# Selected exercises from Abstract Algebra by Dummit and Foote (3rd edition).

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1 Selected exercises from Abstract Algebra by Dummit and Foote (3rd edition). Bryan Félix Abril 12, 2017 Section 2.1 Exercise (6). Let G be an abelian group. Prove that T = {g G g < } 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 T then there exist n < such that x n = i. Then i = x n x 1 = x n x 1 (x 1 ) n = (x n x 1 ) n = ix n since x commutes with itself = x n x n since x n = i = i This shows that x 1 n and therefore x 1 T. iii. T is closed under products. Take arbitrary x and y in T and let x = n and y = m. Then (xy) mn = x mn y mn = (x n ) m (y m ) n = i m i n = i since G is an abelian group Therefore xy mn and xy 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 = x, y x 2 = y 2 = 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 T but (xy) / T. Exercise (7). Fix some n 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

2 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 {0} {0, 1,..., n 1}. 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 ( (0, 1) = n. Hence, the proposed set is not a subgroup. Section 2.2 Exercise (7). Let n Z with n 3. Prove the following: a) C(D 2n ) = {1} if n is odd. b) C(D 2n ) = {1, r n 2 } for n even. Proof. Observe that elements of the form sr i do not commute with r for 0 i < n. Multiplying on the left yields (sr i )r = (s)r i+1 and, on the other hand, right multiplication gives r(sr i ) = s(r i 1 ). These expressions are equal only if r i+1 = r i 1 or equivalently r 2 = 1 which is not the case (since n 3). Now we inspect elements of the form r i. Note that: r i s = sr i sr i = sr i r i = r i 1 = r 2i 1 = (r i ) 2 The only element in D 2n with this property is r n 2 for even n. It is left to check that for n even r n 2 commutes with sr i for 0 < i < n. Observe that: r n 2 (sr i ) = (sr i )r n 2 s(r n 2 r i ) = s(r i r n 2 ) r n 2 r i = r i r n 2 r n 2 = r n 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 H = 2, H = {1, h} where h is an element of order 2. Now let g 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 C(H) as desired. In particular if N(H) = G every element of G commutes with h. Therefore h C(G) and hence H C(G). 2

3 Section 2.3 Exercise (5). Find the number of generators for Z/49000Z. Solution. An element z 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 by mean of the Euler φ function. i.e. φ(49000) =φ( ) =( )( )(7 2 7) =16800 Exercise (16). Assume x = n and y = m. Suppose x and y commute. Prove that xy 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 x and y. Proof. Observe that (xy) lcm(n,m) = 1. Why?! because (xy) lcm(n,m) =x lcm(n,m) y lcm(n,m) since x and y commute =1 1 since n, m lcm(n, m). Therefore xy 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) take, for example, the subgroups r and r 2 in D 12. Clearly r = 6, r 2 = 3, lcm(6, 3) = 6, but r r 2 = r 3 = 2. Exercise (23). Show that (Z/2 n Z) is not cyclic for any n 3. Proof. We follow the hint and we exhibit two distinct subgroups of (Z/2 n Z) with order 2. For n 3 take the subgroups generated by 2 n 1 and 2 n It s easy to see that the former has order 2 since 2 n 1 1 mod 2 n. For the latter note that (2 n 1 + 1) 2 (2 n 1 + 1)(2 n 1 + 1) mod 2 n 2 2n n n Hence 2 n has order 2 as desired. Section 2.4 mod 2 n mod 2 n 1 mod 2 n since 2 n divides 2 2n 2 for n 3 Exercise (11). Prove that SL 2 (F 3 )and S 4 are two nonisomorphic groups of order 24. ( ) 2 1 Proof. Note that SL (F 3 ) has order 6. However, the representative elements of S 4 (up to relabeling) 1 (1 2) (1 2)(3 4) (1 2 3) ( ) have orders 1, 2, 3, and 4. Thus SL 2 (F 3 ) is not isomorphic to S 4. 3

4 Exercise (14). A group H is called finitely generated if there is a finite set A such that H = A. a) Prove that every finite group is finitely generated. Proof. For a finite group G let A = G. Then G = A. b) Prove that Z is finitely generated. Proof. Observe that Z = 1, 1. c) Prove that every finitely generated subgroup of the additive group Q is cyclic. Proof. Let H be a finitely generated subgroup of Q; i.e. H = ± p 1 q 1, ± p 2 q 2,..., ± pn q n where each p i and q i are elements of Z. Observe that p i q i = p i q 1 q 2 q n q1q 2...q n q i where the latter factor is an element of Z. Therefore the group H is a subset (and a subgroup, by construction) of ± 1 q 1 q 2 q n. Since the latter is a cyclic group, it is isomorphic to Z. We have proved earlier 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 exercise Q is cyclic and has a generator p p where p, q Z. We will show that / p. q q+1 q If p p then there exist z Z such that z p = p. Then, z = q p = q. The latter q q q q+1 p q+1 q+1 is irreducible and implies that z / Z. Thus, we arrive at a contradiction. Exercise (15). Exhibit a proper subgroup of Q which is not cyclic. Exhibit. For any prime p let Q p = 1 z Z. It is easy to check that Q p z p is a group. Associativity is inherited form Q. The identity, namely 0 is a member of the set. Inverses have the form ( 1) n p z. Lastly, it is closed under addition ( n p i + m p j = n pj +m p i p i+j ). Exercise (19). A nontrivial group is called divisible if for each element a A and each nonzero integer k there is an element x A such that x k = a, i.e., each element has a k th root in A. a) Prove that the additive group of rational numbers, Q, is divisible. Proof. For any non zero integer k and any q Q let q = q k. Observe that q Q an that kq = q. b) Prove that no finite abelian group is divisible. Proof. Observe that any element in a finite abelian group A has finite order. Let {a i } n 1 be an enumeration of the elements in A and let d = lcm{{ a i n 1}}; i.e. the least common multiple of the order of all the elements in A. We prove that no element (other than the identity) has a d root. We proceed by contradiction. Let a A be a non identity element, and assume there exist â such that â d = a. Since A is abelian â has the form Therefore a p 1 1 a p 2 2 a pn n â d = a dp 1 1 a dp 2 2 a dpn n Since the order of any element divides d we have that a dp i i we arrive at the much desired contradiction. = 1 for all i. Hence â d = 1, and 4

5 Section 2.5 Exercise (8). In each of the following groups find the normalizer of each subgroup a) S 3. Solution. For any of the subgroups generated by a permutation of length 2, the element (1 2 3) does not preserve the group under conjugation, therefore the normalizer can not be the entire group S 3. Hence N( (1 2) ) = (1 2) N( (1 3) ) = (1 3) N( (2 3) ) = (1 2) For the element (1 2 3), we note that the conjugation by (1 2) preserves the group. Therefore (1 2) N( (1 2 3) and the only possible choice is that N( (1 2 3) = S 3. b) Q 8. Solution. We easily check that conjugation by any of the elements i, j, k preserves the group 1, therefore N( 1 ) = Q 8. Similarly we easily check that i N( j ), j N( k ) and k N( i ). Therefore N( i ) = N( j ) = N( k ) = Q 8. 5

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