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 = {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 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 |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 = xmnymn since G is an abelian group = (xn)m(ym)n = imin = i

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 = h x, y|x2 = 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 ∈ 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 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(D2n) = {1} if n is odd.

n b) C(D2n) = {1, r 2 } 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 |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 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 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 |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) =xlcm(n,m)ylcm(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) 2 2 take, for example, the subgroups h r i and h r i in D12. Clearly |r| = 6, |r | = 3, lcm(6, 3) = 6, but |r · r2| = |r3| = 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 ∈ 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 ∈ Z. We will show that q+1 ∈/ h q i. p p p p q p q If q ∈ h q i then there exist z ∈ Z such that z q = q+1 . Then, z = p · q+1 = q+1 . The latter 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. 1 Exhibit. For any prime p let Qep = h pz |z ∈ 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 ∈ A and each nonzero integer k there is an element x ∈ 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 ∈ Q let q = k . Observe that q ∈ 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 {ai}1 be an n enumeration of the elements in A and let d = lcm{{|ai|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 d ba such that ba = a. Since A is abelian ba has the form p1 p2 pn a1 a2 ··· an Therefore d dp1 dp2 dpn ba = a1 a2 ··· an dpi d Since the order of any element divides d we have that ai = 1 for all i. Hence ba = 1, and we arrive at the much desired contradiction.

4 Section 2.5 Exercise (8). In each of the following groups ﬁnd the normalizer of each subgroup a) S3. 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 S3. Hence

N(h (1 2) i) = h (1 2) i N(h (1 3) i) = h (1 3) i N(h (2 3) i) = h (1 2) i

For the element (1 2 3), we note that the conjugation by (1 2) preserves the group. Therefore (1 2) ∈ N(h (1 2 3) i and the only possible choice is that N(h (1 2 3) i = S3. b) Q8. Solution. We easily check that conjugation by any of the elements i, j, k preserves the group h −1 i, therefore N(h −1 i) = Q8. Similarly we easily check that i ∈ N(h j i), j ∈ N(h k i) and k ∈ N(h i i). Therefore

N(h i i) = N(h j i) = N(h k i) = Q8.