MATH 103B Homework 6 - Solutions Due May 17, 2013
... the ring is generated by some element in the ring. Proofs: I. We have proved that Z is an integral domain (cf. Chapter 13, example 1). Let A be an ideal of Z. If A “ t0u then A “ x0y so it is principal. Otherwise, if x P A is negative then 0´x P A as well (by closure under subtraction and since 0 P ...
... the ring is generated by some element in the ring. Proofs: I. We have proved that Z is an integral domain (cf. Chapter 13, example 1). Let A be an ideal of Z. If A “ t0u then A “ x0y so it is principal. Otherwise, if x P A is negative then 0´x P A as well (by closure under subtraction and since 0 P ...
Chapter 2
... These are the only two possible groups for this order. Why? Elements that are not their own inverses must come in at least pairs. Since E is always its own inverse, this leaves 3 other elements. The two groups just developed cover 2 of 3 possibilities. The third possibility has A-1 = B, B-1 = C, and ...
... These are the only two possible groups for this order. Why? Elements that are not their own inverses must come in at least pairs. Since E is always its own inverse, this leaves 3 other elements. The two groups just developed cover 2 of 3 possibilities. The third possibility has A-1 = B, B-1 = C, and ...
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... • Generalizing the last two examples, if R is a ring with multiplicative identity 1, then the units of R (the elements invertible with respect to multiplication) form a group with respect to ring multiplication and with identity element 1. See examples of rings. Most groups of numbers carry natural ...
... • Generalizing the last two examples, if R is a ring with multiplicative identity 1, then the units of R (the elements invertible with respect to multiplication) form a group with respect to ring multiplication and with identity element 1. See examples of rings. Most groups of numbers carry natural ...
1 Principal Ideal Domains
... Theorem 1. Let R be a PID. Let a, b ∈ R be non-zero elements and let d be an element such that (d) = (a, b). Then: a.) d is a GCD of a and b. b.) d = ax + by for some x, y ∈ R. c.) d is unique up to multiplication by a unit in R We’ve already proved all of these - the only thing we needed the Euclid ...
... Theorem 1. Let R be a PID. Let a, b ∈ R be non-zero elements and let d be an element such that (d) = (a, b). Then: a.) d is a GCD of a and b. b.) d = ax + by for some x, y ∈ R. c.) d is unique up to multiplication by a unit in R We’ve already proved all of these - the only thing we needed the Euclid ...
Prelim 2 with solutions
... A couple of comments on Problem 4. Some students mistakenly thought that the conjugacy class C(g) is the same as the subgroup H. If you think about the definition of both, you will see that they are far from the same. H consist of all x that commute with g, i.e., xgx−1 = g. The conjugacy class of g ...
... A couple of comments on Problem 4. Some students mistakenly thought that the conjugacy class C(g) is the same as the subgroup H. If you think about the definition of both, you will see that they are far from the same. H consist of all x that commute with g, i.e., xgx−1 = g. The conjugacy class of g ...
MATH882201 – Problem Set I (1) Let I be a directed set and {G i}i∈I
... (a) Show that G is a closed subset of π. (b) Give G the subspace topology. Show that G is compact and totally disconnected for this topology. (c) Prove that the natural projection maps φi : G → Gi are continuous, and that the (open) subgroups Ki := ker φi for a basis of open neighborhoods of the ide ...
... (a) Show that G is a closed subset of π. (b) Give G the subspace topology. Show that G is compact and totally disconnected for this topology. (c) Prove that the natural projection maps φi : G → Gi are continuous, and that the (open) subgroups Ki := ker φi for a basis of open neighborhoods of the ide ...
LECTURES MATH370-08C 1. Groups 1.1. Abstract groups versus
... x ∈ X are called left H-cosets in G. It is clear (is it?) that they are disjoint and cover the whole group G. The set of all left H-cosets in G is denoted by G/H. (For the set of all right cosets - define them yourself - the notation H\G is used). Lemma 1.1. a) Let gx be any representative of a left ...
... x ∈ X are called left H-cosets in G. It is clear (is it?) that they are disjoint and cover the whole group G. The set of all left H-cosets in G is denoted by G/H. (For the set of all right cosets - define them yourself - the notation H\G is used). Lemma 1.1. a) Let gx be any representative of a left ...
OPEN PROBLEM SESSION FROM THE CONFERENCE
... Is it true if we further suppose that G is rational? Probably one needs the local domain to be regular. The motivation for this problem is that one knows the answer is yes in a similar situation, namely when F is a number field and G is semisimple simply connected. Known cases. (1) Known when all re ...
... Is it true if we further suppose that G is rational? Probably one needs the local domain to be regular. The motivation for this problem is that one knows the answer is yes in a similar situation, namely when F is a number field and G is semisimple simply connected. Known cases. (1) Known when all re ...
Math. 5363, exam 1, solutions 1. Prove that every finitely generated
... Let G be a non-abelian group of order 6. Since G is not abelian, it does not contain any element of order 6. Also, it can’t happen that every element other than 1 is of order 2. Therefore, there is element a ∈ G of order 3. This element generates the subgroup H = {1, a, a2 } ⊆ G of index 2. In parti ...
... Let G be a non-abelian group of order 6. Since G is not abelian, it does not contain any element of order 6. Also, it can’t happen that every element other than 1 is of order 2. Therefore, there is element a ∈ G of order 3. This element generates the subgroup H = {1, a, a2 } ⊆ G of index 2. In parti ...
THE NUMBER OF LATTICE POINTS IN ALCOVES AND THE
... reader who is not familiar with semisimple Lie theory may hop to the beginning of the next paragraph. Let g be a simple complex Lie algebra whose Weyl group is W . (It doesn’t matter whether we take g = so2l+1 (C) or g = sp2l (C) if W is hyperoctahedral.) We choose a Cartan subalgebra h and a Borel ...
... reader who is not familiar with semisimple Lie theory may hop to the beginning of the next paragraph. Let g be a simple complex Lie algebra whose Weyl group is W . (It doesn’t matter whether we take g = so2l+1 (C) or g = sp2l (C) if W is hyperoctahedral.) We choose a Cartan subalgebra h and a Borel ...
Math 75 NOTES on finite fields C. Pomerance Suppose F is a finite
... so K1 = K2 . (Another, equivalent way to see this: count roots of xq − x in L. It has at most q j roots, since no polynomial over a field can have more roots than its degree. But it has q j roots coming from the field K1 , and it has q j roots coming from the field K2 , so these sets of roots must b ...
... so K1 = K2 . (Another, equivalent way to see this: count roots of xq − x in L. It has at most q j roots, since no polynomial over a field can have more roots than its degree. But it has q j roots coming from the field K1 , and it has q j roots coming from the field K2 , so these sets of roots must b ...
Full text
... The first is the unique minimal representation; the last is the unique maximal representation. The others show that representations of any intermediate length need not be unique. It is easy to show that only numbers of the form Fn - 1 have a unique Zeckendorf representation (i.e., one that is maxima ...
... The first is the unique minimal representation; the last is the unique maximal representation. The others show that representations of any intermediate length need not be unique. It is easy to show that only numbers of the form Fn - 1 have a unique Zeckendorf representation (i.e., one that is maxima ...
09 finite fields - Math User Home Pages
... factor of x, all elements of L are roots of xp − x = 0. Thus, with L sitting inside the fixed algebraic closure E of Fp , since a degree pn equation has at most pn roots in E, the elements of L must be just the field K constructed earlier. [5] This proves uniqueness (up to isomorphism). [6] Inside a ...
... factor of x, all elements of L are roots of xp − x = 0. Thus, with L sitting inside the fixed algebraic closure E of Fp , since a degree pn equation has at most pn roots in E, the elements of L must be just the field K constructed earlier. [5] This proves uniqueness (up to isomorphism). [6] Inside a ...
Math330 Fall 2008 7.34 Let g be a non
... g is 2 then we are done. If |g| = 12, 6, 4 then |g 6 | = 2, or |g 3 | = 2, or |g 2 | = 2. So the only way for there not to be an element of order two in G is if all non-identity elements have order 3. Let’s assume that this is true, and get a contradiction. Assume that all 11 non-identity elements o ...
... g is 2 then we are done. If |g| = 12, 6, 4 then |g 6 | = 2, or |g 3 | = 2, or |g 2 | = 2. So the only way for there not to be an element of order two in G is if all non-identity elements have order 3. Let’s assume that this is true, and get a contradiction. Assume that all 11 non-identity elements o ...
14. Isomorphism Theorem This section contain the important
... Similarly, the subalgebra L− generated by all L−α for simple α contains all Lβ for β ∈ Φ− . So the subalgebra L" of L generated by all Lα , L−α contains Lβ for all β ∈ Φ. On the other hand hα ∈ [Lα , L−α ]. So, L" also contains all hα . But the hα generate H since the α are a basis for H ∗ . So, L" ...
... Similarly, the subalgebra L− generated by all L−α for simple α contains all Lβ for β ∈ Φ− . So the subalgebra L" of L generated by all Lα , L−α contains Lβ for all β ∈ Φ. On the other hand hα ∈ [Lα , L−α ]. So, L" also contains all hα . But the hα generate H since the α are a basis for H ∗ . So, L" ...
I can recognize and make equal forms of fractions, decimals, and
... I can apply and understand place values from the millions to the thousandths and place these numbers on a number line. ...
... I can apply and understand place values from the millions to the thousandths and place these numbers on a number line. ...
Group representation theory
... This course is a mix of group theory and linear algebra, with probably more of the latter than the former. You may need to revise your 2nd year vector space notes! In mathematics the word “representation” basically means “structure-preserving function”. Thus—in group theory and ring theory at least— ...
... This course is a mix of group theory and linear algebra, with probably more of the latter than the former. You may need to revise your 2nd year vector space notes! In mathematics the word “representation” basically means “structure-preserving function”. Thus—in group theory and ring theory at least— ...
Polynomials over finite fields
... Theorem 1.1 The cardinality of F is pn where n = [F : Fp] and Fp denotes the prime subfield of F. Proof. The prime subfield Fp of F is isomorphic to the field Z/pZ of integers mod p. Since the field F is an n-dimensional vector space over Fp for some finite n, it is set-isomorphic to Fpn and thus ha ...
... Theorem 1.1 The cardinality of F is pn where n = [F : Fp] and Fp denotes the prime subfield of F. Proof. The prime subfield Fp of F is isomorphic to the field Z/pZ of integers mod p. Since the field F is an n-dimensional vector space over Fp for some finite n, it is set-isomorphic to Fpn and thus ha ...
Complex projective space The complex projective space CPn is the
... compact complex manifold. By definition, CPn is the set of lines in Cn+1 or, equivalently, CPn := (Cn+1 \{0})/C∗, where C∗ acts by multiplication on Cn+1 . The points of CPn are written as (z0 , z1 , ..., zn ). Here, the notation intends to indicate that for λ ∈ C∗ the two points (λz0 , λz1 , ..., λ ...
... compact complex manifold. By definition, CPn is the set of lines in Cn+1 or, equivalently, CPn := (Cn+1 \{0})/C∗, where C∗ acts by multiplication on Cn+1 . The points of CPn are written as (z0 , z1 , ..., zn ). Here, the notation intends to indicate that for λ ∈ C∗ the two points (λz0 , λz1 , ..., λ ...
Take home portion
... 9. Find examples of the following OR explain why no example exists. a. non-cyclic abelian group b. element of order 6 in Z 18 c. element of order 3 in Z 35 d. non-cyclic subgroup of Z 15 e. a cyclic group isomorphic to Z 3 Z 17 f. non-cyclic finite group g. cycle of length 6 in S 4 h. cycle of len ...
... 9. Find examples of the following OR explain why no example exists. a. non-cyclic abelian group b. element of order 6 in Z 18 c. element of order 3 in Z 35 d. non-cyclic subgroup of Z 15 e. a cyclic group isomorphic to Z 3 Z 17 f. non-cyclic finite group g. cycle of length 6 in S 4 h. cycle of len ...
[10.1]
... [10.1] Prove that a finite division ring D (a not-necessarily commutative ring with 1 in which any non-zero element has a multiplicative inverse) is commutative. (This is due to Wedderburn.) (Hint: Check that the center k of D is a field, say of cardinality q. Let D× act on D by conjugation, namely ...
... [10.1] Prove that a finite division ring D (a not-necessarily commutative ring with 1 in which any non-zero element has a multiplicative inverse) is commutative. (This is due to Wedderburn.) (Hint: Check that the center k of D is a field, say of cardinality q. Let D× act on D by conjugation, namely ...
adobe pdf - people.bath.ac.uk
... (b) Suppose that G is finite of order pn where n ≥ 1, Use part (a) to show that |Z(G)| ≡ 0 mod p and deduce that |Z(G)| > 1. 2. Let G be a group of order p2 where p is a prime number. (a) Show that if G is non-abelian, we must have |Z(G)| = p and g p = 1 for every g ∈ G. (b) Suppose that G is non-ab ...
... (b) Suppose that G is finite of order pn where n ≥ 1, Use part (a) to show that |Z(G)| ≡ 0 mod p and deduce that |Z(G)| > 1. 2. Let G be a group of order p2 where p is a prime number. (a) Show that if G is non-abelian, we must have |Z(G)| = p and g p = 1 for every g ∈ G. (b) Suppose that G is non-ab ...
Lecture 1: Lie algebra cohomology
... particular the de Rham cohomology H• (M) has a well-defined multiplication induced from the wedge product. If M is riemannian, compact and orientable one has the celebrated Hodge decomposition theorem stating that in every de Rham cohomology class there is a unique smooth harmonic form. The second e ...
... particular the de Rham cohomology H• (M) has a well-defined multiplication induced from the wedge product. If M is riemannian, compact and orientable one has the celebrated Hodge decomposition theorem stating that in every de Rham cohomology class there is a unique smooth harmonic form. The second e ...
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... prove this for sequences that converges to 0 (instead of an arbitrary element of a) because a is stable subset under addition (one can transform a sequence converging to any element of a into a sequence converging to 0 by subtracting that element to the sequence). Suppose there is a sequence of a th ...
... prove this for sequences that converges to 0 (instead of an arbitrary element of a) because a is stable subset under addition (one can transform a sequence converging to any element of a into a sequence converging to 0 by subtracting that element to the sequence). Suppose there is a sequence of a th ...