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Home / Modern Abstract Algebra / Groups ( Part B)| Modern Abstract Algebra.

Groups ( Part B)| Modern Abstract Algebra.

The order of an element of group: Let a be a element of a group G. a is said to be of order m, if and only if m is the least positive integer such that am = e (the identity).

If there is no positive integer n such that an = e, a is said to be of infinite order or zero order.

The order of the element a is denoted by O(a).

Torsion group: A group G is called a torsion group if every element of G, is of finite order.

Example: The multiplicative group H = {1, -1, i, -i} and the additive group (Z8; +) are torsion.

Torsion free group: A group G is called a torsion free group if no element in G, other than the identity, is of finite order.

Example: The additive group of integers Z is a torsion free group.

Mixed group: A group G is called a mixed group if and only if there exist at least two elements a, b ∊G such that

(i) a ≠ e and O(a) is finite.

(ii) O(b) = ∞.

Cyclic group: A group G is said to be a cyclic group if there exist an element

a ∊ G such that every element of G is expressible as an integral power of a i.e. as ak where k is an integer.

Theorem: Every cyclic group is Abelian group.

Proof: Let G be a cyclic group generated by a.

Let x , y be any two elements in G, then x, y are some integral powers of a. That is x = ar and y = as where r and s some integers.

Now xy = ar as = ar+s = as+r = as ar = yx

This implies that G is abelian.

Theorem: Every finite cyclic group has exactly two generators.

Proof: Let G be an infinite cyclic group and a be a generator of G. Since the inverse of a generator is also a generator, a-1 must be a generator of G. We are to show that a and a-1 are the only generators of G. Since G is finite , a and a-1 are of finite order. This implies that for any two different integers r and s, ar ≠ as and (a-1)r ≠ (a-1)s.

Let m be any integer different from 1 and -1. Then the element am∊ G such that am ≠ a1 and am ≠ a-1.

Now for every integer k,

mk ≠ 1, -1, i.e. (am)k ≠ a and (am)k ≠ a-1.

Thus am fails to generate G.
Hence G has only two generators a and a-1.

Lecture – 4| Abstract Algebra

Problem – 1 : Show that if every element of the group G except the ...