# Subgroup:

If a non-void subset H of a group G is itself a group under the operation of G, we say H is a subgroup of G.

Theorem: - A subset H of a group G is a subgroup of G if:

• the identity element a∈ H.
• H is closed under the operation of G i.e. if a, b∈ H, then a, b∈ H and
• H is closed under inverses, that is if a∈ H then a-1∈ H.

## Cyclic Subgroup:-

A Subgroup K of a group G is said to be cyclic subgroup if there exists an element x∈ G such that every element of K can be written in the form xn for some n ∈Z.

The element x is called generator of K and we write K= <x>

## Cyclic Group:-

In the case when G=, we say G is cyclic and x is a generator of G. That is, a group G is said to be cyclic if there is an element x∈ G such that every element of G can be written in the form xn for the some n∈ Z.

Example: The group G= {1, -1, i,-i} under usual multiplication is a finite cyclic group with i as generator, since i1=i,i2=-1,i3=-i and i4=1

## Abelian Group:

Let us consider an algebraic system (G,*), where * is a binary operation on G. Then the system (G,*) is said to be an abelian group if it satisfies all the properties of the group plus a additional following property:

(1) The operation * is commutative i.e.,
a * b = b * a ∀ a,b ∈G

Example: Consider an algebraic system (G, *), where G is the set of all non-zero real numbers and * is a binary operation defined by Show that (G, *) is an abelian group.

Solution:

Closure Property: The set G is closed under the operation *, since a * b = is a real number. Hence, it belongs to G.

Associative Property: The operation * is associative. Let a,b,c∈G, then we have Identity: To find the identity element, let us assume that e is a +ve real number. Then e * a = a, where a ∈G. Thus, the identity element in G is 4.

Inverse: let us assume that a ∈G. If a-1∈Q, is an inverse of a, then a * a-1=4 Thus, the inverse of element a in G is Commutative: The operation * on G is commutative. Thus, the algebraic system (G, *) is closed, associative, identity element, inverse and commutative. Hence, the system (G, *) is an abelian group.

## Product of Groups:

Theorem: Prove that if (G1,*1)and (G2,*2) are groups, then G = G1 x G2 i.e., (G, *) is a group with operation defined by (a1,b1)*( a2,b2 )=(a1,*1,a2, b1 *2 b2).

Proof: To prove that G1 x G2 is a group, we have to show that G1 x G2 has the associativity operator, has an identity and also exists inverse of every element.

Associativity. Let a, b, c ∈ G1 x G2,then

So,        a * (b * c) = (a1,a2 )*((b1,b2)*(c1,c2))
= (a1,a2 )*(b1 *1 c1,b2 *2 c2)
= (a1 *1 (b1 *1 c1 ),a2 *2 (b2 *2 c2)
= ((a1 *1 b1) *1 c1,( a2 *2 b2) *2 c2)
= (a1 *1 b1,a2 *2 b2)*( c1,c2)
= ((a1,a2)*( b1,b2))*( c1,c2)
= (a * b) * c.

Identity: Let e1 and e2 are identities for G1 and G2 respectively. Then, the identity for G1 x G2 is e=(e1,e2 ).Assume same a ∈ G1 x G2

Then,        a * e = (a1,a2)*( e1,e2)
= (a1 *1 e1,a2 *2 e2)
= (a1,a2)=a

Similarly, we have e * a = a.

Inverse: To determine the inverse of an element in G1 x G2, we will determine it component wise i.e.,
a-1=(a1,a2)-1=(a1-1,a2-1 )

Now to verify that this is the exact inverse, we will compute a * a-1 and a-1*a.

Now,         a * a-1=(a1,a2 )*(a1-1,a2-1 )
= (a1 *1 a1-1,a2 *2 a2-1)=( e1,e2)=e

Similarly, we have a-1*a=e.

Thus, (G1 x G2,*) is a group.

In general, if G1,G2,....Gn are groups, then G = G1 x G2 x.....x Gn is also a group.

## Cosets:

Let H be a subgroup of a group G. A left coset of H in G is a subset of G whose elements may be expressed as xH={ xh | h ∈ H } for any x∈ G. The element x is called a representation of the coset. Similarly, a right coset of H in G is a subset that may be expressed as Hx= {hx | h ∈H } , for any x∈G. Thus complexes xH and Hx are called respectively a left coset and a right coset.

If the group operation is additive (+) then a left coset is denoted as x + H={x+h | h ∈H} and a right coset is denoted by H + x = {h+x | h ∈ H}

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