What is abstract algebra and linear algebra and what is the difference between them?
Well, Algebra has various topics.Mainly algebra can be classified as: Classical algebra, linear algebra, abstract algebra, Boolean algebra.Now, the difference between abstract algebra and linear algebra:Actually, linear algebra is all about the vector spaces, linear mappings.On the other hand, abstract algebra is all about set theory, group, ring, field.Secondly, abstract algebra is independent from linear algebra; but linear algebra is very much dependent on abstract algebra. As for example, one can not access vector spaces before understanding group, ring, field.These are the basic difference.
Abstract Algebra: xax = b iff c^2?
You are missing a quantifier in the question. I think the intended problem is: "prove that there is an element x in G such that xax = b, if and only if there is an element c in G such that ab = c^2." Suppose there is an element x in G with xax = b. Then ab = a(xax) [since xax = b] = (ax)(ax) [by the associative law] = (ax)^2 [by the definition of squaring] showing that ab is the square of the element c = ax in G. This is one half of the problem. Conversely, suppose that there is an element c in G with ab = c^2. Then note that (a^(-1) c) a (a^(-1) c) = a^(-1) c (aa^(-1)) c [by the associative law] = a^(-1) cc [since aa^(-1) is the identity] = a^(-1) c^2 [by the definition of squaring] = a^(-1) ab [as ab = c^2] = b [as a^(-1) a is the identity]. This shows that the element x = a^(-1) c in G has the property that xax = b. This is the second half of the problem.
In Boolean algebra, why does 1 + 1 = 1?
Boolean algebra is an abstract concept. Propositional logic using ({T, F}, ∧, ∨} and set theory using (U, ∩, ∪), where U is the universal set or domain of discourse, are specific examples of the abstract concept.Duality is a fundamental property of Boolean algebras. Among other things, duality indicates that I am free to associate either binary operation in a specific example with either operation defining Boolean algebras as an abstract concept. Some respondents have stated that + means inclusive disjunction and 1 means T (also called True and ⊤); duality tells us that it is equally appropriate and consistent to have + mean conjunction and 1 mean F (False, ⊥).Also, the use of + and · or × is really inappropriate, because these symbols carry baggage looking like addition and multiplication, respectively, and many people use that to argue that conjunction takes precedence over inclusive disjunction—but that would not be the situation if you carried out the swap I suggested in the previous paragraph. The problem is that both binary operations in any Boolean algebra (abstract or applied) behave as multiplication operations; neither behaves as addition, so do not use notation that erroneously conveys a sense of addition.I also do not like expressing the binary operations for the abstract definition as ∧ and ∨--these strongly hint at propositional logic’s conjunction and lattice theory’s meet, and propositional logics inclusive disjunction and lattice theory’s join, respectively. I personally prefer to symbolize the two abstract binary operations as ⧅ and ⧄ — my own notational invention that indicates the inherent symmetry of the two operations expressed in duality, with no hint of addition versus multiplication, nor of existing usage of lattice theory, propositional logic, and set theory notation.Now, why does 1 + 1 = 1 in any Boolean algebra (using the original poster’s notation and assuming 1 is an element of the defining domain)? It is true because every element is idempotent under both binary operations; there is no need for 1 to be an identity element or any other special element.
Can someone help with this Abstract Algebra question?
It is very important that you be able to do this type of problem. It tests basic knowledge of several things. 1) Can you write the definition of a Group 2) Can you write down the definition of Z12 3) Can you write down the definition of a homomorphism. ---- Those are in your text and you cannot start any mathematical proof until you have a precise knowledge of all the terms. I will help with 2) A homomorphism from one group to another is a function such that f(a*b) = f(a)*f(b). To check a function is homomorphism means you need to check that property, In math, if you don't know the definitions you can't get off the starting block
Abstract algebra (about normalizer N_G(P))?
1) let's just write N(P), with the fact that we are normalizing "with respect to G" understood. so we want to show that N(N(P)) = N(P). clearly N(P) is contained in N(N(P)), since N(P) is a subgroup: for n in N(P), nN(P)n^-1 = N(P). now suppose x is in N(N(P)). then xN(P)x^-1 = N(P) (by definition of N(N(P)) itself). now P is a subgroup of N(P), so xPx^-1 is a subgroup of xN(P)x^-1 = N(P). all sylow p-subgroups of N(P) are conjugate in N(P) (and P is a p-sylow subgroup of N(P)), so for some y in N(P), we have xPx^-1 = yPy^-1. but y is in N(P), so yPy^-1 = P. therefore, xPx^-1 = P as well. but this means x is in N(P), so N(N(P)) is contained in N(P), so the two groups must be equal. 2) if |G| = 36, consider a 3-sylow subgroup P, which has order 9. we have 3 choices for |N(P)|: 9, 18 or 36. if |N(P)| = 36, then P is normal. if |N(P)| = 18, then N(P) is normal. if |N(P)| = 9, then we have 4 conjugates of P in G (the number of conjugates of P is the index of N(P)). let G act on the conjugates of P by conjugation. then we get a homomorphism φ of G into S4, thus |ker(φ)| divides 4! = 24 and 36. so |ker(φ)| = 1,2,3 or 6. if |ker(φ)| is not 1, ker(φ) is a non-trivial proper normal subgroup. but |ker(φ)| = 1, means φ(G) is isomorphic to G, so |φ(G)| = 36 > 24, impossible.
Why is Boolean algebra called switching algebra?
Boolean algebra is the branch of algebra in which the values of the variables are the truth values true and false, usually denoted 1 and 0 respectively. Instead of elementary algebra where the values of the variables are numbers, and the main operations are addition and multiplication, the main operations of Boolean algebra are the conjunction and, denoted ∧, the disjunction or, denoted ∨, and the negation not, denoted ¬. It is thus a formalism for describing logical relations in the same way that ordinary algebra describes numeric relations.n the 1930s, while studying switching circuits, Claude Shannon observed that one could also apply the rules of Boole's algebra in this setting, and he introduced switching algebra as a way to analyze and design circuits by algebraic means in terms of logic gates. Shannon already had at his disposal the abstract mathematical apparatus, thus he cast his switching algebra as the two-element Boolean algebra. In circuit engineering settings today, there is little need to consider other Boolean algebras, thus "switching algebra" and "Boolean algebra" are often used interchangeably.Efficient implementation of Boolean functions is a fundamental problem in the design of combinational logiccircuits. Modern electronic design automation tools for VLSI circuits often rely on an efficient representation of Boolean functions known as (reduced ordered) binary decision diagrams (BDD) for logic synthesis and formal verification.Source:Wiki