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Selected Exercises from Abstract Algebra by Dummit and Foote (3Rd Edition)

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Selected Exercises from Abstract Algebra by Dummit and Foote (3Rd Edition) Selected exercises from Abstract Algebra by Dummit and Foote (3rd edition). Bryan F´elix Abril 12, 2017 Section 2.1 Exercise (6). Let G be an abelian group. Prove that T = fg 2 Gjjgj < 1g is a subgroup of G. Give an explicit example where this set is not a subgroup when G is non-abelian. Proof. i. T is non-empty. The identity i of G is an element of T . ii. T is closed under inverses. If x 2 T then there exist n < 1 such that xn = i. Then i = xn x−1 = xnx−1 (x−1)n = (xnx−1)n = ix−n since x commutes with itself = xnx−n since xn = i = i This shows that jx−1j ≤ n and therefore x−1 2 T . iii. T is closed under products. Take arbitrary x and y in T and let jxj = n and jyj = m. Then (xy)mn = xmnymn since G is an abelian group = (xn)m(ym)n = imin = i Therefore jxyj ≤ mn and xy 2 T We conclude thhat T is a subgroup. Now we exhibit an example where T fails to be a subgroup when G is non-abelian. Let G = h x; yjx2 = y2 = i i. Observe that the product xy has infinite order since the product (xy)(xy) is no longer reducible because we lack the commutativity of x with y. Hence, x; y 2 T but (xy) 2= T . Exercise (7). Fix some n 2 Z with n > 1. Find the torsion subgroup of Z×(Z=nZ). Show that the set of elements of infinite order together with the identity is not a subgroup of this direct product. 1 Solution. It easy to see that the only element of finite order in Z is the identity, viz. 0. Furthermore every element of Z=nZ has finite order (since Z=nZ is a finite group). Therefore the torsion group T of Z × (Z=nZ) is given by f0g × f0; 1; : : : ; n − 1g. Now we show the second part of the problem. Observe that (1; 1) and (−1; n) have infinite order. The product (1; 1) + (−1; 0) = (0; 1) however, has finite order (j(0; 1)j = n. Hence, the proposed set is not a subgroup. Section 2.2 Exercise (7). Let n 2 Z with n ≥ 3. Prove the following: a) C(D2n) = f1g if n is odd. n b) C(D2n) = f1; r 2 g for n even. Proof. Observe that elements of the form sri do not commute with r for 0 ≤ i < n . Multiplying on the left yields (sri)r = (s)ri+1 and, on the other hand, right multiplication gives r(sri) = s(ri−1). These expressions are equal only if ri+1 = ri−1 or equivalently r2 = 1 which is not the case (since n ≥ 3). Now we inspect elements of the form ri. Note that: ris = sri , sr−i = sri , r−i = ri , 1 = r2i , 1 = (ri)2 n The only element in D2n with this property is r 2 for even n. It is left to check that for n even n i r 2 commutes with sr for 0 ≤< i < n. Observe that: n i i n r 2 (sr ) = (sr )r 2 −n i i n , s(r 2 r ) = s(r r 2 ) −n i i n , r 2 r = r r 2 −n n , r 2 = r 2 since powers of r commute Which is true by construction of n. Exercise (10). Let H be a group of order 2 in G. Show that N(H) = C(H). Deduce that if N(H) = G then H ≤ C(G). Proof. Since C(H) ≤ N(H) it is only left to show that N(H) ⊂ C(H). First, note that since jHj = 2, H = f1; hg where h is an element of order 2. Now let g 2 N(H), then gHg−1 = H. Since g1g−1 = 1, it must be the case that ghg−1 = h. Therefore g commutes with H and g 2 C(H) as desired. In particular if N(H) = G every element of G commutes with h. Therefore h 2 C(G) and hence H ≤ C(G). 2 Section 2.3 Exercise (5). Find the number of generators for Z=49000Z. Solution. An element z 2 Z=49000Z is a generator of the group if and only if gcd(z; 49000) = 1. Therefore we compute the number of relative prime numbers of 49000 by mean of the Euler φ function. i.e. φ(49000) =φ(23 · 53 · 72) =(23 − 22)(53 − 52)(72 − 7) =16800 Exercise (16). Assume jxj = n and jyj = m. Suppose x and y commute. Prove that jxyj divides the least common multiple of m and n. Need this to be true if x and y do not commute? Give an example of elements x; y such that the order of xy is not equal to the least common multiple of jxj and jyj. Proof. Observe that (xy)lcm(n;m) = 1. Why?! because (xy)lcm(n;m) =xlcm(n;m)ylcm(n;m) since x and y commute =1 · 1 since n; mjlcm(n; m): Therefore jxyj divides lcm(m; n) by Proposition 3. Observe that this is not the case for elements that do not commute. We exhibited such a case at the end of problem 6 in section 2.1. Also, the order of xy need not to be lcm(m; n) 2 2 take, for example, the subgroups h r i and h r i in D12. Clearly jrj = 6, jr j = 3, lcm(6; 3) = 6, but jr · r2j = jr3j = 2. Exercise (23). Show that (Z=2nZ)× is not cyclic for any n ≥ 3. Proof. We follow the hint and we exhibit two distinct subgroups of (Z=2nZ)× with order 2. For n ≥ 3 take the subgroups generated by 2n − 1 and 2n−1 + 1. It's easy to see that the former has order 2 since 2n − 1 ≡ −1 mod 2n. For the latter note that (2n−1 + 1)2 ≡(2n−1 + 1)(2n−1 + 1) mod 2n ≡22n−2 + 2n + 1 mod 2n ≡22n−2 + 1 mod 2n ≡1 mod 2n since 2n divides 22n−2 for n ≥ 3 Hence h 2n−1 + 1 i has order 2 as desired. Section 2.4 Exercise (11). Prove that SL2(F3)and S4 are two nonisomorphic groups of order 24. 2 1 Proof. Note that 2 SL ( ) has order 6. However, the representative elements of S 0 2 2 F3 4 (up to relabeling) 1 (1 2) (1 2)(3 4) (1 2 3) (1 2 3 4) have orders 1, 2, 3, and 4. Thus SL2(F3) is not isomorphic to S4. 3 Exercise (14). A group H is called finitely generated if there is a finite set A such that H = h A i. a) Prove that every finite group is finitely generated. Proof. For a finite group G let A = G. Then G = h A i. b) Prove that Z is finitely generated. Proof. Observe that Z = h −1; 1 i. c) Prove that every finitely generated subgroup of the additive group Q is cyclic. Proof. Let H be a finitely generated subgroup of ; i.e. H = h ± p1 ; ± p2 ;:::; ± pn i where Q q1 q2 qn pi pi q1q2:::qn each pi and qi are elements of . Observe that = · where the latter factor Z qi q1q2···qn qi is an element of Z. Therefore the group H is a subset (and a subgroup, by construction) of h ± 1 i. Since the latter is a cyclic group, it is isomorphic to . We have proved earlier q1q2···qn Z that any subgroup of Z is cyclic. Hence, H is a cyclic group. d) Prove that Q is not finitely generated. Proof. We proceed by contradiction. Assume Q is finitely generated. Then, by the previous p p p exercise Q is cyclic and has a generator q where p; q 2 Z. We will show that q+1 2= h q i. p p p p q p q If q 2 h q i then there exist z 2 Z such that z q = q+1 . Then, z = p · q+1 = q+1 . The latter is irreducible and implies that z2 = Z. Thus, we arrive at a contradiction. Exercise (15). Exhibit a proper subgroup of Q which is not cyclic. 1 Exhibit. For any prime p let Qep = h pz jz 2 Z i. It is easy to check that Qep is a group. Associativity is inherited form Q. The identity, namely 0 is a member of the set. Inverses have n n m n·pj +m·pi the form (−1) · pz . Lastly, it is closed under addition ( pi + pj = pi+j ). Exercise (19). A nontrivial group is called divisible if for each element a 2 A and each nonzero integer k there is an element x 2 A such that xk = a, i.e., each element has a kth root in A. a) Prove that the additive group of rational numbers, Q, is divisible. 0 q 0 Proof. For any non zero integer k and any q 2 Q let q = k . Observe that q 2 Q an that kq0 = q. b) Prove that no finite abelian group is divisible. n Proof. Observe that any element in a finite abelian group A has finite order. Let faig1 be an n enumeration of the elements in A and let d = lcmffjaij1 gg; i.e. the least common multiple of the order of all the elements in A.
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