Algebra I Homework Three

Name:

Instruction: In the following questions, you should work out the solutions in a clear and concise manner. Three questions will be randomly selected and checked for correctness; they count 50% grades of this homework set. The other questions will be checked for completeness; they count the rest 50% grades of the homework set. Staple this sheet of paper as the cover page of your homework set.

1. (Section 2.1) A subset X of an abelian group F is said to be linearly independent if n1x1 + ··· + nrxr = 0 always implies ni = 0 for all i (where ni ∈ Z and x1, ··· , xk are distinct elements of X). (a) X is linearly independent if and only if every nonzero element of the subgroup hXi may be written uniquely in the form n1x1 + ··· + nkxk (ni ∈ Z, ni 6= 0, x1, ··· , xk distinct elements of X).

The set S := {nlx1 + ··· + nkxk | ni ∈ Z, x1, ··· , xk ∈ X, k ∈ N} forms a subgroup of F and it contains X. So hXi ≤ S. Conversely, Every element of S is in hXi. Thus hXi = S. If X is linearly independent, suppose on the contrary, an element a ∈ hXi can be expressed as two diﬀerent linear combinations of elements in X:

0 0 0 0 a = nlx1 + ··· + nkxk = nlx1 + ··· + nkxk, x1, ··· , xk ∈ X, ni, ni ∈ Z, ni 6= 0 or ni 6= 0 for i = 1, ··· , k.

0 0 0 0 0 Then (n1 − n1)x1 + ··· + (nk − nk)xk = 0, and at least one of ni − ni is non-zero (since (n1, ··· , nk) 6= (n1, ··· , nk)). So every element of hXi may be written uniquely in the form nlx1 + ··· + nkxk (ni ∈ Z, ni 6= 0, x1, ··· , xk distinct elements of X).

Conversely, if every element of hXi may be written uniquely in the form nlx1 + ··· + nkxk for ni ∈ Z and xi ∈ X,

then 0 ∈ hXi has only one such expression, namely 0 = 0x1 + ··· + 0xk for any x1, ··· , xk ∈ X. So X is linearly independent. (b) If F is free abelian of ﬁnite rank n, it is not true that every linearly independent subset of n elements is a basis. Consider F = Z, a free abelian group of rank 1. Then every {a} ⊂ Z\{0} is a linearly independent subset of Z, however, {a} is not a basis for a 6= ±1. (c) If F is free abelian, it is not true that every linearly independent subset of F may be extended to a basis of F . Again consider F = Z. Then {a} for a ∈ Z 6= {0, ±1} is a linearly independent subset that cannot be extended to a basis of F . (d) If F is free abelian, it is not true that every generating set of F contains a basis of F . However, if F is also ﬁnitely generated by n elements, F has rank m ≤ n. Consider F = Z, then {2, 3} is a generating set of F , but {2, 3} does not contain a basis of F .

If F is ﬁnitely generated by n elements x1, ··· , xn. Let X = {x1, ··· , xn}; let i : X → F denote the inclusion map; and let F (X) denote the free abelian group on X with a morphism ι : X → F (X). Then by the property of free objects, there exists an abelian group homomorphism p : F (X) → F such that p ◦ ι = i. The assumption that F is generated by X implies that p is an epimorphism. Then F (X)/ ker p ' F . If F has rank m, then it has a basis −1 −1 −1 {y1, ··· , ym}. Thus F (X) = ker p ⊕ hp (y1), p (y2), ··· , p (ym)i. The subgroup ker p of F (X) is also free abelian. Suppose ker p has rank r. Then n = r + m. So m ≤ n. 2 0 1 1 2. (Section 2.1) Let G be the multiplicative group generated by the real matrices a = and b = . If 0 1 0 1 H is the set of all matrices in G whose diagonal entries are 1, then H is an abelian subgroup that is not ﬁnitely generated. The groups n m m 2 2 x 1 2 x G = n, m, x ∈ Z ,H = m, x ∈ Z . 0 1 0 1 1 x Denote M(x) := . The subgroup H = {M(2mx) | m, x ∈ Z} of G is abelian since 0 1

M(x)M(y) = M(y)M(x) = M(x + y).

m1x1 mr xr m0−1 If H is generated by ﬁnitely many elements M(2 ), ··· ,M(2 ), let m0 := min{m1, ··· , mr}, then M(2 ) ∈ H but it is not a ﬁnite product of M(0),M(2m1x1 ), ··· ,M(2mr xr ) and their inverses, which is a contradiction.

3. (Section 2.1) Let F be free abelian of rank n and let H be a subgroup of the same rank. Let {x1, ··· , xn} be a basis Pn of F , let {y1, ··· , yn} be a basis of H, and let yj = i=1 mijxi. Prove that [F : H] = | det[mij]|.

1 We ﬁrst show that | det[mij ]| is independent of the choice of the basis of H. Let {z1, ··· , zn} be another basis of Hwith Pn Pn Pn zj = i=1 rij xi. Suppose that zj = i=1 pij yi and yj = i=1 qij zi for pij , qij ∈ Z. Then

n n n n n n ! X X X X X X zj = rij xi = pkj yk = pkj mikxi = mikpkj xi. i=1 k=1 k=1 i=1 i=1 k=1

Hence the matrix product [rij ] = [mij ][pij ]. Similar argument shows that [pij ][qij ] = In. Then | det[pij ]| · | det[qij ]| = 1. This forces | det[pij ]| = | det[qij ]| = 1 since pij , qij ∈ Z. Therefore, | det[rij ]| = | det[mij ]| · | det[pij ]| = | det[mij ]| as desired. Now F is a free abelian group of rank n and H is a subgroup of F of rank n. There exists a basis, say {x1, ··· , xn}, of F ,

such that H = hw1, ··· , wni where wi = dixi for some integers 1 ≤ d1 | d2 | · · · | dn. Then [G : H] = d1d2 ··· dn = det[mij ]. 4. (Section 2.2) Let G be a ﬁnite abelian group and x an element of maximal order. Show that hxi is a direct summand of G. Use this to obtain another proof of Theorem 2.2.1.

(a) Claim: If y ∈ G\{0}, then |y| divides |x|. Suppose on the contrary, there is y ∈ G whose order does not divide |x|. Let p1, ··· , pk be all prime factors of lcm(|x|, |y|). Then |x| and |y| can be written as

r1 rk s1 sk |x| = p1 ··· pk , |y| = p1 ··· pk , r1, ··· , rk, s1, ··· , sk ∈ N.

max(r1,s1) max(rk,sk) c Thus lcm(|x|, |y|) = p1 ··· pk . Let I := {i | 1 ≤ i ≤ k, ri ≥ si} and I = {1, 2, ··· , k} − I. Let Q ri Q sj a := i∈I pi and b = j∈Ic pj . Then a divides |x|, b divides |y|, gcd(a, b) = 1 and lcm(a, b) = ab = lcm(|x|, |y|). Obviously,

|x| |y| x + y = ab = lcm(|x|, |y|) > |x|. a b This contradicts to the maximality of the order of x. So |y| must divide |x|. (b) Claim: For any y ∈ G\{0}, there is c ∈ N such that hy − cxi ∩ hxi = h0i. If hyi ∩ hxi = h0i, then we are done. Otherwise, hyi ∩ hxi = hbyi for some b ∈ N∗ with b | |y|. Then by = ax for some a ∈ N. Thus |y| |y| |y| 0 = by = ax =⇒ a = |x|` for some ` ∈ N. b b b |x| |x| Then a = b` . Let c := ` . Then c ∈ N and |y| |y|

0 = by − ax = b (y − cx) .

Then hy − cxi ∩ hxi = hb(y − cx)i = h0i as desired. (c) Claim: G is a direct sum of cyclic subgroups in which hxi is a direct summand. We prove the claim by induction on |G|. When |G| = 1, the claim is true. Suppose that the claim is true for all abelian groups of order less than n. Given an abelian group G with |G| = n, the quotient group G/hxi has an order less than n. By induction hypothesis,

t M G/hxi = hy¯ii for some yi ∈ G, wherey ¯i := yi + hxi. i=1

By the preceding claim, for each yi, there is ci ∈ N such that hyi − cixi ∩ hxi = h0i. Let zi := yi − cix. Thenz ¯i =y ¯i. Lt Pt We claim that G = hxi ⊕ i=1hzii. On one hand G = hxi + i=1hzii. On the other hand, if there are d, d1, ··· , dt ∈ Z Pt Pt ¯ such that dx + i=1 dizi = 0, then i=1 diz¯i = 0 in G/hxi, which implies that |y¯i| | di for i = 1, ··· , t. Then for any i,

dizi = di(yi − cix) = diyi − dicix ∈ hyi − cixi ∩ hxi = h0i.

Lt Therefore, dizi = 0 for i = 1, ··· , t, and dx = 0. It shows that G = hxi ⊕ i=1hzii.. So the claim is true for abelian groups of any order. (d) The rest is easy.

5. (Section 2.2) A (sub)group in which every element has order a power of a ﬁxed prime p is called a p-(sub)group (note: |0| = 1 = p0). Let G be an abelian torsion group. Let G(p) = {u ∈ G | |u| = pn for some n ∈ N}.

(a) G(p) is the unique maximum p-subgroup of G. G(p) = {u ∈ G | |u| = pn for some n ≥ 0} is a abelian subgroup of G by Lemma 2.2.5. If H is a p-subgroup of G, then every h ∈ H has an order the power of p, which implies that h ∈ G(p). Therefore G(p) is the unique maximum p-subgroup of G.

2 (b) G = P G(p), where the sum is over all primes p such that G(p) 6= 0. Qr ni + Suppose that an element x of G has the order m = i=1 pi where pi are distinct primes and ni ∈ Z . Let ni ni mi = m/pi for i = 1, ··· , r. The order of mix is pi , so that mix ∈ G(pi). Since gcd(m1, ··· , mr) = 1, there are Pr Pr P P integers c1, ··· , cr such that i=1 cimi = 1. Then x = i=1 ci(mix) ∈ G(pi) ⊆ G(p). This completes the proof.

(c) If H is another abelian torsion group, then G ' H if and only if G(p) ' H(p) for all primes p. If ϕ : G → H is a group homomorphism, then ϕ sends G(p) to H(p) for all primes p. Therefore, G ' H implies that G(p) ' H(p) for all primes p. Conversely, the above sum G = P G(p) is easily seen to be a direct sum. So if G(p) ' H(p) for all primes p, then their coproducts P G(p) ' P H(p), namely G ' H.

6. (Section 2.2)

(a) What are the elementary divisors of the groups Z2 ⊕ Z9 ⊕ Z35? What are its invariant factors? Do the same for Z26 ⊕ Z42 ⊕ Z49 ⊕ Z200 ⊕ Z1000.

The elementary divisors of Z2 ⊕ Z9 ⊕ Z35 are

21, 32, 51, 71.

The invariant factors of Z2 ⊕ Z9 ⊕ Z35 are 21 × 32 × 51 × 71 = 630. The prime decompositions:

26 = 2 · 13, 42 = 2 · 3 · 7, 49 = 72, 200 = 23 · 52, 1000 = 23 · 53.

So the elementary divisors of Z26 ⊕ Z42 ⊕ Z49 ⊕ Z200 ⊕ Z1000 are:

2, 2, 23, 23, 3, 52, 53, 7, 72, 13.

The invariant factors of Z26 ⊕ Z42 ⊕ Z49 ⊕ Z200 ⊕ Z1000 are:

2, 2, 23 · 52 · 7 = 1400, 23 · 3 · 53 · 72 · 13 = 191100.

(b) Determine all abelian groups of order n for n ≤ 20.

n abelian groups of order n 1, 2, 3, 5, 6, 7, 10, 11, Zn 13, 14, 15, 17, 18, 19 4 Z2 × Z2, Z4 8 Z2 × Z2 × Z2, Z2 × Z4, Z8 9 Z3 × Z3, Z9 12 Z2 × Z2 × Z3, Z4 × Z3 16 Z2 × Z2 × Z2 × Z2, Z2 × Z2 × Z4, Z2 × Z8, Z16 20 Z2 × Z2 × Z5, Z4 × Z5

7. (Section 2.3) (a) Z satisﬁes the ACC but not the DCC on subgroups. A subgroup of Z is of the form aZ for some a ∈ N. Furthermore, aZ ≤ bZ if and only if b | a. Given an ascending chain of subgroups: a1Z ≤ a2Z ≤ · · · (ai ∈ N), the integers an+1 is a factor of an and thus a1 ≥ a2 ≥ · · · . There is a + n ∈ Z such that an = an+1 = ··· . Therefore, Z satisﬁes the ACC. However, Z does not satisfy DCC on subgroups. There is a strictly descending chain:

Z > 2Z > 22Z > ··· > 2nZ > ···

(b) For each prime p the group Z(p∞) = {a/b ∈ Q/Z | b = pn for some n ∈ N} satisﬁes the DCC but not the ACC on subgroups. Let H be a subgroup of Z(p∞). If a/pn ∈ H where gcd(a, p) = 1, then ha/pni = h1/pni = {c/bk ∈ Q/Z | k ≤ n}. Therefore, the subgroup diagram of Z(p∞) are:

h0i < h1/p1i < h1/p2i < ··· < h1/pni < ··· < Z(p∞).

Clearly, Z(p∞) satisﬁes the DCC but not the ACC on subgroups.

3 (c) Every finitely generated abelian group satisfies the ACC on subgroups. r Every finitely generated abelian group G is isomorphic to Gτ ⊕ Z , where Gτ is the torsion subgroup of G. As a r finite group, Gτ satisfies the ACC. Also, Z satisfies the ACC. Therefore, the direct sum Gτ ⊕ Z ' G satisfies the ACC.

8. (Section 2.3) If G, H, K and J are groups such that G ' H × K and G ' H × J and G satisﬁes both the ACC and DCC on normal subgroups, then K ' J. Suppose that ϕ : H × K → G and ψ : H × J → G are two isomorphisms. Since G satisﬁes both the ACC and DCC, so is H and K. By Krull-Schmidt Theorem, H and K can be written as direct products of indecomposable subgroups:

H = H1 × H2 × · · · × Hp,K = K1 × K2 × · · · × Kq.

Then

G = ϕ(H1) × · · · × ϕ(Hp) × ϕ(K1) × · · · × ϕ(Kq)

= ψ(H1) × · · · × ψ(Hp) × ψ(J1) × · · · × ψ(Jq).

By Krull-Schmidt Theorem, after reindexing ψ(H1), ··· , ψ(Hp), ψ(J1), ··· , ψ(Jq), say L1, ··· ,Lp+q, we can let ϕ(Hi) ' Li for i = 1, ··· , p and ϕ(Kj ) ' Lp+j for j = 1, ··· , q; moreover, we may replace L1 × · · · × Lp by ϕ(H1) × · · · × ϕ(Hp) so that G = ϕ(H1) × · · · × ϕ(Hp) × Lp+1 × · · · × Lp+q. Then

Lp+1 × · · · × Lp+q ' G/(ϕ(H1) × · · · × ϕ(Hp)) ' ϕ(K1) × · · · × ϕ(Kq) ' K,

Lp+1 × · · · × Lp+q ' G/(L1 × · · · × Lp) ' G/(ψ(H1) × · · · × ψ(Hp)) ' ψ(J1) × · · · × ψ(Jq) ' J.

Therefore K ' J.

9. (Section 2.4) Let G be a group and let In G be the set of all inner automorphisms of G. Show that In G is a normal subgroup of Aut G. −1 The set In G consists of the automorphisms ψx : G → G deﬁned by g 7→ xgx , where x ∈ G and g ∈ G. Then −1 −1 −1 ψx ◦ ψy−1 (g) = ψx(y gy) = xy gyx = ψxy−1 (g) for x, y, g ∈ G. Hence In G is a subgroup of Aut G. If σ ∈ Aut G, then

−1 −1 −1 −1 σ ◦ ψx ◦ σ (g) = σ(xσ (g)x ) = σ(x)gσ(x) = ψσ(x)(g), for x, g ∈ G.

−1 This shows that σ ◦ ψx ◦ σ ∈ In G. Hence In G is a normal subgroup of Aut G. 10. (Section 2.4) If |G| = pn, with p > n, p prime, and H is a subgroup of order p, then H is normal in G. Let S be the set of all subgroups of G of order p: S := {K ≤ G | |K| = p}. Then H has a conjugate action on S: −1 h(K) := hKh for h ∈ H and K ∈ S. The number of elements in the H-orbit of K ∈ S is [H : HK ], a factor of |H| = p,

where HK is the isotropy group of K. So [H : HK ] = 1 or p. If K ∈ S\{H} is ﬁxed by the H-action, i.e. HK = H, then 2 K C hH,Ki = HK and |hH,Ki/K| = p. This implies that |hH,Ki| = p , a contradiction to |G| = np. Therefore, the H-orbit P of every K ∈ S\{K} has p elements. Notice that |S| = K [H : HK ] where K is selected from each H-orbit in S. We have |S| = p` + 1 for some ` ∈ N. Every K ∈ S has p − 1 elements of order p. So G contains (p` + 1)(p − 1) elements of order p. This forces ` = 0 since |G| = pn ≤ p(p − 1). Then |S| = 1 and H is the only subgroup of order p in G. Hence xHx−1 = H for any x ∈ G. Therefore H C G.