Abstract

Download e-book for kindle: Abstract Algebra I by Randall R. Holmes

By Randall R. Holmes

Show description

Read or Download Abstract Algebra I PDF

Similar abstract books

Lie Groups, Lie Algebras, And Representations An Elementary by Brian Hall PDF

This textbook treats Lie teams, Lie algebras and their representations in an trouble-free yet absolutely rigorous type requiring minimum necessities. particularly, the speculation of matrix Lie teams and their Lie algebras is constructed utilizing simply linear algebra, and extra motivation and instinct for proofs is supplied than in such a lot vintage texts at the topic.

New PDF release: Aspects of Sobolev-Type Inequalities

This publication specializes in Poincaré, Nash and different Sobolev-type inequalities and their functions to the Laplace and warmth diffusion equations on Riemannian manifolds. functions coated contain the ultracontractivity of the warmth diffusion semigroup, Gaussian warmth kernel bounds, the Rozenblum-Lieb-Cwikel inequality and elliptic and parabolic Harnack inequalities.

New PDF release: Graded Syzygies

The research of unfastened resolutions is a middle and gorgeous zone in Commutative Algebra. the most aim of this publication is to motivate the readers and advance their instinct approximately syzygies and Hilbert features. Many examples are given with a purpose to illustrate rules and key ideas. A helpful characteristic of the booklet is the inclusion of open difficulties and conjectures; those supply a glimpse of interesting, and sometimes demanding, examine instructions within the box.

Additional info for Abstract Algebra I

Example text

Zm ⊕ Zn ∼ = Zmn if and only if gcd(m, n) = 1. Proof. (⇒) Assume that Zm ⊕ Zn ∼ = Zmn . Then there exists an isomorphism ϕ : Zmn → Zm ⊕ Zn . We have ϕ(1) = (a, b) for some a ∈ Zm and b ∈ Zn . Since the element 1 of Zmn has order mn, the element (a, b) of Zm ⊕ Zn must also have order mn. Let k = gcd(m, n). Then m = km and n = kn for some integers m and n . We have km n (a, b) = (km n a, km n b) = (n ma, m nb) = (0, 0). Since (a, b) has order mn and km n > 0 it follows that mn ≤ km n . So mn ≤ km n ≤ km kn = mn.

Ks ). Put I = {i1 , i2 , . . , ir } and K = {k1 , k2 , . . , ks }. To show that the two functions στ and τ σ are equal, we need to show that (στ )(m) = (τ σ)(m) for all m in their common domain, which is {1, 2, . . , n}. Let m ∈ {1, 2, . . , n}. First assume that m ∈ / I ∪ K (in other words, assume that m does not appear in either cycle). Then σ and τ both fix m giving (στ )(m) = σ(τ (m)) = σ(m) = m = τ (m) = τ (σ(m)) = (τ σ)(m). Now assume that m ∈ I. Then σ(m) ∈ I, as well. Since I and K are disjoint by assumption, m and σ(m) are not in K, so they are fixed by τ .

The following theorem says that disjoint cycles commute. 1 Theorem. If σ and τ are disjoint cycles in Sn , then στ = τ σ. Proof. Let σ and τ be disjoint cycles in Sn . We can write σ = (i1 , i2 , . . , ir ) and τ = (k1 , k2 , . . , ks ). Put I = {i1 , i2 , . . , ir } and K = {k1 , k2 , . . , ks }. To show that the two functions στ and τ σ are equal, we need to show that (στ )(m) = (τ σ)(m) for all m in their common domain, which is {1, 2, . . , n}. Let m ∈ {1, 2, . . , n}. First assume that m ∈ / I ∪ K (in other words, assume that m does not appear in either cycle).

Download PDF sample

Abstract Algebra I by Randall R. Holmes


by Steven
4.1

Rated 4.68 of 5 – based on 48 votes