Here’s some examples of the concept of group homomorphism.
Let , which forms a group under multiplication and the group of all integers under addition, prove that the mapping from onto such that is a homomorphism.
Solution: Since , for all
Hence is a homomorphism.
Show that the mapping of the symmetric group onto the multiplicative group defined by or .
According as is an even or odd permutation in is a homomorphism of onto .
Solution: We know that the product of two permutations both even or both odd is even while the product of one even and one odd permutation is odd. We shall show that
(i) if are both even, then
(ii) if are both odd, then
(iii) if is odd and is even, then
(iv) if is even and is odd, then
Thus . Also obviously is onto .
Therefore is a homomorphism of onto .
Show that a homomorphism from s simple group is either trivial or one-to-one.
Solution: Let be a simple group and be a homomorphism of into another group . Then the kernel is a normal subgroup of . But the only normal subgroups of the simple group are and . Hence either or . If , the image of each element of is the identity of , as such the homomorphism is trivial one. If , the homomorphism is one-to-one.