Here
... The second of these is easy — acb0 d0 = ab0 cd0 = a0 bcd0 = a0 bc0 d = a0 c0 bd. For the first, we have that adb0 d0 = ab0 dd0 = a0 bdd0 = a0 d0 bd, and bcb0 d0 = bb0 cd0 = bb0 c0 d = b0 c0 bd, and adding these gives the required equation. 17. State and prove the factor theorem for the polynomial r ...
... The second of these is easy — acb0 d0 = ab0 cd0 = a0 bcd0 = a0 bc0 d = a0 c0 bd. For the first, we have that adb0 d0 = ab0 dd0 = a0 bdd0 = a0 d0 bd, and bcb0 d0 = bb0 cd0 = bb0 c0 d = b0 c0 bd, and adding these gives the required equation. 17. State and prove the factor theorem for the polynomial r ...
Field Theory
... integers which are equivalent to a modulo n. The set of such integers is written as [a] where [a] := a + nZ = {. . . , a − 2n, a − n, a, a + n, a + 2n, . . .} For example, if a = 12 and n = 7, then the congruence class for 12 modulo 7 would be the set {. . . , −2, 5, 12, 19, 26, . . .}. This means t ...
... integers which are equivalent to a modulo n. The set of such integers is written as [a] where [a] := a + nZ = {. . . , a − 2n, a − n, a, a + n, a + 2n, . . .} For example, if a = 12 and n = 7, then the congruence class for 12 modulo 7 would be the set {. . . , −2, 5, 12, 19, 26, . . .}. This means t ...
Solutions
... 1. The kernel always contains the identity element e, so its non-empty. Let a, b be elements in the kernel. We have that ab−1 is in the kernel since Φ(ab−1 ) = Φ(a)Φ(b)−1 = e · e−1 = e. Therefore the kernel is a subgroup. We need to check normality. Let x be any element of G. The element xax−1 is ma ...
... 1. The kernel always contains the identity element e, so its non-empty. Let a, b be elements in the kernel. We have that ab−1 is in the kernel since Φ(ab−1 ) = Φ(a)Φ(b)−1 = e · e−1 = e. Therefore the kernel is a subgroup. We need to check normality. Let x be any element of G. The element xax−1 is ma ...
Degrees of irreducible polynomials over binary field
... Let G be the Berlekamp matrix with respect to the polynomial f (x) of degree m over F. It is not difficult to show that Gm = Im if and only if f (x) has no square factors, where Im denotes the identity matrix of order m. If f (x) has no square factors then the order o(f (x)) = o(G) of f (x) is defin ...
... Let G be the Berlekamp matrix with respect to the polynomial f (x) of degree m over F. It is not difficult to show that Gm = Im if and only if f (x) has no square factors, where Im denotes the identity matrix of order m. If f (x) has no square factors then the order o(f (x)) = o(G) of f (x) is defin ...
Physics 129B, Winter 2010 Problem Set 1 Solution
... D6 , the dihedral group of an equilateral triangle, is a finite group of order 6 that is not cyclic. So we expect it to be isomorphic to S3 , and the isomorphism is evident when we assign a letter (a, b, or c) to each vertex of the equilateral triangle. Then a rotation by 2π/3 corresponds to a permu ...
... D6 , the dihedral group of an equilateral triangle, is a finite group of order 6 that is not cyclic. So we expect it to be isomorphic to S3 , and the isomorphism is evident when we assign a letter (a, b, or c) to each vertex of the equilateral triangle. Then a rotation by 2π/3 corresponds to a permu ...
(Less) Abstract Algebra
... Definition: For a set G, a binary operation on G is a map ∗ : G × G → G. We say ∗ is associative if ∗(a, ∗(b, c)) = ∗(∗(a, b), c) for all a, b, c ∈ G. If, further there is an element 1 such that ∗(1, a) = ∗(a, 1) = a for every a ∈ G and for each a ∈ G there is an element a−1 such that ∗(a, a−1 ) = ∗ ...
... Definition: For a set G, a binary operation on G is a map ∗ : G × G → G. We say ∗ is associative if ∗(a, ∗(b, c)) = ∗(∗(a, b), c) for all a, b, c ∈ G. If, further there is an element 1 such that ∗(1, a) = ∗(a, 1) = a for every a ∈ G and for each a ∈ G there is an element a−1 such that ∗(a, a−1 ) = ∗ ...
Math 562 Spring 2012 Homework 4 Drew Armstrong
... (because every spanning set contains a basis). We conclude that [K : Q] = 4, and it follows (for general reasons, not yet proved in class) that |Gal(K/Q)| = 4. What could this group be? Recall that there are only two groups of size 4; they are isomorphic to Z/(4) and V := Z/(2) × Z/(2). To prove (c ...
... (because every spanning set contains a basis). We conclude that [K : Q] = 4, and it follows (for general reasons, not yet proved in class) that |Gal(K/Q)| = 4. What could this group be? Recall that there are only two groups of size 4; they are isomorphic to Z/(4) and V := Z/(2) × Z/(2). To prove (c ...
Exam 1 – 02/29/12 SOLUTIONS
... 24. The elements are the equivalence classes [0], [1], . . . , [23]. (b) [5] Does this group have any subgroups of order 7? Why or why not? No. If it did, then Lagrange’s theorem would imply that 7 divides 24, contradiction. (c) [5] Find all elements of order 6 (list them). Note that Z/24Z is a cycl ...
... 24. The elements are the equivalence classes [0], [1], . . . , [23]. (b) [5] Does this group have any subgroups of order 7? Why or why not? No. If it did, then Lagrange’s theorem would imply that 7 divides 24, contradiction. (c) [5] Find all elements of order 6 (list them). Note that Z/24Z is a cycl ...
Final with solutions
... Proof: The group Aut(G) is a subgroup of the group SG of all bijections G → G. This last is isomorphic to Sn , by the argument in the proof of Cayley’s Theorem and its corollary. Therefore, Aut(G) is isomorphic to a subgroup H of Sn . But |Sn | = n!. Therefore, by Lagrange’s Theorem, |Aut(G)| = |H| ...
... Proof: The group Aut(G) is a subgroup of the group SG of all bijections G → G. This last is isomorphic to Sn , by the argument in the proof of Cayley’s Theorem and its corollary. Therefore, Aut(G) is isomorphic to a subgroup H of Sn . But |Sn | = n!. Therefore, by Lagrange’s Theorem, |Aut(G)| = |H| ...
Optimal normal bases Shuhong Gao and Hendrik W. Lenstra, Jr. Let
... and they formulate a conjecture that describes all finite extensions of the field of two elements that admit an optimal normal basis. In [1] this conjecture is extended to all finite fields. In the present paper we confirm the conjecture, and we show that the constructions given in [2] exhaust all o ...
... and they formulate a conjecture that describes all finite extensions of the field of two elements that admit an optimal normal basis. In [1] this conjecture is extended to all finite fields. In the present paper we confirm the conjecture, and we show that the constructions given in [2] exhaust all o ...
Order (group theory)
... The following partial converse is true for finite groups: if d divides the order of a group G and d is a prime number, then there exists an element of order d in G (this is sometimes called Cauchy's theorem). The statement does not hold for composite orders, e.g. the Klein four-group does not have a ...
... The following partial converse is true for finite groups: if d divides the order of a group G and d is a prime number, then there exists an element of order d in G (this is sometimes called Cauchy's theorem). The statement does not hold for composite orders, e.g. the Klein four-group does not have a ...
PDF
... And therefore, the kernel of this map ψ is a subgroup of F?pd of all elements b such that bm = 1. The other coset of this kernel is the set of elements b such that bM = −1. Since these two are cosets, they are of equal size. Hence a randomly chosen b will have bM = 1 with probability 1/2. And theref ...
... And therefore, the kernel of this map ψ is a subgroup of F?pd of all elements b such that bm = 1. The other coset of this kernel is the set of elements b such that bM = −1. Since these two are cosets, they are of equal size. Hence a randomly chosen b will have bM = 1 with probability 1/2. And theref ...
Solutions to Quiz 4
... 2. (4 points) Show that the automorphism group Aut(Z10 ) is isomorphic to a cyclic group Zn . What is n? Aut(Z10 ) ∼ = U (10) ∼ = Z4 3. (6 points) Show that the following pairs of groups are not isomorphic. In each case, explain why. (a) U (12) and Z4 . U (12) is not cyclic, since |U (12)| = 4, but ...
... 2. (4 points) Show that the automorphism group Aut(Z10 ) is isomorphic to a cyclic group Zn . What is n? Aut(Z10 ) ∼ = U (10) ∼ = Z4 3. (6 points) Show that the following pairs of groups are not isomorphic. In each case, explain why. (a) U (12) and Z4 . U (12) is not cyclic, since |U (12)| = 4, but ...
on the defining field of a divisor in an algebraic variety1 797
... U there is a smallest one which is contained in all of them, which we shall call the defining field of the variety U. A d-cycle G in a variety U of dimension r is a finite set of simple subvarieties of dimension d in U, to each of which is assigned an integer called its multiplicity; a cycle is call ...
... U there is a smallest one which is contained in all of them, which we shall call the defining field of the variety U. A d-cycle G in a variety U of dimension r is a finite set of simple subvarieties of dimension d in U, to each of which is assigned an integer called its multiplicity; a cycle is call ...
AN EXTENSION OF YAMAMOTO`S THEOREM
... Our main result is to establish (1.5) in the context of real semisimple Lie groups. 2. Extension of Yamamoto’s theorem Let g = k + p be a fixed Cartan decomposition of a real semisimple Lie algebra g and let G be any connected Lie group having g as its Lie algebra. Let K ⊂ G be the subgroup with Lie ...
... Our main result is to establish (1.5) in the context of real semisimple Lie groups. 2. Extension of Yamamoto’s theorem Let g = k + p be a fixed Cartan decomposition of a real semisimple Lie algebra g and let G be any connected Lie group having g as its Lie algebra. Let K ⊂ G be the subgroup with Lie ...
OPERADS IN ALGEBRAIC TOPOLOGY II Contents The little
... you weren’t very topological. Today we’re going to get into some topology. The little n-disks operad Goal. We’ll consider how to interpolate “up to homotopy” between As and Com. There is ...
... you weren’t very topological. Today we’re going to get into some topology. The little n-disks operad Goal. We’ll consider how to interpolate “up to homotopy” between As and Com. There is ...
PH Kropholler Olympia Talelli
... Proof of the Proposition. For any set &? let [a, z] denote the set of functions a-, Z which take only finitely many distinct values. This is a subgroup of the additive group of all functions from 0 to E. We define 2 to be [G, m]. Then I is a ZG-submodule of the coinduced module CoindyH and contains ...
... Proof of the Proposition. For any set &? let [a, z] denote the set of functions a-, Z which take only finitely many distinct values. This is a subgroup of the additive group of all functions from 0 to E. We define 2 to be [G, m]. Then I is a ZG-submodule of the coinduced module CoindyH and contains ...
1. Prove that the following are all equal to the radical • The union of
... such that rv1 = w1 and rv2 = w2 . Show that R equals Endk (V ). A proof from first principles should be easy to give, but here is a “high level” proof. Since R is 2-transitive, it follows from an observation in §3 of the notes that EndR (V ) = k. But V is also irreducible for R (1transitivity is eno ...
... such that rv1 = w1 and rv2 = w2 . Show that R equals Endk (V ). A proof from first principles should be easy to give, but here is a “high level” proof. Since R is 2-transitive, it follows from an observation in §3 of the notes that EndR (V ) = k. But V is also irreducible for R (1transitivity is eno ...
Lecture 20 1 Point Set Topology
... Theorem. Let φ : A → B be a homomorphism of finitely generated algebras and let f : Y → X be a morphism of their varieties. Then the image of f (Y ) is a constructible set of X. Proof. We will use Noetherian (acc) induction on B or dcc on varieties of Y , this will be the same as induction on dimens ...
... Theorem. Let φ : A → B be a homomorphism of finitely generated algebras and let f : Y → X be a morphism of their varieties. Then the image of f (Y ) is a constructible set of X. Proof. We will use Noetherian (acc) induction on B or dcc on varieties of Y , this will be the same as induction on dimens ...
Section 6: Direct Products One way to build a new groups from two
... One way to build a new groups from two old ones, say G, H, is to take the set G × H of ordered pairs of elements, the first from G and the second from H, and give them the obvious operation: G’s operation on the first coordinates and H’s on the second. This group is called the direct product of G an ...
... One way to build a new groups from two old ones, say G, H, is to take the set G × H of ordered pairs of elements, the first from G and the second from H, and give them the obvious operation: G’s operation on the first coordinates and H’s on the second. This group is called the direct product of G an ...
Garrett 11-04-2011 1 Recap: A better version of localization...
... Claim: The inclusion o/p → OP /Q is an isomorphism. Claim: κ̃ = O/P is normal over κ = o/p, and GP surjects to Aut(κ̃/κ). More named objects: The inertia group: IP is the kernel of GP → Gal(κ̃/κ). The fixed field of IP is the inertia subfield of K. These will not be used much here. p splits complete ...
... Claim: The inclusion o/p → OP /Q is an isomorphism. Claim: κ̃ = O/P is normal over κ = o/p, and GP surjects to Aut(κ̃/κ). More named objects: The inertia group: IP is the kernel of GP → Gal(κ̃/κ). The fixed field of IP is the inertia subfield of K. These will not be used much here. p splits complete ...
CONJUGATION IN A GROUP 1. Introduction A reflection across one
... The conjugates of (12) are in the second row: (12), (13), and (23). Notice the redundancy in the table: each conjugate of (12) arises in two ways. We will see in Theorem 4.4 that in Sn any two transpositions are conjugate. In Appendix A is a proof that the reflections across any two lines in the pla ...
... The conjugates of (12) are in the second row: (12), (13), and (23). Notice the redundancy in the table: each conjugate of (12) arises in two ways. We will see in Theorem 4.4 that in Sn any two transpositions are conjugate. In Appendix A is a proof that the reflections across any two lines in the pla ...
CONJUGATION IN A GROUP 1. Introduction
... The conjugates of (12) are in the second row: (12), (13), and (23). Notice the redundancy in the table: each conjugate of (12) arises in two ways. We will see in Theorem 4.4 that in Sn any two transpositions are conjugate. In Appendix A is a proof that the reflections across any two lines in the pla ...
... The conjugates of (12) are in the second row: (12), (13), and (23). Notice the redundancy in the table: each conjugate of (12) arises in two ways. We will see in Theorem 4.4 that in Sn any two transpositions are conjugate. In Appendix A is a proof that the reflections across any two lines in the pla ...
Representations of su(2) 1 Lie and linear groups
... structure, such that the group operations of multiplication and inversion are differentiable, and that the Lie algebra of a Lie group is the tangent space to the group at the identity. The most commonly used Lie groups are matrix groups, and we will focus on these, disregarding the manifold structur ...
... structure, such that the group operations of multiplication and inversion are differentiable, and that the Lie algebra of a Lie group is the tangent space to the group at the identity. The most commonly used Lie groups are matrix groups, and we will focus on these, disregarding the manifold structur ...
Maximal Elements of Weakly Continuous Relations
... The relation > is said to be lower continuous (lc) if for each I E X the set [ .r E X / x > y } is open, and is said to be weakly lower continuous (wlc) if whenever .X> y there is a neighborhood of y, denoted N(y), which satisfies I 2 N(y), by which notation we mean that s 2 z for every z E N( .Y). ...
... The relation > is said to be lower continuous (lc) if for each I E X the set [ .r E X / x > y } is open, and is said to be weakly lower continuous (wlc) if whenever .X> y there is a neighborhood of y, denoted N(y), which satisfies I 2 N(y), by which notation we mean that s 2 z for every z E N( .Y). ...