Abstract Algebra Theory and Applications - Computer Science ...

13.2 EXAMPLES AND APPLICATIONS 227element of a Sylow 2-subgroup other than the identity must have as its ordera power of 2; and therefore cannot be one of the 48 elements of order 7 inthe Sylow 7-subgroups. Since a Sylow 2-subgroup has order 8, there is onlyenough room for a single Sylow 2-subgroup in a group of order 56. If thereis only one Sylow 2-subgroup, it must be normal.For other groups G it is more difficult to prove that G is not simple.Suppose G has order 48. In this case the technique that we employed in thelast example will not work. We need the following lemma to prove that nogroup of order 48 is simple.Lemma 13.10 Let H and K be finite subgroups of a group G. ThenProof. Recall that|HK| =|H| · |K||H ∩ K| .HK = {hk : h ∈ H, k ∈ K}.Certainly, |HK| ≤ |H| · |K| since some element in HK could be writtenas the product of different elements in H and K. It is quite possible thath 1 k 1 = h 2 k 2 for h 1 , h 2 ∈ H and k 1 , k 2 ∈ K. If this is the case, leta = (h 1 ) −1 h 2 = k 1 (k 2 ) −1 .Notice that a ∈ H ∩ K, since (h 1 ) −1 h 2 is in H and k 2 (k 1 ) −1 is in K; consequently,h 2 = h 1 a −1k 2 = ak 1 .Conversely, let h = h 1 b −1 and k = bk 1 for b ∈ H ∩ K. Then hk = h 1 k 1 ,where h ∈ H and k ∈ K. Hence, any element hk ∈ HK can be written inthe form h i k i for h i ∈ H and k i ∈ K, as many times as there are elementsin H ∩ K; that is, |H ∩ K| times. Therefore, |HK| = (|H| · |K|)/|H ∩ K|.□Example 9. To demonstrate that a group G of order 48 is not simple, wewill show that G contains either a normal subgroup of order 8 or a normalsubgroup of order 16. By the Third Sylow Theorem, G has either one orthree Sylow 2-subgroups of order 16. If there is only one subgroup, then itmust be a normal subgroup.