Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 15 Generated Groups. Direct Product In this section, we discuss two procedures of building (sub)groups, namely, the (sub)groups generated by a subset of a given group and the direct product of two or more groups. 15.1 Finitely and Infinitely Generated Groups The concept of generators can be extended from cyclic groups < a > to more complicated situations where a subgroup is generated by more than one element. It is easy to see that a cyclic group with generator a is the smallest subgroup containing the set S = {a}. Can we extend groups generated by sets with more than one element? The answer is affirmative as suggested by the following theorem. Theorem 15.1 Let G be a group and C be the collection all subgroups of G. Then ∩H∈CH is a subgroup of G. Proof. Let K = ∩H∈CH. Since e belongs to all the subgroups of G then e ∈ K so that K 6= ∅. Now, let a, b ∈ K. Then a ∈ H and b ∈ H for all subgroups H of G. But then a ∈ H and b−1 ∈ H, ∀H ∈ C. Since every H in C is closed under multiplication then ab−1 ∈ H, ∀H ∈ C. That is, ab−1 ∈ K. By Theorem 7.5, K is a subgroup of G. Theorem 15.2 Let G be a group and S ⊆ G. Let C be the collection of all subgroups of G containing S. Then the set < S >= ∩H∈CH satisfies the following: (i) < S > is a subgroup of G containing S. (ii) For every H ∈ C, < S >⊆ H. Thus, < S > is the smallest subgroup of G containing S. 1 Proof. (i) Note that since S ⊆ G then G ∈ C. Thus, C 6= ∅. The fact that < S > is a subgroup of G follows from Theorem 15.1. Since S ⊆ H for all H ∈ C then S ⊆ ∩H∈CH. That is, S ⊆< S > . (ii) Let H ∈ C. If x ∈< S > then x ∈ ∩H∈CH and in particular, x ∈ H. Hence, < S >⊆ H. Theorem 15.3 < S > is the only subgroup of G that satisfies conditions (i) and (ii) of Theorem 15.2 Proof. Suppose that K is a subgroup of G satisfying conditions (i) and (ii) of the previous theorem. We will show that K =< S > . Since < S > is a subgroup containing S then by (ii), we have < S >⊆ K. Also, since K is a subgroup of G containing S then by condition (ii) again, we have K ⊆< S > . Thus, < S >= K. Definition 15.1 If G is a group and S ⊆ G then the subgroup < S > is called the subgroup of G generated by S. If G =< S >, then we say S generates G; and the elements in S are called generators. For S = {a1, a2, ··· , an} we write < S >=< a1, a2, ··· , an > . Example 15.1 Consider the subgroup of S4 generated by S = {(1432), (24)}. Then < S >= {(1), (1432), (24), (14)(23), (1234), (12)(34), (13)(24), (13)}. The following theorem characterizes the elements of < S > . Theorem 15.4 Let G be a group and S ⊆ G. Then < S > consists of finite products of ele- s1 s2 sk ments of S and inverses of elements of S. That is, if K = {a1 a2 ··· ak , ai ∈ S, si = ±1} then < S >= K. Proof. The proof is by double inclusions. By closure, K ⊆< S > . It remains to show that < S >⊆ K. Since each element of S is in K then S ⊆ K. 2 s1 s2 sk We will show that K is a subgroup of G. Indeed, if x = a1 a2 ··· ak and t1 t2 tm y = b1 b2 ··· bm are two elements in K then −1 s1 s2 sk t1 t2 tm −1 xy = (a1 a2 ··· ak )(b1 b2 ··· bm ) s1 s2 sk −tm −t2 −t1 = (a1 a2 ··· ak )(bm ··· b2 b1 ) s1 s2 sk −tm −t2 −t1 = a1 a2 ··· ak bm ··· b2 b1 Thus, xy−1 is a product of elements of S and/or inverses of elements of S. Hence, xy−1 ∈ K. By Theorem 7.5, K is a subgroup of G. By Theo- rem 15.2(ii), < S >⊆ K. Hence, < S >= K. The next theorem provides a condition under which two groups generated by different sets are equal. Theorem 15.5 Let T1 and T2 be subsets of a group G. Then < T1 >=< T2 > if and only if T1 ⊆< T2 > and T2 ⊆< T1 > Proof. Suppose first that < T1 >=< T2 > . Since T1 ⊆< T1 > and < T1 >=< T2 > then T1 ⊆< T2 > . Similarly, T2 ⊆< T1 > . Conversely, suppose that T1 ⊆< T2 > and T2 ⊆< T1 > . By Theorem 15.2(ii), we have < T1 >⊆< T2 > and < T2 >⊆< T1 > . Hence, < T1 >=< T2 > . Example 15.2 Consider (Z, +). Since {3} ⊆< 9, 12 > (3 = 12 + (−9)) and {9, 12} ⊆< 3 > then by the previous theorem we have < 9, 12 >=< 3 > . Example 15.3 In S4 we have that (124) = (123)(12)(34)(123) (234) = (132)(12)(34) = (123)−1(12)(34) (123) = (124)(234) (12)(34) = (234)(124)−1 Thus, we conclude that {(124), (134)} ⊆< (123), (12)(34) > and {(123), (12)(34)} ⊆< (124), (134) > . Hence, by Theorem 15.5, we have < (124), (234) >=< (123), (12)(34) > . 3 15.2 Direct Product of Groups. In this section we keep building examples of groups. By defining a binary operation on the Cartesian product of two groups we obtain the group known as the direct product of the two groups. Let H and K be two arbitrary groups and let H × K be the Cartesian product of H and K. In set-builder notation H × K = {(h, k): h ∈ H and k ∈ K}. Equality in H × K is defined by (h, k) = (h0, k0) if and only if h = h0 and k = k0. Define multiplication on H × K as follows: (h1, k1)(h2, k2) = (h1h2, k1k2) The set H × B is a group under the above multiplication as shown in the next theorem. Theorem 15.6 Multiplication on H × K satisfies the following properties: (i) H × K is closed under multiplication. (ii) Multiplication is associative. (iii) (eH , eK ) is the identity element. (iv) For each (h, k) ∈ H × K we have (h, k)−1 = (h−1, k−1). Proof. (i) We must show that the rule on (H ×K)×(H ×K) −→ H ×K defined by 0 0 0 0 0 0 ((h, k), (h , k )) −→ (hh , kk ) is a mapping. Indeed, if ((h1, k1), (h1, k1)) = 0 0 0 0 0 0 ((h2, k2), (h2, k2)) then (h1, k1) = (h2, k2) and (h1, k1) = (h2, k2). Thus, h1 = 0 0 0 0 0 0 0 0 h2, k1 = k2, h1 = h2, and k1 = k2. Therefore, h1h1 = h2h2 and k1k1 = k2k2. 0 0 0 0 Hence, (h1h1, k1k1) = (h2h2, k2k2). So multiplication is a binary operation. (ii) Multiplication on H × K is associative since multiplication is associative as operation on H and K. (h1, k1)[(h2, k2)(h3, k3)] = (h1, k1)(h2h3, k2k3) = (h1(h2h3), k1(k2k3)) = ((h1h2)h3, (k1k2)k3) = (h1h2, k1k2)(h3, k3) = [(h1, k1)(h2, k2)](h3, k3) 4 (iii) Let (h, k) ∈ H × K. Since heH = eH h = h and keK = eK k = k then (heH , keK ) = (eH h, eK k) = (h, k). Hence, (h, k)(eH , eK ) = (eH , eK )(h, k) = (h, k). That is, (eH , eK ) is the identity element of H × K. −1 −1 −1 −1 (iv) Let (h, k) ∈ H × K. Since hh = h h = eH and kk = k k = eK −1 −1 −1 −1 then (hh , kk ) = (eH , eK ). That is, (h, k)(h , k ) = (eH , eK ). Similarly, −1 −1 −1 −1 (h , k )(h, k) = (eH , eK ) so that (h, k) is invertible with inverse (h , k ) Definition 15.2 By the above theorem, H × K is a group, called the direct product of the groups H and K. Example 15.4 If Z3 = {[0], [1], [2]} and S2 = {(1), (12)} then Z3 × S2 = {([0], (1)), ([1], (1)), ([2], (1)), ([0], (12)), ([1], (12)), ([2], (12))} For example, ([1], (12))([2], (1)) = ([1] ⊕ [2], (12)(1)) = ([0], (12)). Some of the subgroups of H × K can be constructed as follows. Theorem 15.7 The sets H × {eK } = {(h, eK ): h ∈ H} and {eH } × K = {(eH , k)) : k ∈ K} are subgroups of H × K. Proof. We prove that H × {eK } is a subgroup of H × K and we leave the second part of the theorem as an exercise for the reader. See Exercise ?? Since (eH , eK ) ∈ H × {eK } then H × {eK } 6= ∅. Let (h1, eK ) and (h2, eK ) be two elements of H × {eK }. Then −1 −1 −1 (h1, eK )(h2, eK ) = (h1, eK )(h2 , eK ) = (h1h2 , eK ) ∈ H × {eK } −1 since h1h2 ∈ H. Hence, by Theorem 7.5, H × {eK } is a subgroup of H × K Remark 15.1 By the Principle of Counting, if H and K are finite sets then |H × K| = |H| · |K|. 5.