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# Modern 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 ...

## 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 ...

## 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 ...

## 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 ...

## Correspondence theorem on homomorphism

Let f be a homomorphism of a group G onto a group Gʹ with Kernel K(=Kerf). Let Hʹ be a subgroup of Gʹ and H = {x ∊G | f(x) ∊ Hˊ} = f -1(Hʹ). Then H is a subgroup of G containing K. If Hʹ is a normal subgroup ...

## Homomorphism and Isomorphism

Definition of Homomorphism: Let (G, *) and (Gʹ, 0) be any two groups. Then a mapping f : G → Gʹ is said to be a homomorphism if f(a * b) = f(a) o f(b), ∀ a, b ∈ G. Definition: A homomorphism f of a group G to a ...

## A subgroup H of a group G is a normal subgroup if and only if g-1 Hg = H, ∀ g ∈ G

Proof: Let H be a normal subgroup og G. Then g-1hg ∈ H, ∀ g ∈ G and ∀ h ∈ H. Since g-1 Hg = {g-1 hg | g ∈ h and h ∈ H}, g-1 Hg ⊂ H   ————————(i) Again for every h ∈ H and for every ...

## A subgroup H of group G is a normal subgroup if and only if aH = Ha, ∀ a ∈ G

i.e., Every left coset of H in G is a right coset of H in G. Proof: Let H be a normal subgroup of G. Then g-1hg ∈ H, ∀ g ∈ G and ∀ h ∈ H. Let a ∈ G and h ∈ H. Then a-1 ha ∈ ...