Homomorphisms

Definition (Homomorphism) Let and be groups and let be a mapping such that then is called an homomorphism of and and in such a case we say and are homomorphic to each other. The set is called the homomorphic image of

Proposition (Homomorphism) Let and be groups and suppose is an homomorphism. Then

    (i)
    
    (ii) for each
    
    (iii) for each and for each integer
    
    (iv) is a subgroup of and
    
    (v) if is one-to-one, then

    Proof.  This proof is left as an exercise.

Definition (Isomorphism) Let and be groups and let be a one-to-one and onto mapping such that then is called an isomorphism of and and in such a case we say and are isomorphic to each other and denote this by

Example (Isomorphism) Consider the subgroup of and with the following Cayley tables



The mapping defined by shows that and are isomorphic because is a well-defined, one-to-one, and onto mapping that preserves the group operation, namely

Example (Isomorphism) Which, if any of the groups of order 8 are isomorhpic: and quaternion group:
    


The Cayley table for the symmetry group of the square is  

Proposition (Isomorphism is an Equivalence Relation) Isomorphism is an equivalence relation on the class of all groups.

    Proof. The identity mapping shows that and so is reflexive. If is an isomorphism then so is because given any with and we have and implying Therefore, as desired. The rest of the proof is left as an exercise.

Proposition (Isomorphisms) Let and be groups and suppose is an isomorphism. Then

    (i)
    
    (ii) for each
    
    (iii) for each and for each integer
    
    (iv) is Abelian H is Abelian,
    
    (v) is cyclic H is cyclic,
    
    (vi) has a subgroup of order H has a subgroup of order ,
    
    (vii) every element in is its own inverse every element in is its own inverse, and
    
    (viii) every element in has finite order every element in has finite order.

Proposition (Group Stucture)

    (i) Any group of prime order is isomorphic to
    
    (iii) If is the standard factorization of then
    
    Proof. (i) Let be a group of prime order then is a subgroup and is either or by Lagrange's Theorem. Thus, is cyclic say Define by and notice that is well-defined and one-to-one since if and only if if and only if if and only if Also, is onto and
    (ii) This proof is left as an exercise.

Proposition (Cayley's Theorem) Every group is isomorphic to a permutation group on its set of elements. In particular, every group of finite order is isomorphic to a subgroup of

    Proof. Let be any group.  We need to map every element in to a permutation of To do so given we define by for all Then is onto since the equation has a solution for each and it is one-to-one since this solution is unique, so is a permutation of This shows that the function defined by is well-defined.
    Next we want to show that is a subgroup of which are both nonempty because of the identity mapping. Let Then there exists such that defined by   for all and    defined by   for all are permutations of Then which is is a permutation of and indeed Therefore,   is a subgroup of
    To show that preserves the group operation, let then is a permutation say with and in fact and are permutations also. It follows that and so Since is one-to-one and onto as desired.
    Finally, we restrict our attention to the case when is finite. Then will assign to every element in (as shown) a permutation which corresponds to a unique element of

Example (Cayley's Theorem)

    (i) Find the permutations associated with every element in as in the proof of Cayley's Theorem.
    
    (ii) Find the permutations associated with every element in the symmetry group of the square as in the proof of Cayley's Theorem.
    
    (iii) Find the permutations associated with every element of as in the proof of Cayley's Theorem.

Definition (Kernel) If is a homomorphism then the kernel of is defined by

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