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:
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 x^{n} for some n ∈Z. The element x is called generator of K and we write K= <x> Cyclic Group:-In the case when G= Example: The group G= {1, -1, i,-i} under usual multiplication is a finite cyclic group with i as generator, since i^{1}=i,i^{2}=-1,i^{3}=-i and i^{4}=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., 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 (G_{1},*_{1})and (G_{2},*_{2}) are groups, then G = G_{1} x G_{2} i.e., (G, *) is a group with operation defined by (a_{1},b_{1})*( a_{2},b_{2} )=(a_{1},*_{1},a_{2}, b_{1} *_{2} b_{2}). Proof: To prove that G_{1} x G_{2} is a group, we have to show that G_{1} x G_{2} has the associativity operator, has an identity and also exists inverse of every element. Associativity. Let a, b, c ∈ G_{1} x G_{2},then So, a * (b * c) = (a_{1},a_{2} )*((b_{1},b_{2})*(c_{1},c_{2})) Identity: Let e_{1} and e_{2} are identities for G_{1} and G_{2} respectively. Then, the identity for G_{1} x G_{2} is e=(e_{1},e_{2} ).Assume same a ∈ G_{1} x G_{2} Then, a * e = (a_{1},a_{2})*( e_{1},e_{2}) Similarly, we have e * a = a. Inverse: To determine the inverse of an element in G_{1} x G_{2}, we will determine it component wise i.e., Now to verify that this is the exact inverse, we will compute a * a^{-1} and a^{-1}*a. Now, a * a^{-1}=(a_{1},a_{2} )*(a_{1}^{-1},a_{2}^{-1} ) Similarly, we have a^{-1}*a=e. Thus, (G_{1} x G_{2},*) is a group. In general, if G_{1},G_{2},....G_{n} are groups, then G = G_{1} x G_{2} x.....x G_{n} 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} Next TopicNormal Subgroup |