Abstract Algebra Theory and Applications - Computer Science ...

158 CHAPTER 9 HOMOMORPHISMS AND FACTOR GROUPSExample 10. The kernel of the group homomorphism φ : R → C ∗ definedby φ(θ) = cos θ + i sin θ is {2πn : n ∈ Z}. Notice that ker φ ∼ = Z. Example 11. Suppose that we wish to determine all possible homomorphismsφ from Z 7 to Z 12 . Since the kernel of φ must be a subgroup ofZ 7 , there are only two possible kernels, {0} and all of Z 7 . The image ofa subgroup of Z 7 must be a subgroup of Z 12 . Hence, there is no injectivehomomorphism; otherwise, Z 12 would have a subgroup of order 7, which isimpossible. Consequently, the only possible homomorphism from Z 7 to Z 12is the one mapping all elements to zero.Example 12. Let G be a group. Suppose that g ∈ G and φ is the homomorphismfrom Z to G given by φ(n) = g n . If the order of g is infinite,then the kernel of this homomorphism is {0} since φ maps Z onto the cyclicsubgroup of G generated by g. However, if the order of g is finite, say n,then the kernel of φ is nZ.Simplicity of A nOf special interest are groups with no nontrivial normal subgroups. Suchgroups are called simple groups. Of course, we already have a wholeclass of examples of simple groups, Z p , where p is prime. These groups aretrivially simple since they have no proper subgroups other than the subgroupconsisting solely of the identity. Other examples of simple groups are notso easily found. We can, however, show that the alternating group, A n , issimple for n ≥ 5. The proof of this result requires several lemmas.Lemma 9.5 The alternating group A n is generated by 3-cycles for n ≥ 3.Proof. To show that the 3-cycles generate A n , we need only show thatany pair of transpositions can be written as the product of 3-cycles. Since(ab) = (ba), every pair of transpositions must be one of the following:(ab)(ab) = id(ab)(cd) = (acb)(acd)(ab)(ac) = (acb).Lemma 9.6 Let N be a normal subgroup of A n , where n ≥ 3. If N containsa 3-cycle, then N = A n .□