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Home / Tag Archives: Abstract algebra

# Tag Archives: Abstract algebra

## Lecture – 4| Abstract Algebra

Problem – 1 : Show that if every element of the group G except the identity element is of order 2, then G is abelian. Solution: Let a, b ∈ G such that a ≠ e, b ≠ e Then a2 = e, b2 = e. Also ab ∈ G ...

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## Every group is isomorphic to a group of permutations.

PROOF:  To prove this, let G be any group. We must find a group of permutations that we believe is isomorphic to G. Since G is all we have to work with, we will have to use it to construct . For any g in G, define a function Tg ...

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## PDF| Division Algorithm |Abstract algebra

Division Algorithm

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## Division Algorithm on abstract algebra

Division Algorithm on abstract algebra: Let a and b be integers, with b > 0.Then there exist unique integers q and r such that a = bq + r                 where 0 ≤ r < b. Proof: Suppose that the numbers q and r actually exist. Then we must show that ...

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## Principle of Well-Ordering

Principle of Well-Ordering: Every nonempty subset of the natural numbers is well-ordered. The Principle of Well-Ordering is equivalent to the Principle of Mathematical Induction. Lemma: The Principle of Mathematical Induction implies that 1 is the least positive natural number. Proof: Let S = {n ∊ N : n ≥1}. Then ...

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## Show that 0(b) = 1 or 0(b) = 31

Problem: Let a, b be two elements of a group G such that a5 = e and ab a-1 = b2, where e is the identity of G. Show that 0(b) = 1 or 0(b) = 31. Solution: Given that ab a-1 = b2 ————————–(i) (ab a-1)2 = (b2)2 ⟹ ...

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## Show that if every element of the group G execept the identity element is of order 2, then G is abelian.

Solution: Let a, b∈G such that a ≠ e, b ≠ e. Then according to the question, a2 = e, b2 = e. Also ab∈G and So (ab)2 = e Now (ab)2 = e ⟹ (ab) (ab) = e ⟹ a(ab ab) b = a e b ⟹ a2bab2 ...

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## show that G is abelian.

If G is a group in which (ab)1 = a1 b1 for three consecutive integers I for all a, b ∈G, show that G is abelian. Solution: We have (ab)1 = a1 b1 ——————————-(i) (ab)i+1 = ai+1 bi+1 ————————-(ii) (ab)i+2 = ai+2 bi+2 ————————-(iii) From (ii), we have (ab)i+1 = ...

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## Prove that a^n = e iff m | n( m divides n).

Theorem: Let G be a group with identity e. If a ∊ G and O(a) = m, then for some positive integer n, an = e if and only if m | n (m divides n). Proof: O(a) = m ⇒ m is the least positive integer such that am ...

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## Groups| Modern Abstract Algebra

Definition of group: A group is a non empty set G together with a binary operation multiplication (*) such that the following four properties are satisfied (i) Closure Property: a*b∊ G, ∀ a, b ∊G (ii) Associative property: (a * b) * c = a * ( b* c), ∀ ...

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