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F1.3YR1 ABSTRACT ALGEBRA Lecture Notes: Part 7 1 Finitely

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F1.3YR1 ABSTRACT ALGEBRA Lecture Notes: Part 7 1 Finitely Generated Abelian Groups If A is a ¯nitely generated abelian group, then we have seen that there is a natural n » n number n and a subgroup K of Z such that A = Z =K. Moreover, K is free abelian of rank m · n. A basis for K can be expressed as the set of rows of an m £ n matrix: such a matrix is known as a presentation matrix for A. Example. µ ¶ 300 040 3 is a presentation matrix for A = Z3 £ Z4 £ Z, since the subgroup K of Z generated by the rows of the matrix is K = f(3m; 4n; 0); m; n 2 Zg =3Z£4Z£0Z and so 3 Z Z £ Z £ Z Z Z Z » = = £ £ = Z3 £ Z4 £ Z: K 3Z £ 4Z £ 0Z 3Z 4Z 0Z » Of course, since Z3 £ Z4 = Z12, it follows that ¡ ¢ 12 0 » is another presentation for A = Z12 £ Z. Lemma 1 Let the m £ n matrix M be a presentation matrix for a ¯nitely generated abelian group A.Ifm=nthen A is ¯nite, of order j det(M)j, where M is any presentation matrix for A. Otherwise there exists a surjective homomorphism A ! Z. Proof. Since the rows of M represent a basis for K, they are linearly independent. Suppose ¯rst that m = n. Then the euclidean algorithm will transform M into a new presentation matrix M 0 which is upper triangular in the sense that the entries below the diagonal are all 0: mi;j = 0 for i>j. The elementary row operations used in the euclidean algorithm do not alter the fact that the rows are linearly independent. Moreover, the only e®ect they have on the 0 determinant is to mulitply it by 1 or ¡1, so j det(M)j = j det(M )j = j¹1 ¢¢¢¹nj, where 0 ¹1;:::;¹n are the diagonal entries in M . Let B be the cyclic subgroup of A generated by en =(0;:::;0;1). Then B has order 0 j¹nj in A, since (0;:::;0;¹n)=¹nen is one of the rows of the presentation matrix M . Moreover, A=B is generated by e1;:::;en¡1, and we can get a presentation matrix M" for A=B by removing the nth row and the nth column from M 0. By induction on n we can assume that A=B is ¯nite of order j¹1 ¢¢¢¹n¡1j, and it follows that A is ¯nite of order jA=Bj¢jBj=j¹1¢¢¢¹nj. n Now assume that m<n. For any vector v =(v1;:::;vn) 2 Z , we can de¯ne a n homomorphism Áv : Z ! Z by Áv(u)=u:v = u1v1 + ¢¢¢+unvn. Suppose that u:v =0 whenever u is one of the rows of the presentation matrix M for A. Then each row of n M belongs to Ker(Áv). Hence the subgroup K of Z generated by the rows of M is also n contained in Ker(Áv), and we may then de¯ne a homomorphism θv : A = Z =K ! Z by θv(u + K)=u:v = Áv(u). n n We have an n-paramter family Áv, v =(v1;:::;vn)2Z of homomorphisms Z ! Z, and a family of m linear equations u:v = 0 that determine which of the Áv give rise to homomorphisms θv : A ! Z. Since m<n, there is at least one nonzero solution to this family of simultaneous equations, and so at least one nonzero homomrohphism θv : A ! Z. Of course, the homomorphism θv is not necessarily surjective. However, since it is nonzero, its image Im(θv) has the form kZ for some k>0, so θ : A ! Z de¯ned by θ(u)=θv(u)=k is surjective. ¤ Example. Let µ ¶ 43 M = 52 be a presentation matrix for a ¯nitely generated abelian group A. Then det(M)= » 8¡5=¡7, so A has order 7, which is a prime number, and hence A = Z7. To see this isomorphism more directly, we can use row operations to put M into row echelon form, for example: µ ¶ µ ¶ µ ¶µ ¶ ¡ ¡ 43 ! 52 ! 1 1 !11 52 43 43 07 Then the subgroup K generated by the rows of M is also generated by the rows of the 2 new matrix, (1; ¡1) = e1 ¡ e2, and (0; 7) = 7e2. The group A = Z =K is generated by two elements, a1 = e1 + K and a2 = e2 + K. But the fact that e1 ¡ e2 2 K means that a1 ¡ a2 =0inA,iea1 =a2,soAis cyclic. The fact that 7e2 2 K means that 7a2 =0 in A,soAis cyclic of order dividing 7. On the other hand, the order of A must be greater than 1, since no Z-linear combination of the rows of M contains e2,soe2 2=K, » so a2 =0in6 A.ThusA=Z7, as promised by the Lemma. Example. Let µ ¶ 143 M = 052 be a presentation matrix for a ¯nitely generated abelian group A. Since M has fewer rows than columns, the Lemma tells us that there must be a surjective homomorphism f : A ! Z.NowA=Z3=K, where K is the subgroup generated by the rows of M, 2 and so A is generated by a1 = e1 + K, a2 = e2 + K and a3 = e3 + K. To ¯nd a homomorphism f, we need to de¯ne f(a1)=u1,f(a2)=u2 and f(a3)=u3, where u1;u2;u3 2Z are solutions of the simultaneous linear equations 1:u1 +4:u2 +3:u3 =0; 0:u1 +5:u2 +2:u3 =0: A nonzero solution to the second equation can be obtained by putting u2 = 2 and u3 = ¡5. We may than subsitute these values back into the ¯rst equation to get u1 =7. Lemma 2 Let A be an abelian group, and let θ : A ! Z be a surjective homomorphism. » Then A = Ker(θ) £ Z. Proof. Choose a 2 A such that θ(a)=12Z. De¯ne Á : Ker(θ) £ Z ! A by Á(x; n)=x+na. Then Á is a homomorphism, since Á((x1;n1)+(x2;n2)) = Á(x1 + x2;n1 +n2)=x1+x2+(n1 +n2)a =(x1+n1a)+(x2 +n2a)=Á(x1;n1)+Á(x2;n2): Moreover, if (x; n) 2 Ker(Á) then n = θ(na)=θ(x)+θ(na)=θ(x+na)=θ(Á(x; n)) = θ(0) = 0; and hence also x = Á(x; 0) = Á(x; n) = 0. Thus ker(Á)=f(0; 0)g and so Á is injective. Finally, if ® 2 A, let m = θ(®) 2 Z, and let y = ® ¡ ma. Then θ(y)=θ(®)¡ mθ(a)=m¡m=0,soy2Ker(θ). Moreover, Á(y; m)=y+ma = ®. Hence Á is also surjective. Thus Á is an isomorphism from Ker(θ) £ Z to A, as required. ¤ Corollary 1 Let A be a ¯nitely generated abelian group. Then there is a ¯nite subgroup » m A0 of A and an integer m ¸ 0 such that A = A0 £ Z . » n n Proof. We know that A = Z =K for some integer n ¸ 0 and some subgroup K ½ Z , where K is free abelian of rank r · n.Ifr=nthen we know from Lemma 1 that A is ¯nite, so we can take A0 = A. Otherwise we know { also from Lemma 1 { that there exists a surjective homomorphism f : A ! Z. Composing f with the surjective quotient map Zn ! A, we also get a surjective homomorphism Á : Zn ! Z. » n » By the last Lemma, A = Ker(f) £ Z, and similarly Z = Ker(Á) £ Z. Since we also know that Ker(Á) ½ Zn is free abelian of rank t · n, it follows that t = n ¡ 1. Also » » n¡1¡r Ker(f) = Ker(Á)=K. By induction on n¡r, we may assume that Ker(f) = A0 £Z » » n¡r for some ¯nite subgroup A0, and hence A = Ker(f) £ Z = A0 £ Z . ¤ 3 2 Finite Abelian Groups Corollary 1 reduces the classi¯cation of ¯nitely generated abelian groups to that of ¯nite abelian groups. The next step is to simplify further. We start with a result similar to the Chinese Remainder Theorem. Lemma 3 Let A be a ¯nite abelian group of order mn, where m; n are positive integers » which are coprime { in other words hcf(m; n)=1. Then A = mA £ nA. Proof. The map f : A ! mA £ nA de¯ned by f(a)=(ma; na) is a homomorphism, since f(a+b)=(m(a+b);n(a+b)) = (ma+mb; na+nb)=(ma; na)+(mb; nb)=f(a)+f(b): It is also injective, because hcf(m; n) = 1 means that m; n generate Z, which means that mp + nq = 1 for some integers p; q, and so if a 2 A with (ma; na)=(0;0) then a =(mp + nq)a = p(ma)+q(na)=0. Finally, if a 2 A then mna = 0 by Lagrange's Theorem, since jAj = mn.Ifp; q 2 Z with mp + nq = 1, then f(mpa)=(m2pa; mnpa)=(ma ¡ mnqa; mnpa)=(ma; 0). Similarly, if a0 2 A, then f(nqa0)=(0;na0). But then an arbitrary element of mA£nA has the form (ma; na0)=f(mpa + nqa0) for some a; a0 2 A, and so f is surjective. Hence f : A ! mA £ nA is an isomorphism. ¤ De¯nition. Let p be a prime number. A p-group is a ¯nite group whose order is a power of p. Lemma 4 A ¯nite abelian group A is a p-group if and only if the order of each element a 2 A is a power of p.
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