Exercises (§1.5, p.47)

Basic Algebra (Solutions)

by Huah Chu

Exercises (§1.5, p.47)

1. As in section 1.4, let C(A) denote the centralizer of the subset A of a monoid M (or a group G). Note that C(C(A)) ⊃ A and if A ⊂ B then C(A) ⊃ C(B). Show that these imply that C(C(C(A))) = C(A). Without using the explicit form of the elements of hAi show that C(A) = C(hAi). (Hint: Note that if c ∈ C(A) then A ⊂ C(c) and hence hAi ⊂ C(c).) Use the last result to show that if a monoid (or a group) is generated by a set of elements A which pairwise commute, then the monoid (group) is commutative.

Proof. (1) C(C(C(A))) = C(A):

¿From C(C(A)) ⊃ A, we have C(C(C(A))) ⊂ C(A). On the other hand, replacing A in C(C(A)) by C(A), we have C(C(C(A))) ⊃ C(A). Hence these two sets are equal.

(2) C(A) = C(hAi):

For any c ∈ C(A), A ⊂ C(c). Thus hAi ⊂ C(c) since C(c) is a monoid. hAi commute with every element of C(A), hence C(A) ⊂ C(hAi).

(3) If a monoid M is generated by S and st = ts for all s, t ∈ S, then M is commutative:

S ⊂ C(S) since the elements of S are commutative. Applying (2), we have C(S) = C(hSi) = C(M ) and hence S ⊂ C(M). Thus M = hSi ⊂ C(M) and M

is commutative. ¤

2. Let M be a monoid generated by a set S and suppose every element of S is invertible.

Show that M is a group.

Proof. Every element of M has the form s₁s₂· · · s_r, s_i ∈ S, which is invertible with inverse s⁻¹_r s⁻¹_r−1· · · s⁻¹₁ ∈ M. Hence M is a group. ¤ 3. Let G be a finitely generated abelian group which is periodic in the sense that all of its elements have finite order. Show that G is finite.

Proof. Suppose G is an abelian group generated by {g₁, . . . , g_n} and o(g_i) = `<sub>i</sub>. Then any element of G has the form g<sub>1</sub><sup>α1</sup> · · · g<sub>n</sub><sup>αn</sup> with 0 ≤ αi < `i. Hence |G| ≤ Q_n

i=1`i. ¤ 4. Show that if g is an element of a group and o(g) = n then g^k, k 6= 0, has order [n, k]/k = n/(n, k). Show that the number of generators of hgi is the number of positive

1

integers < n which are relatively prime to n. This number is denoted as φ(n) and φ is called the Euler φ-function.

Proof. (1) Let g ∈ G with o(g) = n and o(g^k) = h. It’s clear that (g^k)^[n,k]/k = 1. Thus h|[n, k]/k and hk|[n, k]. On the other hand, (g^k)^h = 1 implies n|hk. Hence hk is a common multiple of n and k and divides [n, k]. Therefore hk = [n, k], h = [n, k]/k = n/(n, k).

(2) g^k ∈ hgi is a generator of hgi if and only if o(g^k) = n. By the result of (1), o(g^k) = n/(n, k). Hence g^k is a generator if and only if (n, k) = 1. ¤

5. Show that any finitely generated subgroup of the additive group of rationals (Q, +, 0) is cyclic. Use this to prove that this group is not isomorphic to the direct product of two copies of it.

Proof. (1) Let H be the subgroup of (Q, +) generated by {_p^q1₁, . . . ,^q_pⁿ

n}. Then H is contained in the cyclic group h_p ¹

1···pni. Hence H is cyclic.

(2) Let H be the subgroup of Q ⊕ Q which is generated by (1,0) and (0,1). We shall show that H is not cyclic. Thus Q ⊕ Q 6' Q.

Suppose that H is cyclic with generator (a, b) 6= (0, 0). Then (1, 0) = n(a, b) and (0, 1) = m(a, b) for some n, m 6= 0. Which implies a = b = 0 and leads to a

contradiction. ¤

Remark. Given a finitely generated subgroup H = h_p^q1

1, . . . ,^q_pⁿ

ni, (p_i, q_i) = 1, of (Q, +), the reader is urged to construct a generator of H explicitly.

Exercise: Show that if o(a) = n = rs where (r, s) = 1, then hai ' hbi × hci where o(b) = r and o(c) = s. Hence prove that any finite cyclic group is isomorphic to a direct product of cyclic groups of prime power order.

2