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Introduction to group theory/Part 2 Subgroups and cyclic groups

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Introduction

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We will learn briefly about subgroups and cyclic groups. As per the name suggest "sub" means a part of i.e; heading-subheading ,set- subset etc etc.

So subgroup is a part of a group which forms under the same binary operation as per the group. [ Note: all elements of subgroup must be elements of groups]

Definition

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A non-empty subset H of a group G is said to be subgroup of G if , under the product in G, H itself forms a group.

i.e;

It should follow all the properties of groups.

  • Closure
  • Associative
  • Indentity
  • Inverse
  • Commutative (optional)

The following remark is clear: if H is a subgroup of G and K is a subgroup of H , then K is a subgroup of G.

LEMMA

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A non empty subset H of the group G is a subgroup of G if and only if

Proof

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If H is a subgroup of G , then it is obvious that (1) and (2) must hold.

In order to establish that H is subgroup, all that is needed is to verify that and that the associative law holds for elements of H .

Since associative law does hold for G, it holds all more so for H, which is a subset of G.

if , by part 2, and so by part 1 ,

LEMMA

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A non-empty finite subset H of the group G is a subgroup of G if and only if

1.

Proof

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The Proof is similar to the one for the previous lemma and left as an exercise for the reader.

Cyclic Group

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A group is said to be a cyclic group , if there exist an element such that every element of G can be expressed as some power of a.

If G is a group generated by 'a' . we can say that a is a generator of G and all the elements of G can form by some power of a.

G = (a) { Here a is the generator}

Notes

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  • A cyclic group may have more that one generator.
  • Every group has two trivial subgroups.
    • The group containing all elements.
    • The group containing identity element only.

Congruent Modulo of a Subgroup

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Let H is a subgroup of G.( )

if

LEMMA

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is an equivalence relation.

proof

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An equivalence relation must follows 3 properties.

  1. Reflexive
  2. Symmetric
  3. Transitive

Reflexive

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let

as

Symmetric

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let such that

we have to show that

[ Subgroup follows closure law]

Transitive

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let such that

and

and

[Closure]

[Associative]

[Identity]

modulo relation is an equivalence relation.