F : H K G by: F (h, k) = hk. We shall show that F is an isomorphism. By Proposition 1.1, the function F is injective, and its image is HK = G by assumption (ii). Thus F is a bijection. To see that F is an isomorphism, note that F ((h 1, k 1 )(h 2, k 2 )) = F (h 1 h 2, k 1 k 2 ) = (h 1 h 2 )(k 1 k 2 ) = h 1 (h 2 k 1 )k 2 = h 1 (k 1 h 2 )k 2 = (h 1 k 1 )(h 2 k 2 ) Thus F is an isomorphism. = F (h 1, k 1 )F (h 2, k 2 ). Example 2.2. The abelian group C contains two subgroups, R >0 and U(1), the subgroup of positive real numbers (under multiplication). Clearly R >0 U(1) = {1}, since a positive real number of absolute value 1 is 1. Also, every nonzero complex number z can be written as z = z z z, where z R >0 and z/ z U(1). Hence C = R >0 U(1). By the theorem, C = R >0 U(1). Let us give an application to finite groups where we use this criterion: Proposition 2.3. Let p be a prime and let G be a group of order p 2. The either G = Z/p 2 Z or G = (Z/pZ) (Z/pZ). Proof. We have seen previously that G is abelian. By Lagrange s theorem, every element of G, not the identity, has order p or p 2. If there exists an element of G whose order is p 2, then G is cyclic of order p 2 and hence G = Z/p 2 Z. Thus we may assume that every element of G, not the identity, has order p. Choose x G, x 1. Then x has order p. In particular, x G. Choose an element y G, y / x. Then y has order p as well. Since x and y are two different subgroups of G, by Proposition 1.3, x y = {1}. By Corollary 1.2, as #( x )#( y ) = p 2 = #(G), x y = G. Thus Conditions (i) and (ii) above are satisfied, and Condition (iii) is automatic since G is abelian. Thus G = x y = (Z/pZ) (Z/pZ). 4
5 In general, the condition above that, for all h H and k K, hk = kh (i.e. the condition that every element of H commutes with every element of K), or equivalently the condition that G = H K, is too restrictive, and we want to relax it a bit and as a by-product describe a new class of interesting groups. To motivate the construction, consider the following set of assumptions on a pair of subgroups H and K of a group G: Assumption. Let G be a group and let H and K be subgroups of G such that: (i) H K = {1}; (ii) HK = G; (iii) H is a normal subgroup of G. Note: We do not require that K is also a normal subgroup of G. In fact, if K is also a normal subgroup of G, then, as we have seen in the homework, for all h H and k K, hk = kh (by considering the product hkh 1 k 1 which lies in both H and K), so we are back in the situation where G is isomorphic to the direct (Cartesian) product H K of H and K. Using the data above, let us see how to understand the group structure on H. By Proposition 1.1 and the assumptions (i) and (ii) above, we see that every element of G can be written uniquely as hk for h H and k K, i.e. the function F : H K G defined by F (h, k) = hk is a bijection. To understand the group law on G, or equivalently on H K, note that (h 1 k 1 )(h 2 k 2 ) = (h 1 (k 1 h 2 k 1 1 ))(k 1k 2 ), which is of the form h k for some (in fact unique) h H, k K since H is normal. To rewrite this operation in another way, define a function f : K Aut H by f(k) = i k, where i k denotes conjugation by k, i.e. f(k) is the automorphism of H defined by f(k)(h) = khk 1. Then we see that multiplication in G is given by the rule (h 1 k 1 )(h 2 k 2 ) = (h 1 f(k 1 )(h 2 ))(k 1 k 2 ). We can ask if it is possible to reverse this process. Concretely, given two abstract groups H and K and a homomorphism f : K Aut H, define a binary operation on the Cartesian product H K by the rule (h 1, k 1 ) (h 2, k 2 ) = (h 1 f(k 1 )(h 2 ), k 1 k 2 ). 5
6 Note that this operation depends on the choice of f. In particular, is the usual product operation in H K f is the trivial homomorphism from K to Aut H, i.e. f(k) = Id for all k K. Proposition 2.4. Let H and K be two groups and let f : K Aut H be a homomorphism. Define a binary operation on H K by the formula (h 1, k 1 ) (h 2, k 2 ) = (h 1 f(k 1 )(h 2 ), k 1 k 2 ). Then H K, together with this operation, is a group H K, the semi-direct product of H and K. Under this operation, H {1} is a normal subgroup of H K isomorphic to H, {1} K is a subgroup of H K isomorphic to K, and the conjugation action of K = {1} K on H {1} = H can be identified with the action given by the automorphism f. The notation is meant to suggest both the product and the fact that H corresponds to a normal subgroup of H K. Of course, the definition of the group operation depends not just on the groups H and K but also on the homomorphism f : K Aut H. To make this dependence explicit, we sometimes write H f K for the semi-direct product. Proof of Proposition 2.4. As usual, the hardest part to check is associativity: ((h 1, k 1 ) (h 2, k 2 )) (h 3, k 3 ) = (h 1 f(k 1 )(h 2 ), k 1 k 2 ) (h 3, k 3 ) = (h 1 f(k 1 )(h 2 ) f(k 1 k 2 )(h 3 ), k 1 k 2 k 3 ) = (h 1 f(k 1 )(h 2 ) f(k 1 )(f(k 2 )(h 3 )), k 1 k 2 k 3 ), where we have freely used associativity in G and the fact that f is a homomorphism. On the other hand, (h 1, k 1 ) ((h 2, k 2 ) (h 3, k 3 )) = (h 1, k 1 ) (h 2 f(k 2 )(h 3 ), k 2 k 3 ) = (h 1 f(k 1 )(h 2 f(k 2 )(h 3 )), k 1 k 2 k 3 ) = (h 1 f(k 1 )(h 2 ) f(k 1 )(f(k 2 )(h 3 )), k 1 k 2 k 3 ), where we have also used the fact that f(k 1 ) is an automorphism of H. Hence is associative. Clearly (1, 1) (h, k) = (1 f(1)(h), 1 k) = (Id(h), k) = (h, k), and (h, k) (1, 1) = (h f(k)(1), k 1) = (h, k), 6
7 since f(k) is an automorphism and hence f(k)(1) = 1. Finally, it is straightforward to check that the inverse of (h, k) is (f(k) 1 (h 1 ), k 1 ), where the element f(k) 1 (h 1 ) is the inverse function of the automorphism f(k) (which is f(k 1 ) since f is a homomorphism) applied to the element k 1 of K: whereas (h, k) (f(k) 1 (h 1 ), k 1 ) = (h f(k)(f(k) 1 (h 1 )), kk 1 ) = (hh 1, kk 1 ) = (1, 1), (f(k) 1 (h 1 ), k 1 ) (h, k) = (f(k) 1 (h 1 ) f(k 1 )(h), k 1 k) = (f(k) 1 (h 1 h), 1) = ((f(k) 1 (1), 1) = (1, 1), where we have again used the fact that f(k 1 ) is an automorphism, as well as f(k 1 ) = f(k) 1. Note that, given h 1 and h 2 in H, (h 1, 1) (h 2, 1) = (h 1 f(1)(h 2 ), 1 1) = (h 1 Id(h 2 ), 1) = (h 1 h 2, 1). Hence the function g 1 : H H K defined by g 1 (h) = (h, 1) is a homomorphism from H to H K whose image is clearly H {1}, showing that H {1} is a subgroup of H K. Since g 1 is clearly injective, it induces an isomorphism from H to H {1}. Likewise, since, given k 1, k 2 K, (1, k 1 ) (1, k 2 ) = (1 f(k 1 )(1), k 1 k 2 ) = (1, k 1 k 2 ), the function g 2 : K H K is an injective homomorphism whose image is {1} K, so that {1} K is a subgroup of H K isomorphic to K. Note that (h, 1) (1, k) = (h f(1)(1), 1 k) = (h, k), so that every element of H K can be (uniquely) written as a product of an element in H {1} times an element in {1} K. The same statement is true in the other order, but it is more complicated since (1, k) (h, 1) = (f(k)(h), k). We then have the formula (1, k) (h, 1) (1, k) 1 = (f(k)(h), k) (1, k 1 ) = (f(k)(h) f(k)(1), kk 1 ) = (f(k)(h), 1). This says that conjugation by an element of {1} K takes H {1} to itself, and since H {1} is a subgroup, conjugation by an element of H {1} also takes H {1} to itself. Since every element of H K is a product of 7
8 an element in H {1} and an element in {1} K, H {1} is a normal subgroup of H K. Moreover, identifying H {1} with H via g1 1 and identifying {1} K with K via g2 1, the conjugation action of {1} K on H {1} is identified with the action of K on H via the homomorphism f. This concludes the proof of Proposition 2.4. Then the remarks before the statement of Proposition 2.4 show the following: Theorem 2.5. Let G be a group and let H and K be two subgroups of G such that (i) H K = {1}; (ii) HK = G; (iii) H is a normal subgroup of G. Then there is a homomorphism from K to Aut H defined by conjugation. Using this homomorphism to define H K, there is an isomorphism F : G H K such that F (H) is the normal subgroup H {1} and F (K) is the subgroup {1} K. Remark 2.6. If G is isomorphic to a semi-direct product of H and K, then identifying H with the corresponding subgroup of G, H is a normal subgroup of G and G/H = K by the second isomorphism theorem, Theorem 1.5 (or directly). This says that, in the case where G contains a normal subgroup H such that G/H = K, and moreover G is isomorphic to the semi-direct product H K, then we can reconstruct G from the smaller building blocks H and K, together with the knowledge of the homomorphism f : K Aut H. However, while the construction of a semi-direct product is important in group theory (and we shall give many examples below), one should not be too optimistic for its applicability. To say that a group G is a semi-direct product, one needs a normal subgroup H and another subgroup K such that the quotient homomorphism π induces an isomorphism from K to G/H. This is often expressed as follows: we can lift the quotient group G/H to a subgroup of G. This usually turns out to be impossible, even in very simple cases. For example, the quotient of Z/4Z by the subgroup H = 2 = Z/2Z is isomorphic to Z/2Z, but there is no subgroup K of Z/4Z which maps isomorphically onto the quotient Z/2Z, and Z/4Z is not isomorphic to a product, semi-direct or otherwise, of Z/2Z with Z/2Z. There is a gadget 8
9 (group cohomology) which classifies (in a certain sense) all groups G which have a normal subgroup H and such that the quotient is isomorphic to K, in terms of H and K and a homomorphism f : K Aut H, provided that H is abelian, but there is no mechanism for handling the general case. Example 2.7. (1) Every direct product H K is a semi-direct product with f : K Aut H the trivial homomorphism (f(k) = Id for all k K). (2) The dihedral group D n of order 2n is a semi-direct product. Here H = ρ is the cyclic subgroup of order n consisting of the rotations, and K = τ is the order 2 subgroup generated by a reflection. Via the discussion of HW 6, Problem 2, H = A 2π/n and K = R. Since H has index 2, it is normal, and τ ρ τ 1 = τ ρ τ = ρ 1. In terms of the matrices A 2kπ/n and R, since R has order 2 and hence R 1 = R, we have RA 2kπ/n R 1 = RA 2kπ/n R = A 2kπ/n. Thus D n = Z/nZ Z/2Z, where the homomorphism f : Z/2Z Aut(Z/nZ) is given by f(0) = Id and f(1) = Id. (3) Similarly, the group O 2 contains ( the ) normal subgroup SO 2 and the 1 0 subgroup R, where R = B 0 =, in the notation of Problem 2 of 0 1 HW 3. Since R has order 2 and, and, for all A SO 2, RAR 1 = A 1, we see that O 2 is a semi-direct product SO 2 {±1} = U(1) {±1}. Here, the corresponding homomorphism f : {±1} Aut U(1) is defined by f(1) = Id and f( 1) is the automorphism of U(1) sending z to z 1. (4) Let G be the group of linear-affine functions l: R R under function composition. Here a linear affine function is a function of the form l(x) = ax+b with a 0. There are two very natural subgroups of G: the translation subgroup H consisting of all functions of the form t b (x) = x + b, and the similitudes K, consisting of all functions of the form m a (x) = ax, a 0. Note that H = R and K = R. Clearly H K = {Id} and, if l(x) = ax + b, then l = t b m a. Finally, a calculation shows that m a t b m 1 a = t ab. Hence H is normal, and the action of K = R on H = R is by sending m a to the automorphism of the (additive) group R given by multiplication by a. Thus G = R R using the action described above. A very similar argument works for the group G of affine-linear transformations of R n ; here L: R n R n is affine-linear if it is of the form L(x) = Ax + b, where A GL n (R) and b R n, with L 1 (x) = A 1 x A 1 b. Then G has a subgroup isomorphic to R n (the translation subgroup) and a subgroup isomorphic to GL n (R). In this case, if T p is translation by p, 9
10 so that T p (x) = x + p, and L(x) = Ax + b is as above, then a calculation shows that (L T p L 1 )(x) = x + Ap = T Ap (x). Hence the translation subgroup is a normal subgroup isomorphic to R n, GL n (R) is a (non-normal) subgroup of G, and G = R n GL n (R), where the action of GL n (R) on R n is the usual one (matrix multiplication). Finally, one often (especially in physics, and for n = 3) considers the Euclidean group, the subgroup E of G, where A SO n, so an element L of E is of the form L(x) = Ax + b, A SO n. Thus A is a rigid motion of R n fixing the origin followed by a translation. In case n = 4 replacing SO n by the Lorentz group gives a group of symmetries well-adapted to electromagnetism and special relativity. (5) As in the homework, let B be the group of upper triangular 2 2 matrices: {( ) } a b B = : a, b, d R, ad 0. 0 d Inside B there are two natural subgroups: the subgroup T of strictly upper triangular matrices, {( ) } 1 b T = : b R, 0 1 and the subgroup D of diagonal matrices D: {( ) } a 0 D = : a, d R, ad 0. 0 d Clearly T D = {Id}, and since ( ) ( ) 1 bd 1 a 0 = d ( ) a b, 0 d we see that T D = B. We have seen in the homework that T B. Hence B = T D. In this case, the action of D = (R ) 2 on T = R is given by: (a, d) (R ) 2 is sent to the automorphism of R given by multiplication by ad 1. (6) Let G be a group of order pq, where p and q are distinct primes and p < q. If P is a p-sylow subgroup and Q is the q-sylow subgroup, then we have seen in class that Q G and P Q = {1} since their orders are relatively prime. Finally, QP = G since #(G) = #(Q)#(P ), by Corollary 1.2. Hence G = Q P. 10
11 The additional piece of information is the homomorphism f : P = Z/pZ Aut Q = Aut(Z/qZ). Clearly f is trivial G is abelian P is normal there is a unique p-sylow subgroup. Otherwise, f corresponds to a non-trivial homomorphism from Z/pZ to Aut(Z/qZ). As we have seen, Aut(Z/qZ) = (Z/qZ) is a group of order q 1, and since Z/pZ is simple, f is non-trivial f is injective. This is only possible if p divides q 1, by Lagrange s theorem, since the image of f is then a subgroup of (Z/qZ) of order p. Of course, by part (3) of the Sylow theorem, we know that the number of p-sylow subgroups is a divisor of pq and 1 mod p, so if this number is not 1, it must be q and we must have q 1 mod p, i.e. p divides q 1. In fact, using the fact that (Z/qZ) is cyclic, one can show that every two nonabelian groups of order pq are isomorphic. 11
Teddy Einstein Math 4320
Teddy Einstein Math 4320 HW4 Solutions Problem 1: 2.92 An automorphism of a group G is an isomorphism G G. i. Prove that Aut G is a group under composition. Proof. Let f, g Aut G. Then f g is a bijective
More informationFINITE 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
More informationSimple 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
More informationMath 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
More informationMA441: 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
More informationA 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
More informationDefinitions. 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
More informationElements of solution for Homework 5
Elements of solution for Homework 5 General remarks How to use the First Isomorphism Theorem A standard way to prove statements of the form G/H is isomorphic to Γ is to construct a homomorphism ϕ : G Γ
More informationAlgebra SEP Solutions
Algebra SEP Solutions 17 July 2017 1. (January 2017 problem 1) For example: (a) G = Z/4Z, N = Z/2Z. More generally, G = Z/p n Z, N = Z/pZ, p any prime number, n 2. Also G = Z, N = nz for any n 2, since
More informationMath 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
More informationCOURSE SUMMARY FOR MATH 504, FALL QUARTER : MODERN ALGEBRA
COURSE SUMMARY FOR MATH 504, FALL QUARTER 2017-8: MODERN ALGEBRA JAROD ALPER Week 1, Sept 27, 29: Introduction to Groups Lecture 1: Introduction to groups. Defined a group and discussed basic properties
More informationSolutions of exercise sheet 4
D-MATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 4 The content of the marked exercises (*) should be known for the exam. 1. Prove the following two properties of groups: 1. Every
More informationMath 581 Problem Set 8 Solutions
Math 581 Problem Set 8 Solutions 1. Prove that a group G is abelian if and only if the function ϕ : G G given by ϕ(g) g 1 is a homomorphism of groups. In this case show that ϕ is an isomorphism. Proof:
More informationProblem 1.1. Classify all groups of order 385 up to isomorphism.
Math 504: Modern Algebra, Fall Quarter 2017 Jarod Alper Midterm Solutions Problem 1.1. Classify all groups of order 385 up to isomorphism. Solution: Let G be a group of order 385. Factor 385 as 385 = 5
More informationWritten Homework # 2 Solution
Math 516 Fall 2006 Radford Written Homework # 2 Solution 10/09/06 Let G be a non-empty set with binary operation. For non-empty subsets S, T G we define the product of the sets S and T by If S = {s} is
More informationMath 546, Exam 2 Information.
Math 546, Exam 2 Information. 10/21/09, LC 303B, 10:10-11:00. Exam 2 will be based on: Sections 3.2, 3.3, 3.4, 3.5; The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/546fa09/546.html)
More informationANALYSIS OF SMALL GROUPS
ANALYSIS OF SMALL GROUPS 1. Big Enough Subgroups are Normal Proposition 1.1. Let G be a finite group, and let q be the smallest prime divisor of G. Let N G be a subgroup of index q. Then N is a normal
More informationGroup 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
More informationJohns 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
More informationSchool 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
More informationCourse 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.......................
More informationExtra exercises for algebra
Extra exercises for algebra These are extra exercises for the course algebra. They are meant for those students who tend to have already solved all the exercises at the beginning of the exercise session
More informationCONSEQUENCES 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.
More informationMATH 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
More informationLecture 7 Cyclic groups and subgroups
Lecture 7 Cyclic groups and subgroups Review Types of groups we know Numbers: Z, Q, R, C, Q, R, C Matrices: (M n (F ), +), GL n (F ), where F = Q, R, or C. Modular groups: Z/nZ and (Z/nZ) Dihedral groups:
More informationMATH 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
More informationYale University Department of Mathematics Math 350 Introduction to Abstract Algebra Fall Midterm Exam Review Solutions
Yale University Department of Mathematics Math 350 Introduction to Abstract Algebra Fall 2015 Midterm Exam Review Solutions Practice exam questions: 1. Let V 1 R 2 be the subset of all vectors whose slope
More informationBASIC GROUP THEORY : G G G,
BASIC GROUP THEORY 18.904 1. Definitions Definition 1.1. A group (G, ) is a set G with a binary operation : G G G, and a unit e G, possessing the following properties. (1) Unital: for g G, we have g e
More informationNOTES 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
More informationAlgebra 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
More informationGroups 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
More informationDIHEDRAL GROUPS II KEITH CONRAD
DIHEDRAL GROUPS II KEITH CONRAD We will characterize dihedral groups in terms of generators and relations, and describe the subgroups of D n, including the normal subgroups. We will also introduce an infinite
More informationExercises 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
More informationits 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
More informationABSTRACT 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
More informationMath 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
More informationPRACTICE FINAL MATH , MIT, SPRING 13. You have three hours. This test is closed book, closed notes, no calculators.
PRACTICE FINAL MATH 18.703, MIT, SPRING 13 You have three hours. This test is closed book, closed notes, no calculators. There are 11 problems, and the total number of points is 180. Show all your work.
More informationMath 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
More informationMath 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
More informationFall 2014 Math 122 Midterm 1
1. Some things you ve (maybe) done before. 5 points each. (a) If g and h are elements of a group G, show that (gh) 1 = h 1 g 1. (gh)(h 1 g 1 )=g(hh 1 )g 1 = g1g 1 = gg 1 =1. Likewise, (h 1 g 1 )(gh) =h
More informationMath 210A: Algebra, Homework 6
Math 210A: Algebra, Homework 6 Ian Coley November 13, 2013 Problem 1 For every two nonzero integers n and m construct an exact sequence For which n and m is the sequence split? 0 Z/nZ Z/mnZ Z/mZ 0 Let
More informationISOMORPHISMS KEITH CONRAD
ISOMORPHISMS KEITH CONRAD 1. Introduction Groups that are not literally the same may be structurally the same. An example of this idea from high school math is the relation between multiplication and addition
More informationINVERSE 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
More informationABSTRACT ALGEBRA 1 COURSE NOTES, LECTURE 11: SYLOW THEORY.
ABSTRACT ALGEBRA 1 COURSE NOTES, LECTURE 11: SYLOW THEORY. ANDREW SALCH Here s a quick definition we could have introduced a long time ago: Definition 0.1. If n is a positive integer, we often write C
More informationMATH 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.
More informationMath 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
More informationALGEBRA HOMEWORK SET 2. Due by class time on Wednesday 14 September. Homework must be typeset and submitted by as a PDF file.
ALGEBRA HOMEWORK SET 2 JAMES CUMMINGS (JCUMMING@ANDREW.CMU.EDU) Due by class time on Wednesday 14 September. Homework must be typeset and submitted by email as a PDF file. (1) Let H and N be groups and
More information120A 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
More informationIIT Mumbai 2015 MA 419, Basic Algebra Tutorial Sheet-1
IIT Mumbai 2015 MA 419, Basic Algebra Tutorial Sheet-1 Let Σ be the set of all symmetries of the plane Π. 1. Give examples of s, t Σ such that st ts. 2. If s, t Σ agree on three non-collinear points, then
More informationFall /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
More information1.1 Definition. A monoid is a set M together with a map. 1.3 Definition. A monoid is commutative if x y = y x for all x, y M.
1 Monoids and groups 1.1 Definition. A monoid is a set M together with a map M M M, (x, y) x y such that (i) (x y) z = x (y z) x, y, z M (associativity); (ii) e M such that x e = e x = x for all x M (e
More informationExercises 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
More informationSemidirect products are split short exact sequences
CHAPTER 16 Semidirect products are split short exact sequences Chit-chat 16.1. Last time we talked about short exact sequences G H K. To make things easier to read, from now on we ll write L H R. The L
More informationφ(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 =
More information5 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
More informationEXERCISES ON THE OUTER AUTOMORPHISMS OF S 6
EXERCISES ON THE OUTER AUTOMORPHISMS OF S 6 AARON LANDESMAN 1. INTRODUCTION In this class, we investigate the outer automorphism of S 6. Let s recall some definitions, so that we can state what an outer
More informationA Primer on Homological Algebra
A Primer on Homological Algebra Henry Y Chan July 12, 213 1 Modules For people who have taken the algebra sequence, you can pretty much skip the first section Before telling you what a module is, you probably
More informationSolutions to Assignment 4
1. Let G be a finite, abelian group written additively. Let x = g G g, and let G 2 be the subgroup of G defined by G 2 = {g G 2g = 0}. (a) Show that x = g G 2 g. (b) Show that x = 0 if G 2 = 2. If G 2
More informationABSTRACT ALGEBRA 1, LECTURE NOTES 5: HOMOMORPHISMS, ISOMORPHISMS, SUBGROUPS, QUOTIENT ( FACTOR ) GROUPS. ANDREW SALCH
ABSTRACT ALGEBRA 1, LECTURE NOTES 5: HOMOMORPHISMS, ISOMORPHISMS, SUBGROUPS, QUOTIENT ( FACTOR ) GROUPS. ANDREW SALCH 1. Homomorphisms and isomorphisms between groups. Definition 1.1. Let G, H be groups.
More information1 Chapter 6 - Exercise 1.8.cf
1 CHAPTER 6 - EXERCISE 1.8.CF 1 1 Chapter 6 - Exercise 1.8.cf Determine 1 The Class Equation of the dihedral group D 5. Note first that D 5 = 10 = 5 2. Hence every conjugacy class will have order 1, 2
More informationSupplement. 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
More informationMATH 436 Notes: Homomorphisms.
MATH 436 Notes: Homomorphisms. Jonathan Pakianathan September 23, 2003 1 Homomorphisms Definition 1.1. Given monoids M 1 and M 2, we say that f : M 1 M 2 is a homomorphism if (A) f(ab) = f(a)f(b) for all
More information3. 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
More informationMATH 430 PART 2: GROUPS AND SUBGROUPS
MATH 430 PART 2: GROUPS AND SUBGROUPS Last class, we encountered the structure D 3 where the set was motions which preserve an equilateral triangle and the operation was function composition. We determined
More informationHomework Problems, Math 200, Fall 2011 (Robert Boltje)
Homework Problems, Math 200, Fall 2011 (Robert Boltje) Due Friday, September 30: ( ) 0 a 1. Let S be the set of all matrices with entries a, b Z. Show 0 b that S is a semigroup under matrix multiplication
More informationGroup 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
More information0 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
More informationPseudo 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
More informationMath 120 HW 9 Solutions
Math 120 HW 9 Solutions June 8, 2018 Question 1 Write down a ring homomorphism (no proof required) f from R = Z[ 11] = {a + b 11 a, b Z} to S = Z/35Z. The main difficulty is to find an element x Z/35Z
More informationTheorems 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
More information1.5 Applications Of The Sylow Theorems
14 CHAPTER1. GROUP THEORY 8. The Sylow theorems are about subgroups whose order is a power of a prime p. Here is a result about subgroups of index p. Let H be a subgroup of the finite group G, and assume
More informationAbstract Algebra II Groups ( )
Abstract Algebra II Groups ( ) Melchior Grützmann / melchiorgfreehostingcom/algebra October 15, 2012 Outline Group homomorphisms Free groups, free products, and presentations Free products ( ) Definition
More informationGROUPS AND THEIR REPRESENTATIONS. 1. introduction
GROUPS AND THEIR REPRESENTATIONS KAREN E. SMITH 1. introduction Representation theory is the study of the concrete ways in which abstract groups can be realized as groups of rigid transformations of R
More informationMath 122 Midterm 2 Fall 2014 Solutions
Math 122 Midterm 2 Fall 2014 Solutions Common mistakes i. Groups of order pq are not always cyclic. Look back on Homework Eight. Also consider the dihedral groups D 2n for n an odd prime. ii. If H G and
More informationMath 3140 Fall 2012 Assignment #3
Math 3140 Fall 2012 Assignment #3 Due Fri., Sept. 21. Remember to cite your sources, including the people you talk to. My solutions will repeatedly use the following proposition from class: Proposition
More informationMATH 28A MIDTERM 2 INSTRUCTOR: HAROLD SULTAN
NAME: MATH 28A MIDTERM 2 INSTRUCTOR: HAROLD SULTAN 1. INSTRUCTIONS (1) Timing: You have 80 minutes for this midterm. (2) Partial Credit will be awarded. Please show your work and provide full solutions,
More informationPROBLEMS 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.
More informationTCC Homological Algebra: Assignment #3 (Solutions)
TCC Homological Algebra: Assignment #3 (Solutions) David Loeffler, d.a.loeffler@warwick.ac.uk 30th November 2016 This is the third of 4 problem sheets. Solutions should be submitted to me (via any appropriate
More informationProfinite Groups. Hendrik Lenstra. 1. Introduction
Profinite Groups Hendrik Lenstra 1. Introduction We begin informally with a motivation, relating profinite groups to the p-adic numbers. Let p be a prime number, and let Z p denote the ring of p-adic integers,
More informationCosets, 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
More informationAutomorphism Groups Definition. An automorphism of a group G is an isomorphism G G. The set of automorphisms of G is denoted Aut G.
Automorphism Groups 9-9-2012 Definition. An automorphism of a group G is an isomorphism G G. The set of automorphisms of G is denoted Aut G. Example. The identity map id : G G is an automorphism. Example.
More informationFrank Moore Algebra 901 Notes Professor: Tom Marley Direct Products of Groups:
Frank Moore Algebra 901 Notes Professor: Tom Marley Direct Products of Groups: Definition: The external direct product is defined to be the following: Let H 1,..., H n be groups. H 1 H 2 H n := {(h 1,...,
More informationAlgebraic structures I
MTH5100 Assignment 1-10 Algebraic structures I For handing in on various dates January March 2011 1 FUNCTIONS. Say which of the following rules successfully define functions, giving reasons. For each one
More informationMAT534 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)
More informationCosets and Normal Subgroups
Cosets and Normal Subgroups (Last Updated: November 3, 2017) These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from
More informationENTRY 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.
More informationDISCRETE 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
More informationSUMMARY OF GROUPS AND RINGS GROUPS AND RINGS III Week 1 Lecture 1 Tuesday 3 March.
SUMMARY OF GROUPS AND RINGS GROUPS AND RINGS III 2009 Week 1 Lecture 1 Tuesday 3 March. 1. Introduction (Background from Algebra II) 1.1. Groups and Subgroups. Definition 1.1. A binary operation on a set
More informationRings and Fields Theorems
Rings and Fields Theorems Rajesh Kumar PMATH 334 Intro to Rings and Fields Fall 2009 October 25, 2009 12 Rings and Fields 12.1 Definition Groups and Abelian Groups Let R be a non-empty set. Let + and (multiplication)
More information2) e = e G G such that if a G 0 =0 G G such that if a G e a = a e = a. 0 +a = a+0 = a.
Chapter 2 Groups Groups are the central objects of algebra. In later chapters we will define rings and modules and see that they are special cases of groups. Also ring homomorphisms and module homomorphisms
More informationMath 4400, Spring 08, Sample problems Final Exam.
Math 4400, Spring 08, Sample problems Final Exam. 1. Groups (1) (a) Let a be an element of a group G. Define the notions of exponent of a and period of a. (b) Suppose a has a finite period. Prove that
More informationAlgebra 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
More informationAlgebra. Travis Dirle. December 4, 2016
Abstract Algebra 2 Algebra Travis Dirle December 4, 2016 2 Contents 1 Groups 1 1.1 Semigroups, Monoids and Groups................ 1 1.2 Homomorphisms and Subgroups................. 2 1.3 Cyclic Groups...........................
More information23.1. Proof of the fundamental theorem of homomorphisms (FTH). We start by recalling the statement of FTH introduced last time.
23. Quotient groups II 23.1. Proof of the fundamental theorem of homomorphisms (FTH). We start by recalling the statement of FTH introduced last time. Theorem (FTH). Let G, Q be groups and ϕ : G Q a homomorphism.
More informationMA441: 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 :
More informationHigher Algebra Lecture Notes
Higher Algebra Lecture Notes October 2010 Gerald Höhn Department of Mathematics Kansas State University 138 Cardwell Hall Manhattan, KS 66506-2602 USA gerald@math.ksu.edu This are the notes for my lecture
More informationSolutions of exercise sheet 8
D-MATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 8 1. In this exercise, we will give a characterization for solvable groups using commutator subgroups. See last semester s (Algebra
More informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #24 12/03/2013
18.78 Introduction to Arithmetic Geometry Fall 013 Lecture #4 1/03/013 4.1 Isogenies of elliptic curves Definition 4.1. Let E 1 /k and E /k be elliptic curves with distinguished rational points O 1 and
More informationHOMEWORK 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
More informationThe Gordon game. EWU Digital Commons. Eastern Washington University. Anthony Frenk Eastern Washington University
Eastern Washington University EWU Digital Commons EWU Masters Thesis Collection Student Research and Creative Works 2013 The Gordon game Anthony Frenk Eastern Washington University Follow this and additional
More information