DOCSLIB.ORG

Is a Normal Subgroup of GL(2,R). Apply the One-Step Subgroup Test to N = SL(2,R)

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Is a Normal Subgroup of GL(2,R). Apply the One-Step Subgroup Test to N = SL(2,R) Handout #9 Math 545 Fall, 2010 1. Prove that SL(2,R) is a normal subgroup of GL(2,R). Apply the One-Step Subgroup Test to N = SL(2,R). N 6= ∅ since the identity matrix of G has its determinant equal to 1 and therefore belongs to N. Let A, B ∈ N. Show AB−1 ∈ N. It suffices to show that det(AB−) = 1. det(AB−1) = det A det(B−1) (since the function det is multiplicative) 1 1 = det A det B = 1. 1 (since A, B ∈ N) = 1. Now show that N is a normal subgroup of GL(2,R). Let A ∈ GL(2, R) and B ∈ N. Show ABA−1 ∈ N. It suffices to show that det(ABA−) = 1. −1 −1 1 det(ABA ) = det A det B det(A ) (since the function det is multiplicative) = det A det B det A = 1 det A.1. det A (since B ∈ N) = 1. 2. Viewing < 3 > and < 12 > as subgroups of Z, list all elements of the factor group < 3 > / < 12 >. Let m ∈< 3 >. Then m = 12q + r, where r = 0, 3, 6 or 9. =⇒ m+ < 12 >= r + 12q+ < 12 >= r+ < 12 > (since 12q ∈< 12 >). Thus < 3 > / < 12 >= {m+ < 12 > |m ∈< 3 >} = {r+ < 12 > |r = 0, 3, 6, 9} = {< 12 >, 3+ < 12 >, 6+ < 12 >, 9+ < 12 >} 3. What is the order of the element 14+ < 8 > of the factor group Z24/ < 8 >? Z24 = {0, 1, 2,..., 23} and < 8 >= {0, 8, 16}. Z24/ < 8 >= {< 8 >, 1+ < 8 >, 2+ < 8 > , 3+ < 8 >, 4+ < 8 >, 5+ < 8 >, 6+ < 8 >, 7+ < 8 >}. Note 14+ < 8 >= 6 + (8+ < 8 >). Thus 14+ < 8 >= 6+ < 8 >(since 8 ∈ < 8 >). Since 1(6+ < 8 >) = 6+ < 8 >6=< 8 >; 2(6+ < 8 >) = 12+ < 8 >= 4+ < 8 >6=< 8 >; 3(6+ < 8 >) = 18+ < 8 >= 2+ < 8 >6=< 8 > and 4(6+ < 8 >) = 24+ < 8 >=< 8 > (because 24 ≡24 0 ∈< 8 >). Hence |6+ < 8 > | = 4 and since 14+ < 8 >= 6+ < 8 >, we have |6+ < 8 > | = 4. 4. Suppose that G is a non-Abelian group of order p3 (where p is prime) and Z(G) 6= {e}. Prove that |Z(G)| = p. Since Z(G) is a subgroup of G and |G| = p3. =⇒ |Z(G)| = 1, p, p2, or p3 (Lagrange). Now prove: |Z(G)| = p2 and |Z(G)| = p3 are impossible. |Z(G)| = p3 =⇒ |Z(G)| = |G| =⇒ G is abelian, 2 |G| p3 a contradiction since G is non-Abelian and |Z(G)| = p =⇒ |G/Z(G)| = |Z(G)| = p2 = p =⇒ G/Z(G) is of prime order p and therefore cyclic =⇒ G is abelian, a contradiction since G is non-abelian. Hence |Z(G)| = p (since |Z(G)|= 6 1 from the given). 5. If |G| = pq, where p and q are not necessarily distinct primes, prove that |Z(G)| = 1 or pq. |G| = pq =⇒ |Z(G)| = 1, p, q, or pq. Prove: |Z(G)| = p and |Z(G)| = q are impossible. If |G| pq |Z(G)| = p then |G/Z(G)| = |Z(G)| = p = q. But then, since G/Z(G) is cyclic of prime order, |G/Z(G)| = 1. Thus q = 1, a contradiction since q > 1. Similarly, |Z(G)| = q =⇒ p = 1, a contradiction to the definition of p. 6. If M and N are a normal subgroups of G, then prove that M ∩ N is a normal subgroup of G. Since M and N are subgroups of G, M∩N is a subgroup of G (done in class). Need to prove M∩N is normal in G. Let g ∈ G and let n ∈ M ∩ N. Show gng−1 ∈ M ∩ N. Now n ∈ M ∩ N =⇒ n ∈ M and n ∈ N. Then gng−1 ∈ M and gng−1 ∈ N (since n ∈ M, n ∈ N and g ∈ G and M and N are normal in G). Thus gng−1 ∈ M ∩ N. 7. If N and M are normal subgroups of G, prove that NM is also a normal subgroup of G. First note that e = ee ∈ NM, since e ∈ N and e ∈ M. So NM 6= ∅. Let n1m1, n2m2 ∈ NM, where n1, n2 ∈ N and m1, m2 ∈ M. −1 −1 −1 −1 −1 Then (n1m1)(n2m2) = n1m1(m2) (n2) = n1n3m1(m2) (since N is normal in G, m1(m2) N = −1 −1 −1 −1 −1 Nm1(m2) =⇒ m1(m2) (n2) = n3m1(m2) , where n3 ∈ N) Thus (n1m1)(n2m2) = 1 −1 −1 n1n3m1(m2) ∈ NM, since n1n3 ∈ N and m1(m2) ∈ M. Hence, by One-Step Subgroup Test, NM is a subgroup of G. Now show NM is a normal subgroup of G. Let nm ∈ NM and g ∈ G. Prove g(nm)g−1 ∈ NM. g(nm)g−1 = g(nem)g−1 = g(ng−1gm)g−1 = (gng−1)(gmg−1) ∈ NM (since N and M are normal in G, gng−1 ∈ N and gmg−1 ∈ M). 8. List all elements of Z/ < 6 >. Z/ < 6 >= {a+ < 6 > |a ∈ Z}. Let a ∈ Z. Then, by the division algorithm, ∃q, r ∈ Z such that a = 6q + r, where r = 0, 1, 2, 3, 4 or 5. =⇒ a+ < 6 >= r + 6q+ < 6 >= r+ < 6 > (since 6q ∈< 6 >). Thus Z/ < 6 >= {a+ < 6 > |a ∈ Z} = {r+ < 6 > |r = 0, 1, 2, 3, 4, 5} = {< 6 > , 1+ < 6 >, 2+ < 6 >, 3+ < 6 >, 4+ < 6 >, 5+ < 6 >} 2 3 4 5 2 3 4 5 9. Consider the dihedral group of order 12, D6 = {e, a, a , a , a , a , b, ab, a b, a b, a b, a b}, and its 3 center, Z(D6) = {e, a }. Find the order of the element aZ(D6) in the factor group D6/Z(D6). 4 aZ(D6) = {a, a }. 2 2 2 5 (aZ(D6)) = a Z(D6) = {a , a }. 3 3 3 6 (aZ(D6)) = a Z(D6) = {a , a = e} = Z(D6). Thus |aZ(D6)| = 3. 10. Prove that a factor group of a cyclic group is cyclic. Let G be a cyclic group, say G =< a > and let H be a subgroup of G. Then H =< at >, where t ∈ Z (since every subgroup of a cyclic group is cyclic). Prove: G/H =< aH >. Let gH ∈ G/H. Then g ∈ G =⇒ g = as, for some s ∈ Z. Thus gH = asH = (aH)s. =⇒ gH ∈< aH >. 11. List all generators of the group Z/ < 9 >. Z/ < 9 >=< 1+ < 9 >> and |1+ < 9 > | = 9 . Now k+ < 9 >∈ Z/ < 9 > is a gen- erator of Z/ < 9 > ⇐⇒ gcd(k,9) = 1. Since 1 ≤ k ≤ 9, the generators of Z/ < 9 > are 1+ < 9 >, 2+ < 9 >, 4+ < 9 >, 5+ < 9 >, 7+ < 9 >, and 8+ < 9 >. 12. Let G =< a > be a cyclic group of order 30 and H =< a5 >. List all generators of the group G/H. G/H =< a + H > and |a + H| = 5 (see problem 7 above). Now ak + H ∈ G/H is a generator of G/H ⇐⇒ gcd(k,5) = 1. Since 1 ≤ k ≤ 5, the generators of G/H are a + H, a2 + H, a3 + H, and a4 + H. 2.
Load more

Recommended publications
  • On Abelian Subgroups of Finitely Generated Metabelian
    J. Group Theory 16 (2013), 695–705 DOI 10.1515/jgt-2013-0011 © de Gruyter 2013 On abelian subgroups of finitely generated metabelian groups Vahagn H. Mikaelian and Alexander Y. Olshanskii Communicated by John S. Wilson To Professor Gilbert Baumslag to his 80th birthday Abstract. In this note we introduce the class of H-groups (or Hall groups) related to the class of B-groups defined by P. Hall in the 1950s. Establishing some basic properties of Hall groups we use them to obtain results concerning embeddings of abelian groups. In particular, we give an explicit classification of all abelian groups that can occur as subgroups in finitely generated metabelian groups. Hall groups allow us to give a negative answer to G. Baumslag’s conjecture of 1990 on the cardinality of the set of isomorphism classes for abelian subgroups in finitely generated metabelian groups. 1 Introduction The subject of our note goes back to the paper of P. Hall [7], which established the properties of abelian normal subgroups in finitely generated metabelian and abelian-by-polycyclic groups. Let B be the class of all abelian groups B, where B is an abelian normal subgroup of some finitely generated group G with polycyclic quotient G=B. It is proved in [7, Lemmas 8 and 5.2] that B H, where the class H of countable abelian groups can be defined as follows (in the present paper, we will call the groups from H Hall groups). By definition, H H if 2 (1) H is a (finite or) countable abelian group, (2) H T K; where T is a bounded torsion group (i.e., the orders of all ele- D ˚ ments in T are bounded), K is torsion-free, (3) K has a free abelian subgroup F such that K=F is a torsion group with trivial p-subgroups for all primes except for the members of a finite set .K/.
    [Show full text]
  • On Finite Groups Whose Every Proper Normal Subgroup Is a Union
    Proc. Indian Acad. Sci. (Math. Sci.) Vol. 114, No. 3, August 2004, pp. 217–224. © Printed in India On finite groups whose every proper normal subgroup is a union of a given number of conjugacy classes ALI REZA ASHRAFI and GEETHA VENKATARAMAN∗ Department of Mathematics, University of Kashan, Kashan, Iran ∗Department of Mathematics and Mathematical Sciences Foundation, St. Stephen’s College, Delhi 110 007, India E-mail: ashrafi@kashanu.ac.ir; geetha [email protected] MS received 19 June 2002; revised 26 March 2004 Abstract. Let G be a finite group and A be a normal subgroup of G. We denote by ncc.A/ the number of G-conjugacy classes of A and A is called n-decomposable, if ncc.A/ = n. Set KG ={ncc.A/|A CG}. Let X be a non-empty subset of positive integers. A group G is called X-decomposable, if KG = X. Ashrafi and his co-authors [1–5] have characterized the X-decomposable non-perfect finite groups for X ={1;n} and n ≤ 10. In this paper, we continue this problem and investigate the structure of X-decomposable non-perfect finite groups, for X = {1; 2; 3}. We prove that such a group is isomorphic to Z6;D8;Q8;S4, SmallGroup(20, 3), SmallGroup(24, 3), where SmallGroup.m; n/ denotes the mth group of order n in the small group library of GAP [11]. Keywords. Finite group; n-decomposable subgroup; conjugacy class; X-decompo- sable group. 1. Introduction and preliminaries Let G be a finite group and let NG be the set of proper normal subgroups of G.
    [Show full text]
  • Classification of Finite Abelian Groups
    Math 317 C1 John Sullivan Spring 2003 Classification of Finite Abelian Groups (Notes based on an article by Navarro in the Amer. Math. Monthly, February 2003.) The fundamental theorem of finite abelian groups expresses any such group as a product of cyclic groups: Theorem. Suppose G is a finite abelian group. Then G is (in a unique way) a direct product of cyclic groups of order pk with p prime. Our first step will be a special case of Cauchy’s Theorem, which we will prove later for arbitrary groups: whenever p |G| then G has an element of order p. Theorem (Cauchy). If G is a finite group, and p |G| is a prime, then G has an element of order p (or, equivalently, a subgroup of order p). ∼ Proof when G is abelian. First note that if |G| is prime, then G = Zp and we are done. In general, we work by induction. If G has no nontrivial proper subgroups, it must be a prime cyclic group, the case we’ve already handled. So we can suppose there is a nontrivial subgroup H smaller than G. Either p |H| or p |G/H|. In the first case, by induction, H has an element of order p which is also order p in G so we’re done. In the second case, if ∼ g + H has order p in G/H then |g + H| |g|, so hgi = Zkp for some k, and then kg ∈ G has order p. Note that we write our abelian groups additively. Definition. Given a prime p, a p-group is a group in which every element has order pk for some k.
    [Show full text]
  • Mathematics 310 Examination 1 Answers 1. (10 Points) Let G Be A
    Mathematics 310 Examination 1 Answers 1. (10 points) Let G be a group, and let x be an element of G. Finish the following definition: The order of x is ... Answer: . the smallest positive integer n so that xn = e. 2. (10 points) State Lagrange’s Theorem. Answer: If G is a finite group, and H is a subgroup of G, then o(H)|o(G). 3. (10 points) Let ( a 0! ) H = : a, b ∈ Z, ab 6= 0 . 0 b Is H a group with the binary operation of matrix multiplication? Be sure to explain your answer fully. 2 0! 1/2 0 ! Answer: This is not a group. The inverse of the matrix is , which is not 0 2 0 1/2 in H. 4. (20 points) Suppose that G1 and G2 are groups, and φ : G1 → G2 is a homomorphism. (a) Recall that we defined φ(G1) = {φ(g1): g1 ∈ G1}. Show that φ(G1) is a subgroup of G2. −1 (b) Suppose that H2 is a subgroup of G2. Recall that we defined φ (H2) = {g1 ∈ G1 : −1 φ(g1) ∈ H2}. Prove that φ (H2) is a subgroup of G1. Answer:(a) Pick x, y ∈ φ(G1). Then we can write x = φ(a) and y = φ(b), with a, b ∈ G1. Because G1 is closed under the group operation, we know that ab ∈ G1. Because φ is a homomorphism, we know that xy = φ(a)φ(b) = φ(ab), and therefore xy ∈ φ(G1). That shows that φ(G1) is closed under the group operation.
    [Show full text]
  • The General Linear Group
    18.704 Gabe Cunningham 2/18/05 [email protected] The General Linear Group Definition: Let F be a field. Then the general linear group GLn(F ) is the group of invert- ible n × n matrices with entries in F under matrix multiplication. It is easy to see that GLn(F ) is, in fact, a group: matrix multiplication is associative; the identity element is In, the n × n matrix with 1’s along the main diagonal and 0’s everywhere else; and the matrices are invertible by choice. It’s not immediately clear whether GLn(F ) has infinitely many elements when F does. However, such is the case. Let a ∈ F , a 6= 0. −1 Then a · In is an invertible n × n matrix with inverse a · In. In fact, the set of all such × matrices forms a subgroup of GLn(F ) that is isomorphic to F = F \{0}. It is clear that if F is a finite field, then GLn(F ) has only finitely many elements. An interesting question to ask is how many elements it has. Before addressing that question fully, let’s look at some examples. ∼ × Example 1: Let n = 1. Then GLn(Fq) = Fq , which has q − 1 elements. a b Example 2: Let n = 2; let M = ( c d ). Then for M to be invertible, it is necessary and sufficient that ad 6= bc. If a, b, c, and d are all nonzero, then we can fix a, b, and c arbitrarily, and d can be anything but a−1bc. This gives us (q − 1)3(q − 2) matrices.
    [Show full text]
  • On Some Generation Methods of Finite Simple Groups
    Introduction Preliminaries Special Kind of Generation of Finite Simple Groups The Bibliography On Some Generation Methods of Finite Simple Groups Ayoub B. M. Basheer Department of Mathematical Sciences, North-West University (Mafikeng), P Bag X2046, Mmabatho 2735, South Africa Groups St Andrews 2017 in Birmingham, School of Mathematics, University of Birmingham, United Kingdom 11th of August 2017 Ayoub Basheer, North-West University, South Africa Groups St Andrews 2017 Talk in Birmingham Introduction Preliminaries Special Kind of Generation of Finite Simple Groups The Bibliography Abstract In this talk we consider some methods of generating finite simple groups with the focus on ranks of classes, (p; q; r)-generation and spread (exact) of finite simple groups. We show some examples of results that were established by the author and his supervisor, Professor J. Moori on generations of some finite simple groups. Ayoub Basheer, North-West University, South Africa Groups St Andrews 2017 Talk in Birmingham Introduction Preliminaries Special Kind of Generation of Finite Simple Groups The Bibliography Introduction Generation of finite groups by suitable subsets is of great interest and has many applications to groups and their representations. For example, Di Martino and et al. [39] established a useful connection between generation of groups by conjugate elements and the existence of elements representable by almost cyclic matrices. Their motivation was to study irreducible projective representations of the sporadic simple groups. In view of applications, it is often important to exhibit generating pairs of some special kind, such as generators carrying a geometric meaning, generators of some prescribed order, generators that offer an economical presentation of the group.
    [Show full text]
  • Math 412. Simple Groups
    Math 412. Simple Groups DEFINITION: A group G is simple if its only normal subgroups are feg and G. Simple groups are rare among all groups in the same way that prime numbers are rare among all integers. The smallest non-abelian group is A5, which has order 60. THEOREM 8.25: A abelian group is simple if and only if it is finite of prime order. THEOREM: The Alternating Groups An where n ≥ 5 are simple. The simple groups are the building blocks of all groups, in a sense similar to how all integers are built from the prime numbers. One of the greatest mathematical achievements of the Twentieth Century was a classification of all the finite simple groups. These are recorded in the Atlas of Simple Groups. The mathematician who discovered the last-to-be-discovered finite simple group is right here in our own department: Professor Bob Greiss. This simple group is called the monster group because its order is so big—approximately 8 × 1053. Because we have classified all the finite simple groups, and we know how to put them together to form arbitrary groups, we essentially understand the structure of every finite group. It is difficult, in general, to tell whether a given group G is simple or not. Just like determining whether a given (large) integer is prime, there is an algorithm to check but it may take an unreasonable amount of time to run. A. WARM UP. Find proper non-trivial normal subgroups of the following groups: Z, Z35, GL5(Q), S17, D100.
    [Show full text]
  • Janko's Sporadic Simple Groups
    Janko’s Sporadic Simple Groups: a bit of history Algebra, Geometry and Computation CARMA, Workshop on Mathematics and Computation Terry Gagen and Don Taylor The University of Sydney 20 June 2015 Fifty years ago: the discovery In January 1965, a surprising announcement was communicated to the international mathematical community. Zvonimir Janko, working as a Research Fellow at the Institute of Advanced Study within the Australian National University had constructed a new sporadic simple group. Before 1965 only five sporadic simple groups were known. They had been discovered almost exactly one hundred years prior (1861 and 1873) by Émile Mathieu but the proof of their simplicity was only obtained in 1900 by G. A. Miller. Finite simple groups: earliest examples É The cyclic groups Zp of prime order and the alternating groups Alt(n) of even permutations of n 5 items were the earliest simple groups to be studied (Gauss,≥ Euler, Abel, etc.) É Evariste Galois knew about PSL(2,p) and wrote about them in his letter to Chevalier in 1832 on the night before the duel. É Camille Jordan (Traité des substitutions et des équations algébriques,1870) wrote about linear groups defined over finite fields of prime order and determined their composition factors. The ‘groupes abéliens’ of Jordan are now called symplectic groups and his ‘groupes hypoabéliens’ are orthogonal groups in characteristic 2. É Émile Mathieu introduced the five groups M11, M12, M22, M23 and M24 in 1861 and 1873. The classical groups, G2 and E6 É In his PhD thesis Leonard Eugene Dickson extended Jordan’s work to linear groups over all finite fields and included the unitary groups.
    [Show full text]
  • Arxiv:Hep-Th/0510191V1 22 Oct 2005
    Maximal Subgroups of the Coxeter Group W (H4) and Quaternions Mehmet Koca,∗ Muataz Al-Barwani,† and Shadia Al-Farsi‡ Department of Physics, College of Science, Sultan Qaboos University, PO Box 36, Al-Khod 123, Muscat, Sultanate of Oman Ramazan Ko¸c§ Department of Physics, Faculty of Engineering University of Gaziantep, 27310 Gaziantep, Turkey (Dated: September 8, 2018) The largest finite subgroup of O(4) is the noncrystallographic Coxeter group W (H4) of order 14400. Its derived subgroup is the largest finite subgroup W (H4)/Z2 of SO(4) of order 7200. Moreover, up to conjugacy, it has five non-normal maximal subgroups of orders 144, two 240, 400 and 576. Two groups [W (H2) × W (H2)]×Z4 and W (H3)×Z2 possess noncrystallographic structures with orders 400 and 240 respectively. The groups of orders 144, 240 and 576 are the extensions of the Weyl groups of the root systems of SU(3)×SU(3), SU(5) and SO(8) respectively. We represent the maximal subgroups of W (H4) with sets of quaternion pairs acting on the quaternionic root systems. PACS numbers: 02.20.Bb INTRODUCTION The noncrystallographic Coxeter group W (H4) of order 14400 generates some interests [1] for its relevance to the quasicrystallographic structures in condensed matter physics [2, 3] as well as its unique relation with the E8 gauge symmetry associated with the heterotic superstring theory [4]. The Coxeter group W (H4) [5] is the maximal finite subgroup of O(4), the finite subgroups of which have been classified by du Val [6] and by Conway and Smith [7].
    [Show full text]
  • Orders on Computable Torsion-Free Abelian Groups
    Orders on Computable Torsion-Free Abelian Groups Asher M. Kach (Joint Work with Karen Lange and Reed Solomon) University of Chicago 12th Asian Logic Conference Victoria University of Wellington December 2011 Asher M. Kach (U of C) Orders on Computable TFAGs ALC 2011 1 / 24 Outline 1 Classical Algebra Background 2 Computing a Basis 3 Computing an Order With A Basis Without A Basis 4 Open Questions Asher M. Kach (U of C) Orders on Computable TFAGs ALC 2011 2 / 24 Torsion-Free Abelian Groups Remark Disclaimer: Hereout, the word group will always refer to a countable torsion-free abelian group. The words computable group will always refer to a (fixed) computable presentation. Definition A group G = (G : +; 0) is torsion-free if non-zero multiples of non-zero elements are non-zero, i.e., if (8x 2 G)(8n 2 !)[x 6= 0 ^ n 6= 0 =) nx 6= 0] : Asher M. Kach (U of C) Orders on Computable TFAGs ALC 2011 3 / 24 Rank Theorem A countable abelian group is torsion-free if and only if it is a subgroup ! of Q . Definition The rank of a countable torsion-free abelian group G is the least κ cardinal κ such that G is a subgroup of Q . Asher M. Kach (U of C) Orders on Computable TFAGs ALC 2011 4 / 24 Example The subgroup H of Q ⊕ Q (viewed as having generators b1 and b2) b1+b2 generated by b1, b2, and 2 b1+b2 So elements of H look like β1b1 + β2b2 + α 2 for β1; β2; α 2 Z.
    [Show full text]
  • 324 COHOMOLOGY of DIHEDRAL GROUPS of ORDER 2P Let D Be A
    324 PHYLLIS GRAHAM [March 2. P. M. Cohn, Unique factorization in noncommutative power series rings, Proc. Cambridge Philos. Soc 58 (1962), 452-464. 3. Y. Kawada, Cohomology of group extensions, J. Fac. Sci. Univ. Tokyo Sect. I 9 (1963), 417-431. 4. J.-P. Serre, Cohomologie Galoisienne, Lecture Notes in Mathematics No. 5, Springer, Berlin, 1964, 5. , Corps locaux et isogênies, Séminaire Bourbaki, 1958/1959, Exposé 185 6. , Corps locaux, Hermann, Paris, 1962. UNIVERSITY OF MICHIGAN COHOMOLOGY OF DIHEDRAL GROUPS OF ORDER 2p BY PHYLLIS GRAHAM Communicated by G. Whaples, November 29, 1965 Let D be a dihedral group of order 2p, where p is an odd prime. D is generated by the elements a and ]3 with the relations ap~fi2 = 1 and (ial3 = or1. Let A be the subgroup of D generated by a, and let p-1 Aoy Ai, • • • , Ap-\ be the subgroups generated by j8, aft, • * • , a /3, respectively. Let M be any D-module. Then the cohomology groups n n H (A0, M) and H (Aif M), i=l, 2, • • • , p — 1 are isomorphic for 1 l every integer n, so the eight groups H~ (Df Af), H°(D, M), H (D, jfef), 2 1 Q # (A Af), ff" ^. M), H (A, M), H-Wo, M), and H»(A0, M) deter­ mine all cohomology groups of M with respect to D and to all of its subgroups. We have found what values this array takes on as M runs through all finitely generated -D-modules. All possibilities for the first four members of this array are deter­ mined as a special case of the results of Yang [4].
    [Show full text]
  • Dihedral Groups Ii
    DIHEDRAL GROUPS II KEITH CONRAD We will characterize dihedral groups in terms of generators and relations, and describe the subgroups of Dn, including the normal subgroups. We will also introduce an infinite group that resembles the dihedral groups and has all of them as quotient groups. 1. Abstract characterization of Dn −1 −1 The group Dn has two generators r and s with orders n and 2 such that srs = r . We will show every group with a pair of generators having properties similar to r and s admits a homomorphism onto it from Dn, and is isomorphic to Dn if it has the same size as Dn. Theorem 1.1. Let G be generated by elements x and y where xn = 1 for some n ≥ 3, 2 −1 −1 y = 1, and yxy = x . There is a surjective homomorphism Dn ! G, and if G has order 2n then this homomorphism is an isomorphism. The hypotheses xn = 1 and y2 = 1 do not mean x has order n and y has order 2, but only that their orders divide n and divide 2. For instance, the trivial group has the form hx; yi where xn = 1, y2 = 1, and yxy−1 = x−1 (take x and y to be the identity). Proof. The equation yxy−1 = x−1 implies yxjy−1 = x−j for all j 2 Z (raise both sides to the jth power). Since y2 = 1, we have for all k 2 Z k ykxjy−k = x(−1) j by considering even and odd k separately. Thus k (1.1) ykxj = x(−1) jyk: This shows each product of x's and y's can have all the x's brought to the left and all the y's brought to the right.
    [Show full text]