contact email: donsen2 at hotmail.com Contemporary abstract
... Suppose that x ∈ Zn is not a zero-divisor. Then power(xk just say k) of x is not a zero-divisor, too. if not there is a y ∈ Zn such that xk · y = x · xk−1 y = 0. x is turned out to be a zero-divisor. Invertibility of elements is only that is left to prove. Let’s consider the set of {xk |k ∈ Z}. Sinc ...
... Suppose that x ∈ Zn is not a zero-divisor. Then power(xk just say k) of x is not a zero-divisor, too. if not there is a y ∈ Zn such that xk · y = x · xk−1 y = 0. x is turned out to be a zero-divisor. Invertibility of elements is only that is left to prove. Let’s consider the set of {xk |k ∈ Z}. Sinc ...
N AS AN AEC Very Preliminary We show the concept of an Abstract
... 0; we will use this condition when studying Abelian groups below. While the class of cotorsion modules is complex, the torsion free cotorsion Abelian groups are more fully understood; they are the pure-injective Abelian groups. The Whitehead groups, ⊥ Z do not form an AEC with ≺N for N = Z; this can ...
... 0; we will use this condition when studying Abelian groups below. While the class of cotorsion modules is complex, the torsion free cotorsion Abelian groups are more fully understood; they are the pure-injective Abelian groups. The Whitehead groups, ⊥ Z do not form an AEC with ≺N for N = Z; this can ...
Abelian group
... a third integer, addition is associative, zero is the additive identity, every integer n has an additive inverse, −n, and the addition operation is commutative since m + n = n + m for any two integers m and n. • Every cyclic group G is abelian, because if x, y are in G, then xy = aman = am + n = an ...
... a third integer, addition is associative, zero is the additive identity, every integer n has an additive inverse, −n, and the addition operation is commutative since m + n = n + m for any two integers m and n. • Every cyclic group G is abelian, because if x, y are in G, then xy = aman = am + n = an ...
*These are notes + solutions to herstein problems(second edition
... 1)If A,B are groups,PT A X B isomorphic to B X A (a,b)->(b,a) 2)G,H,I are groups.PT (G X H) X I isomorphic to G X H X I ((g,h),i) -> (g,h,i) 3)T = G1 X G2…X Gn.PT for all i there exists an onto homomorphism h(i) from T to Gi What is the kernel of h(i)? h(i) : (g1,g2..gn) -> gi Kernel of h(i) = {(g1, ...
... 1)If A,B are groups,PT A X B isomorphic to B X A (a,b)->(b,a) 2)G,H,I are groups.PT (G X H) X I isomorphic to G X H X I ((g,h),i) -> (g,h,i) 3)T = G1 X G2…X Gn.PT for all i there exists an onto homomorphism h(i) from T to Gi What is the kernel of h(i)? h(i) : (g1,g2..gn) -> gi Kernel of h(i) = {(g1, ...
ABELIAN GROUPS THAT ARE DIRECT SUMMANDS OF EVERY
... If, in particular, R consists of the rational integers, then the hypotheses of Corollary 2 are satisfied. In this case the sufficiency of the condition of the Corollary 2 has been known for a long time. 3 An abelian group G over the ring R is termed complete, if it is ikf-complete for every ideal M ...
... If, in particular, R consists of the rational integers, then the hypotheses of Corollary 2 are satisfied. In this case the sufficiency of the condition of the Corollary 2 has been known for a long time. 3 An abelian group G over the ring R is termed complete, if it is ikf-complete for every ideal M ...
Nilpotent groups
... Q Pi are all normal and every maximal subgroup of G has the form N × i6=j Pi where N < Pj is maximal. But we proved that maximal subgroups of p-groups are normal (with factor group Z/p). ...
... Q Pi are all normal and every maximal subgroup of G has the form N × i6=j Pi where N < Pj is maximal. But we proved that maximal subgroups of p-groups are normal (with factor group Z/p). ...
Homework #3 Solutions (due 9/26/06)
... 3.4 a) Let G be a group and a, b ∈ G. Then (aba−1 )n = abn a−1 , for all n ∈ Z. Proof. For n = 0 this is clear since e = (aba−1 )0 = ab0 a−1 = aa−1 . For n > 0, the idea is that (aba−1 )n = (aba−1 )(aba−1 ) · · · (aba−1 )(aba−1 ) = ab(aa−1 )b(aa−1 ) · · · b(aa−1 )ba−1 = abb · · · ba−1 = abn a−1 . Th ...
... 3.4 a) Let G be a group and a, b ∈ G. Then (aba−1 )n = abn a−1 , for all n ∈ Z. Proof. For n = 0 this is clear since e = (aba−1 )0 = ab0 a−1 = aa−1 . For n > 0, the idea is that (aba−1 )n = (aba−1 )(aba−1 ) · · · (aba−1 )(aba−1 ) = ab(aa−1 )b(aa−1 ) · · · b(aa−1 )ba−1 = abb · · · ba−1 = abn a−1 . Th ...
Mat 247 - Definitions and results on group theory Definition: Let G be
... (1) If F is a field and n is a postive integer, let GLn (F ) = { A ∈ Mn×n (F ) | det(A) 6= 0 }. Then GLn (F ) is a group under the operation of matrix multiplication. (2) Let V be a finite-dimensional vector space over a field F . Let G = { T ∈ L(V ) | T is invertible }. Then GL(V ) is a group under ...
... (1) If F is a field and n is a postive integer, let GLn (F ) = { A ∈ Mn×n (F ) | det(A) 6= 0 }. Then GLn (F ) is a group under the operation of matrix multiplication. (2) Let V be a finite-dimensional vector space over a field F . Let G = { T ∈ L(V ) | T is invertible }. Then GL(V ) is a group under ...
1.5.4 Every abelian variety is a quotient of a Jacobian
... affine subset U = Spec(R) ⊂ Y , the inverse image f −1 (U ) ⊂ X is an affine open subset Spec(B) with B a finitely generated R-module. Finite morphisms have finite fibers, but not conversely.) We assume this lemma and deduce the theorem. Let J be the Jacobian of C; by the universal property of Jacob ...
... affine subset U = Spec(R) ⊂ Y , the inverse image f −1 (U ) ⊂ X is an affine open subset Spec(B) with B a finitely generated R-module. Finite morphisms have finite fibers, but not conversely.) We assume this lemma and deduce the theorem. Let J be the Jacobian of C; by the universal property of Jacob ...
Chapter 7: Infinite abelian groups For infinite abelian
... • Let Hp = i∈N Zp , where the elements are infinite sequences of elements of Zp where all but finitely many are 0. L • Hp = i∈N Zp is an infinite dimensional, countable vector space over the field GF (p), the Galois field of order p. • Let H = {Z} ∪ {Hp : p a prime}. • the groups Hp are not finitely ...
... • Let Hp = i∈N Zp , where the elements are infinite sequences of elements of Zp where all but finitely many are 0. L • Hp = i∈N Zp is an infinite dimensional, countable vector space over the field GF (p), the Galois field of order p. • Let H = {Z} ∪ {Hp : p a prime}. • the groups Hp are not finitely ...
Math 594. Solutions 3 Book problems §5.1: 14. Let G = A1 × A2
... kh = hk for all k ∈ K and h ∈ H. Let φ : H × K → HK be given by (h, k) 7→ hk. We claim that φ is a homomorphism, where we consider H × K as a group with the usual component-wise structure. For we have φ((h1 , k1 )(h2 , k2 )) = φ(h1 h2 , k1 k2 ) = h1 h2 k1 k2 = (h1 k1 )(h2 k1 ) = φ(h1 , k1 )φ(h2 , k2 ...
... kh = hk for all k ∈ K and h ∈ H. Let φ : H × K → HK be given by (h, k) 7→ hk. We claim that φ is a homomorphism, where we consider H × K as a group with the usual component-wise structure. For we have φ((h1 , k1 )(h2 , k2 )) = φ(h1 h2 , k1 k2 ) = h1 h2 k1 k2 = (h1 k1 )(h2 k1 ) = φ(h1 , k1 )φ(h2 , k2 ...
on end0m0rpb3sms of abelian topological groups
... proved that if G is a compact connected Hausdorff abelian group of countable weight then every countable family * of nontrivial endomorphisms of G is small. He asked if "compact" can be replaced by "complete". In this note the answer is given in the negative, but it is shown that "compact" can be re ...
... proved that if G is a compact connected Hausdorff abelian group of countable weight then every countable family * of nontrivial endomorphisms of G is small. He asked if "compact" can be replaced by "complete". In this note the answer is given in the negative, but it is shown that "compact" can be re ...
FREE GROUPS - Stanford University
... sj:XjS such that for every object Y in the category and every collection of morphisms fj:XjY, there EXISTS a UNIQUE morphism f:SY such that fsj = fj:XjY. For example, maybe the category being studied is abelian groups, or left R-modules where R is a ring, or topological spaces, or commutative ri ...
... sj:XjS such that for every object Y in the category and every collection of morphisms fj:XjY, there EXISTS a UNIQUE morphism f:SY such that fsj = fj:XjY. For example, maybe the category being studied is abelian groups, or left R-modules where R is a ring, or topological spaces, or commutative ri ...
MATH 436 Notes: Finitely generated Abelian groups.
... If |X| < ∞ then | ⊕x∈X Z/2Z| = 2|X| . Thus | ⊕y∈Y Z/2Z| is also finite and so |Y | < ∞ and 2|Y | = 2|X| which yields |X| = |Y | as desired. So it only remains to consider the case |X| = |Y | = ∞. Let Pf inite (X) = {S ⊂ X||S| < ∞} be the set of finite subsets of X. Then there is a bijection Θ : Pf i ...
... If |X| < ∞ then | ⊕x∈X Z/2Z| = 2|X| . Thus | ⊕y∈Y Z/2Z| is also finite and so |Y | < ∞ and 2|Y | = 2|X| which yields |X| = |Y | as desired. So it only remains to consider the case |X| = |Y | = ∞. Let Pf inite (X) = {S ⊂ X||S| < ∞} be the set of finite subsets of X. Then there is a bijection Θ : Pf i ...
9 Direct products, direct sums, and free abelian groups
... In such situation we say that (L, R) is an adjoint pair of functors. 11.2 Note. 1) The collection of morphisms {ηc }c∈C is called the unit of adjunction of (L, R). 2) For any adjoint pair (L, R) we also have morphisms {εd : LR(d) → d}d∈D satisfying analogous conditions as {ηc }c∈C . This collection ...
... In such situation we say that (L, R) is an adjoint pair of functors. 11.2 Note. 1) The collection of morphisms {ηc }c∈C is called the unit of adjunction of (L, R). 2) For any adjoint pair (L, R) we also have morphisms {εd : LR(d) → d}d∈D satisfying analogous conditions as {ηc }c∈C . This collection ...
2. For each binary operation ∗ defined on a set below, determine
... The philosophy here is that (G, ∗) inherits all necessary characteristics from (G, ·). We proceed by checking the four axioms on page 91: Closure: Let a, b ∈ G. Then a ∗ b = b · a, and b · a ∈ G because (G, ·) is a group and thus satisfies closure. Associativity: Let a, b and c ∈ G. Then a ∗ (b ∗ c) ...
... The philosophy here is that (G, ∗) inherits all necessary characteristics from (G, ·). We proceed by checking the four axioms on page 91: Closure: Let a, b ∈ G. Then a ∗ b = b · a, and b · a ∈ G because (G, ·) is a group and thus satisfies closure. Associativity: Let a, b and c ∈ G. Then a ∗ (b ∗ c) ...
Math 400 Spring 2016 – Test 3 (Take
... K is a subgroup of G as well.) If K C H and H C G is it necessarily true that K C G? If so, prove it. If not, give a counterexample (and show that it is, in fact, a counterexample.) Solution: It’s false. The smallest (lowest order) counterexample is G = D4 , H = {R0 , R180 , v, h}, K = {R0 , v}. (Pr ...
... K is a subgroup of G as well.) If K C H and H C G is it necessarily true that K C G? If so, prove it. If not, give a counterexample (and show that it is, in fact, a counterexample.) Solution: It’s false. The smallest (lowest order) counterexample is G = D4 , H = {R0 , R180 , v, h}, K = {R0 , v}. (Pr ...
Homework 9 - Material from Chapters 9-10
... Solution: This question is asking how many times the coset 14 + h8i must be added to itself in order to get the identity coset 0 + h8i. To equal the identity coset we must have a + h8i where a ∈ h8i, so a = 0, 8, or 16. We can just add 14 to itself in Z24 until we get 0 8, or 16, and see how many ti ...
... Solution: This question is asking how many times the coset 14 + h8i must be added to itself in order to get the identity coset 0 + h8i. To equal the identity coset we must have a + h8i where a ∈ h8i, so a = 0, 8, or 16. We can just add 14 to itself in Z24 until we get 0 8, or 16, and see how many ti ...
RSA: 1977--1997 and beyond
... algebraic structure in crypto. Mathematical group G = (S,*): binary operation * defined on (finite) set S: associative, identity, inverses, perhaps abelian. Example: Zn* (running example). Computational group [G] implements a mathematical group G. Each element x in G has one or more representati ...
... algebraic structure in crypto. Mathematical group G = (S,*): binary operation * defined on (finite) set S: associative, identity, inverses, perhaps abelian. Example: Zn* (running example). Computational group [G] implements a mathematical group G. Each element x in G has one or more representati ...
Finitely generated abelian groups, abelian categories
... Comment: One can give an abstract definition of an abelian category. See §II.1 of Hartshorne’s “Algebraic Geometry” or ”Homological Algebra” by Hilton and Stammbach. Our first definition of abelian categories then becomes a Theorem of Peter Freyd. One requires first that M orA(C, D) is an abelian g ...
... Comment: One can give an abstract definition of an abelian category. See §II.1 of Hartshorne’s “Algebraic Geometry” or ”Homological Algebra” by Hilton and Stammbach. Our first definition of abelian categories then becomes a Theorem of Peter Freyd. One requires first that M orA(C, D) is an abelian g ...
Example sheet 4
... 12. Show that a complex number α is an algebraic integer if and only if the additive group of the ring Z[α] is finitely generated (i.e. Z[α] is a finitely generated Z-module). Furthermore if α and β are algebraic integers show that the subring Z[α, β] of C generated by α and β also has a finitely ge ...
... 12. Show that a complex number α is an algebraic integer if and only if the additive group of the ring Z[α] is finitely generated (i.e. Z[α] is a finitely generated Z-module). Furthermore if α and β are algebraic integers show that the subring Z[α, β] of C generated by α and β also has a finitely ge ...
selected solutions to Homework 6
... 8. (Section 3.2, Problem 20) Prove or disprove that every group of order 3 is abelian. Solution. This statement is true. Proof. Let G = {a, b, e} be an arbitrary group of order 3. We already know that ea = ae = a and eb = be = b. All to show now is that ab = e = ba. For sake of contradiction, assume ...
... 8. (Section 3.2, Problem 20) Prove or disprove that every group of order 3 is abelian. Solution. This statement is true. Proof. Let G = {a, b, e} be an arbitrary group of order 3. We already know that ea = ae = a and eb = be = b. All to show now is that ab = e = ba. For sake of contradiction, assume ...
Algebra 411 Homework 5: hints and solutions
... h(0, 2)i ∼ = Z2 , h(1, 1)i = h(1, 3)i ∼ = Z4 , h(1, 2)i ∼ = Z2 This is eight elements in G, but two pairs of them give the same cyclic group, so we only get six different subgroups. I have also stated above what each subgroup is isomorphic to, although this was not part of the question. There is al ...
... h(0, 2)i ∼ = Z2 , h(1, 1)i = h(1, 3)i ∼ = Z4 , h(1, 2)i ∼ = Z2 This is eight elements in G, but two pairs of them give the same cyclic group, so we only get six different subgroups. I have also stated above what each subgroup is isomorphic to, although this was not part of the question. There is al ...
p-Groups - Brandeis
... A p-group is a finite group P of order pk where k ≥ 0. Note that every subgroup of a p-group is a p-group. When we want to exclude the trivial case k = 0 we say that P is a nontrivial p-group. One of the most important properties of p-groups is that they have nontrivial centers: Theorem 3.1. Every n ...
... A p-group is a finite group P of order pk where k ≥ 0. Note that every subgroup of a p-group is a p-group. When we want to exclude the trivial case k = 0 we say that P is a nontrivial p-group. One of the most important properties of p-groups is that they have nontrivial centers: Theorem 3.1. Every n ...