Abstract Algebra Theory and Applications - Computer Science ...

13.2 EXAMPLES AND APPLICATIONS 225in G. Hence, G cannot be simple. Furthermore, if q ≢ 1 (mod p), then Gis cyclic.Proof. We know that G contains a subgroup H of order q. The number ofconjugates of H divides pq and is equal to 1 + kq for k = 0, 1, . . .. However,1+q is already too large to divide the order of the group; hence, H can onlybe conjugate to itself. That is, H must be normal in G.The group G also has a Sylow p-subgroup, say K. The number of conjugatesof K must divide q and be equal to 1 + kp for k = 0, 1, . . .. Since qis prime, either 1 + kp = q or 1 + kp = 1. If 1 + kp = 1, then K is normalin G. In this case, we can easily show that G satisfies the criteria, given inChapter 8, for the internal direct product of H and K. Since H is isomorphicto Z q and K is isomorphic to Z p , G ∼ = Z p × Z q∼ = Zpq by Theorem 8.10.□Example 3. Every group of order 15 is cyclic. This is true because 15 = 5·3and 5 ≢ 1 (mod 3).Example 4. Let us classify all of the groups of order 99 = 3 2 · 11 up toisomorphism. First we will show that every group G of order 99 is abelian.By the Third Sylow Theorem, there are 1 + 3k Sylow 3-subgroups, each oforder 9, for some k = 0, 1, 2, . . .. Also, 1 + 3k must divide 11; hence, therecan only be a single normal Sylow 3-subgroup H in G. Similarly, there are1 + 11k Sylow 11-subgroups and 1 + 11k must divide 9. Consequently, thereis only one Sylow 11-subgroup K in G. By Corollary 12.5, any group oforder p 2 is abelian for p prime; hence, H is isomorphic either to Z 3 × Z 3or to Z 9 . Since K has order 11, it must be isomorphic to Z 11 . Therefore,the only possible groups of order 99 are Z 3 × Z 3 × Z 11 or Z 9 × Z 11 up toisomorphism.To determine all of the groups of order 5 · 7 · 47 = 1645, we need thefollowing theorem.Theorem 13.9 Let G ′ = 〈aba −1 b −1 : a, b ∈ G〉 be the subgroup consistingof all finite products of elements of the form aba −1 b −1 in a group G. ThenG ′ is a normal subgroup of G and G/G ′ is abelian.The subgroup G ′ of G is called the commutator subgroup of G. Weleave the proof of this theorem as an exercise.Example 5. We will now show that every group of order 5 · 7 · 47 = 1645is abelian, and cyclic by Corollary 8.11. By the Third Sylow Theorem, Ghas only one subgroup H 1 of order 47. So G/H 1 has order 35 and must