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Algebra I Homework Three 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 2 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 2 Z, ni 6= 0, x1; ··· ; xk distinct elements of X). The set S := fnlx1 + ··· + nkxk j ni 2 Z; x1; ··· ; xk 2 X; k 2 Ng 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 2 hXi can be expressed as two different linear combinations of elements in X: 0 0 0 0 a = nlx1 + ··· + nkxk = nlx1 + ··· + nkxk; x1; ··· ; xk 2 X; ni; ni 2 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 2 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 2 Z and xi 2 X, then 0 2 hXi has only one such expression, namely 0 = 0x1 + ··· + 0xk for any x1; ··· ; xk 2 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 fag ⊂ Znf0g is a linearly independent subset of Z, however, fag 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 fag for a 2 Z 6= f0; ±1g 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 f2; 3g is a generating set of F , but f2; 3g does not contain a basis of F . If F is finitely generated by n elements x1; ··· ; xn. Let X = fx1; ··· ; xng; 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 fy1; ··· ; ymg. 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 2 Z ;H = m; x 2 Z : 0 1 0 1 1 x Denote M(x) := . The subgroup H = fM(2mx) j m; x 2 Zg 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 := minfm1; ··· ; mrg, then M(2 ) 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 fx1; ··· ; xng be a basis Pn of F , let fy1; ··· ; yng be a basis of H, and let yj = i=1 mijxi. Prove that [F : H] = j det[mij]j. 1 We first show that j det[mij ]j is independent of the choice of the basis of H. Let fz1; ··· ; zng 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 2 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 j det[pij ]j · j det[qij ]j = 1. This forces j det[pij ]j = j det[qij ]j = 1 since pij ; qij 2 Z. Therefore, j det[rij ]j = j det[mij ]j · j det[pij ]j = j det[mij ]j 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 fx1; ··· ; xng, of F , such that H = hw1; ··· ; wni where wi = dixi for some integers 1 ≤ d1 j d2 j · · · j 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 2 Gnf0g, then jyj divides jxj. Suppose on the contrary, there is y 2 G whose order does not divide jxj. Let p1; ··· ; pk be all prime factors of lcm(jxj; jyj). Then jxj and jyj can be written as r1 rk s1 sk jxj = p1 ··· pk ; jyj = p1 ··· pk ; r1; ··· ; rk; s1; ··· ; sk 2 N: max(r1;s1) max(rk;sk) c Thus lcm(jxj; jyj) = p1 ··· pk . Let I := fi j 1 ≤ i ≤ k; ri ≥ sig and I = f1; 2; ··· ; kg − I. Let Q ri Q sj a := i2I pi and b = j2Ic pj . Then a divides jxj, b divides jyj, gcd(a; b) = 1 and lcm(a; b) = ab = lcm(jxj; jyj). Obviously, jxj jyj x + y = ab = lcm(jxj; jyj) > jxj: a b This contradicts to the maximality of the order of x. So jyj must divide jxj. (b) Claim: For any y 2 Gnf0g, there is c 2 N such that hy − cxi \ hxi = h0i. If hyi \ hxi = h0i, then we are done. Otherwise, hyi \ hxi = hbyi for some b 2 N∗ with b j jyj. Then by = ax for some a 2 N. Thus jyj jyj jyj 0 = by = ax =) a = jxj` for some ` 2 N: b b b jxj jxj Then a = b` . Let c := ` . Then c 2 N and jyj jyj 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 jGj. When jGj = 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 jGj = n, the quotient group G=hxi has an order less than n. By induction hypothesis, t M G=hxi = hy¯ii for some yi 2 G; wherey ¯i := yi + hxi: i=1 By the preceding claim, for each yi, there is ci 2 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 2 Z Pt Pt ¯ such that dx + i=1 dizi = 0; then i=1 diz¯i = 0 in G=hxi, which implies that jy¯ij j di for i = 1; ··· ; t. Then for any i, dizi = di(yi − cix) = diyi − dicix 2 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: j0j = 1 = p0). Let G be an abelian torsion group. Let G(p) = fu 2 G j juj = pn for some n 2 Ng: (a) G(p) is the unique maximum p-subgroup of G.
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