Mathematics 352 – Abstract Algebra II Solutions for Problem Set 1 January 21, 2008

Section 8.2

1. Let G be an abelian group. By definition, G(p) = {a ∈ G : p^{ℓ} · a = 0 for some ℓ ≥ 0}. Let a, b ∈ G(p). Then p^{ℓ} · a = 0 and p^{m} · b = 0 for some ℓ, m ≥ 0. Let n = max{ℓ, m}. Then p^{n} · a = p^{n} · b = 0 so

p^{n} · (a − b) = p^{n} · a − p^{n} · b = 0 − 0 = 0.

This shows that a − b ∈ G(p). Hence G(p) is a subgroup of G.

2. Let G be an abelian group and let pG = {p · x : x ∈ G}. Let a = p · x and b = p · y be elements of pG, where x, y ∈ G. Then a − b = p · x − p · y = p · (x − y) and x − y ∈ G since G is a group under +. Hence a − b ∈ pG, so pG is a subgroup.

3. (d) We have 72 = 2^{3}·3^{2}. As in the example on p. 256 in Hungerford or the examples done in class last Friday, the structure theorem implies every abelian group of order 72 has elementary divisors 2^{3}, 3^{2} or 2^{3}, 3, 3, or 2^{2}, 2, 3^{2}, or 2^{2}, 2, 3, 3, or 2, 2, 2, 3^{2}, or 2, 2, 2, 3, 3. The group is isomorphic to one of

ℤ_{8} ⊕ ℤ_{9},  ℤ_{8} ⊕ ℤ_{3} ⊕ ℤ_{3},
ℤ_{4} ⊕ ℤ_{2} ⊕ ℤ_{9},  ℤ_{4} ⊕ ℤ_{2} ⊕ ℤ_{3} ⊕ ℤ_{3},
ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{9}, or ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{3} ⊕ ℤ_{3}.

(h) Similarly, 1160 = 2^{3} · 5 · 29, so every abelian group of order 1160 is isomorphic to one of

ℤ_{8} ⊕ ℤ_{5} ⊕ ℤ_{29}, ℤ_{4} ⊕ ℤ_{2} ⊕ ℤ_{5} ⊕ ℤ_{29}, or ℤ_{2} ⊕ Z_{2} ⊕ ℤ_{2} ⊕ ℤ_{5} ⊕ ℤ_{29}.

5 (b) We can use the fact that if (m, n) = 1, then ℤ_{mn} ≃ ℤ_{m} ⊕ ℤ_{n} (Lemma 8.8). Applying this,

ℤ_{6} ⊕ ℤ_{12} ⊕ ℤ_{18} ≃ (ℤ_{2} ⊕ ℤ_{3}) ⊕ (ℤ_{4} ⊕ ℤ_{3}) ⊕ (ℤ_{2} ⊕ ℤ_{9}).

Since the elementary divisors are unique, this shows the elementary divisors of ℤ_{6} ⊕ ℤ_{12} ⊕ ℤ_{18} are 2, 2, 4, 3, 3, 9 (reordering to put powers in increasing order).

Comment: Note that this also gives the invariant factors of this group, which is isomorphic to ℤ_{6} ⊕ ℤ_{6} ⊕ ℤ_{36}.

8. By definition,

G(2) = \{[\frac{m}{n}] ∈ ℚ/ℤ : 2^{ℓ} · [\frac{m}{n}] = [0] for some ℓ ≥ 0\} .

This is equivalent to saying that 2^{ℓ} \frac{m}{n} is an integer. If \frac{m}{n} is written in lowest terms, then n = 2^{ℓ} for some ℓ ≥ 0, and hence G(2) is the set of cosets of rational numbers of the form \frac{m}{2^{ℓ}} in lowest terms. Similarly, G(p) is the set of cosets of rational numbers of the form \frac{m}{p^{ℓ}} in lowest terms.

9. (a) Saying G is a p-group means by definition that every element of G has order p^{ℓ} for some ℓ ≥ 0. If G is also a finite group, then there is some element m ∈ G of maximal order p^{ℓ_{0}} where ℓ_{0} ≥ 1. Now consider the group pG as in Exercise 2 above. If x has order p^{ℓ} with ℓ ≥ 1, then p^{ℓ−1} · (p · x) = p^{ℓ} · x = 0, but p^{k} · (p · x) = p^{k+1} · x ≠ 0 if k < ℓ − 1. Hence the order of p · x is ℓ − 1. This shows that pG cannot contain any elements of order ℓ_{0}. Hence pG ≠ G.

(b) Let G = ℚ/ℤ as in Exercise 8. G(2) is the subgroup consisting of cosets of rational numbers of the form \frac{m}{2^{ℓ}} for all ℓ ≥ 0. Note that G(2) is not a finite group because ℓ can be arbitrarily large. And moreover, if x = [\frac{m}{2^{ℓ}}] ∈ G(2) then x = 2 · [\frac{m}{2^{ℓ+1}} ] and y = [\frac{m}{2^{ℓ+1}} ] ∈ G(2). This shows that G(2) ⊆ 2 · G(2). The other inclusion is clear, so in fact G(2) = 2 · G(2).

11. By the structure theorem (Theorem 8.7), G is isomorphic to a direct sum of subgroups, each of order q^{e} for some primes q and exponents e ≥ 1. If G is a p-group, then only powers of one prime can appear among the elementary divisors, and

G ≃ ℤ_{p^{e_{1}}} ⊕ ⋯ ⊕ ℤ_{p^{es}}  (1)

for some e_{i} ≥ 1. What we must show is that if pG = {0}, then all the e_{i} = 1. Assume not. That is, suppose e_{i} ≥ 2 for some i. Then let a be a generator of Z_{p^{e_{i}}}, so |a| = p^{e_{i}} > p. In the direct sum we have the element x = (0, …, 0, a, 0, …, 0) (a in the ith factor). Then p · x = (0, …, 0, p · a, 0, …, 0) ≠ (0, …, 0). Hence the group pG ≠ {0}. By contraposition, if pG = {0}, then all e_{i} = 1 in Eq. (1).

12. Let G be a finite abelian group and let p | |G|. By the structure theorem (Theorem 8.7), G is isomorphic to a direct sum of cyclic groups of prime power orders. Each of those orders must divide |G| and hence at least one must be a power of p (since otherwise the order of G would be a product of powers of primes different from p and that would contradict p | |G|). Hence

G ≃ ℤ_{p^{k}} ⊕ H_{2} ⊕ ⋯ ⊕ H_{s},

for some k ≥ 1 and some cyclic groups H_{2}, …, H_{s} of prime power orders. Hence G contains a subgroup isomorphic to ℤ_{p^{k}} for some k ≥ 1. Let a be a generator of this subgroup, so |a| = p^{k}. By the same sort of argument used in Exercise 9 (a) above, we have |p^{k−1} · a| = p (also see part (4) of Theorem 7.8). Hence G contains an element of order p.

14. We are given that

|G| = p^{t}m  where (p, m) = 1.  (2)

By Theorem 8.5, we also have G ≃ G(p) ⊕ H where H is the direct sum of the G(q) for the primes q ≠ p dividing |G|, which implies |G| = |G(p)| · |H|. By Lemma 8.6, it follows that |G(p)| = p^{s} for some s and |H| is a product of powers of primes q ≠ p. Hence

|G| = p^{s}m′  where (p, m′) = 1.  (3)

By unique factorization in ℤ, comparing Eq. (2) and Eq. (3), we see that t = s, and m = m′. Therefore, |G(p)| = p^{t} which is what we wanted to show.

15. If |G| = p^{t}m with (p, m) = 1, then Exercise 14 shows that G contains the subgroup G(p) of order p^{t}. In other words, the statement to be proved is true for n = t, the largest power of p dividing |G|. The statement is also clearly true with n = 0, since {0} is a subgroup of G.

What remains to be proved is that the same statement is true for all n with 1 ≤ n < t as well. It suffices to show that G(p) contains a subgroup of order p^{n} since a subgroup of G(p) is also a subgroup of G. We know

G(p) ≃ ℤ_{p^{e_{1}}} ⊕ ⋯ ⊕ ℤ_{p^{es}}

for some e_{i} ≥ 1 satisfying e_{1} + ⋯ + e_{s} = t. The cyclic group ℤ_{p^{e}} contains elements of orders 1, p, p^{2}, …, p^{e} since if a is a generator with order p^{e}, then p^{e−j} · a has order p^{j} whenever 0 ≤ j ≤ e. Now, if 1 ≤ n < t, then we can always write n = f_{1} + ⋯ + f_{s} where 0 ≤ f_{i} ≤ e_{i} for all i. (In fact, we always can do this in many different ways when s > 1). If a_{i} is a generator of ℤ_{p^{e_{i}}}, then that summand contains a subgroup K_{i} = ⟨p^{e_{i}−f_{i}} · a_{i}⟩ of order p^{f_{i}} by the observation above. Hence G(p) contains the subgroup K = K_{1} ⊕ ⋯ ⊕ K_{s} which has order p^{f_{1}+···+f_{s}} = p^{n}.

16. We claim that this is equivalent to saying that n = p_{1}p_{2} ⋯ p_{s} for some distinct primes p_{1}, …, p_{s}. Such integers are called square-free numbers. If n has this form, then there is just one possibility for the elementary divisors of G, namely the set p_{1}, p_{2}, …, p_{s}. By Lemma 8.8 (applied repeatedly), we have

G ≃ ℤ_{p_{1}} ⊕ ⋯ ⊕ ℤ_{p_{s}} ≃ ℤ_{p_{1}···p_{s}} = ℤ_{n},

and there is just one abelian group of order n up to isomorphism. Conversely, suppose there is just one abelian group of order n up to isomorphism. If any prime p satisfies p^{2} | |G|, then there are at least two different possibilities for the elementary divisors of G and hence there are at least two nonisomorphic abelian groups of order n. Hence for each prime p that divides |G|, we must have p^{2} |̸ |G|. Hence n is square-free.