28 Finitely Generated Abelian Groups

Transcription

1 8 Fntely Generated Abelan Groups In ths last paragraph of Chapter, we determne the structure of fntely generated abelan groups A complete classfcaton of such groups s gven Complete classfcaton theorems are very rare n mathematcs and, n general, they requre sophstcated machnery However, the man theorems n ths paragraph are proved by qute elementary methods, chefly by nducton! Ths s due to the fact that commutatvty s a very strong condton Ths paragraph s not needed n the sequel 8 Lemma: Let G be an abelan group We wrte T(G) := {g G: o(g) s fnte} () T(G) s a subgroup of G (called the torson subgroup of G) () In G/T(G), every nondentty element s of nfnte order Proof: () Snce o() =, T(G) and T(G) Suppose now a,b are n T(G), say o(a) = n, o(b) = m (n,m ) Then (ab) nm = a nm b nm = =, so o(ab) nm, thus ab T(G); and o(a ) = n, thus a T(G) By the subgroup crteron, T(G) G () Snce G s abelan, we can buld the factor group G/T(G) If T(G)x n G/T(G) has fnte order, say n, then (T(G)x) n = T(G), so T(G)x n = T(G), so x n T(G), so o(x n ) s fnte Let o(x n ) = m Then x nm = (x n ) m =, so o(x) nm Thus o(x) s fnte and x T(G) It follows that T(G)x = T(G) s the dentty element of G/T(G) Hence every nondentty element of G/T(G) has nfnte order 8 Defnton: A group G s called a torson group f every element of G has fnte order A group s sad to be wthout torson, or torson-free f every nondentty element of G has nfnte order 30

2 Thus s the only group whch s both a torson group and torson-free Every fnte group s a torson group, but there are also nfnte torson groups, for example / In vew of Lemma 8, we are led to nvestgate two classes of abelan groups: torson abelan groups and torson-free abelan groups When ths s done, we wll know the structure of T(G) and G/T(G), where G s an abelan group We must then nvestgate how T(G) and G/T(G) are combned to buld G We cannot expect to carry out ths ambtous program wthout mposng addtonal condtons on G We wll assume that G s fntely generated (Defnton 44) Under ths assumpton, T(G) turns out to be a fnte group (Theore85) The study of fnte abelan groups reduces to the study of fnte abelan p-groups, p beng a prme number, whose structures are descrbed n Theore80 After that, we turn our attenton to torson-free abelan groups (Theore83) The next step n our program s to put the peces T(G) and G/T(G) together n the approprate way to form G The approprate way proves to be the smplest way: G s somorphc to the drect product of T(G) and G/T(G) The structure of G wll be completely determned by a set of ntegers 83 Defnton: Let G be an abelan group and let S = { } be a fnte, nonempty subset of G If, for any ntegers a,a,,a r, the relaton a a a r = a mples that a = = a = g r r =, then S s sad to be ndependent If S s ndependent and generates G, and f S, then S s called a bass of G In the followng lemma, we wll prove, among other thngs, that S = { } s a bass of G f and only f G s the drect product of the cyclc groups,,, g r Lemma 84() s of especal mportance: t states that a fntely generated abelan torson group s n fact a fnte group 303

3 84 Lemma: Let G be an abelan group and be fntely many elements of G, not necessarly dstnct (r ) Let B G () = () If each g has fnte order, then o( )o( ) o(g r ) (3) If G = and : G A s a homomorphsm onto A, then A = (4) If G =, then G/B = B,B,,Bg r (5) If G/B = B,B,,Bg r, then G = B If, n addton, b,,b s B and B = b,,b s, then G = b,,b s, (6) If B = and G/B = B,,Bg r, then G = (7) { } s an ndependent subset of G and G = f and only f G = In partcular, n case are all dstnct fro, the subset { } s a bass of G f and only f G = Proof: () Certanly { } G by repeated use of Lemma 94(3), and so by the defnton of Also, any element of, necessarly of the form m m m r wth sutable ntegers,,,m r, s n by Lemma 4 and so Hence = () Suppose o(g ) = k for each =,,,r If g, then, by part (), g = m r wth sutable ntegers m Dvdng m by k, we may wrte m = k q + t, where q,t and 0 t k Then g m = (g k ) q g t = g t and g = t t t r Thus and { t t t r : 0 t k for all =,,,r} k k k r (3) If a A, then a = g for some g G snce s onto and g = m r wth sutable ntegers m snce G = Thus a = g = ( m r ) = ( ) ( ) (g r ) m r and A (4) Ths follows from part (3) when we take A to be G/B and to be the natural homomorphsm : G G/B 304

4 (5) Suppose G/B = B,B,,Bg r Let g G Then Bg G/B and, by part () wth G/B n place of G and Bg n place of g, we have Bg = (B ) (B ) (Bg r ) m r = B m r for some ntegers m Hence g = b m r for some b B and g B So G = B If, n addton, B = b,,b s, then G = b,,b s = b b s = b,,b s, (6) Ths follows from part (5) wth a slght change n notaton (7) Snce G s abelan, G = f and only f every element of G can be expressed n the form u u u r, where u unque manner (Theore5) g, n a Every element of G has at least one such representaton f and only f G =, that s, f and only f G = We want to show that every element of G has at most one such representaton f and only f { } s ndependent Equvalently, we wll prove that there s an element n G wth two dfferent representatons f and only f { } s not ndependent Indeed, there s an element n G wth two dfferent representatons f and only f m r = n n n r for some ntegers such that g m {,,,r} The latter condton holds f and only f n n m r n r =, g n for at least one where not all of n, n,, g r m r n r are equal to, that s, f and only f { } s not ndependent 85 Lemma: Let G be a group and elements of G Let B = and suppose o(g ) = o(bg ) for =,,r () If {B,,Bg r } s an ndependent subset of G/B, then { } s an ndependent subset of G () Assume are all dstnct fro If G/B = B Bg r, then G = 305

5 Proof: () If,,,m r are ntegers such that ( * ) m r =, then m B = B m m r = (B ) (B ) (Bg r ) m r = (B ) (Bg r ) m r, so (B ) = = (Bg r ) m r = B snce {B,,Bg r } s ndependent Thus o(g ) = o(bg ) dvdes m n case o(g ) s fnte and m = 0 n case o(g ) = o(bg ) s m nfnte ( =,,r) In both cases g = ( =,,r), and, because of ( * ), m = as well Hence { } s ndependent () If G/B = B Bg r, then G/B = B,,Bg r and {B,,Bg r } s ndependent (Lemma 84(7)), G = (Lemma 84(6)), { } s ndependent (Lemma 85()), G = (Lemma 84(7)) We now examne the structure of fnte abelan groups A fnte abelan group s a drect product of ts Sylow p-subgroups Ths follows mmedately f the exstence of Sylow p-subgroups s granted In order to keep ths paragraph ndependent of 6, we gve another proof, from whch the exstence of Sylow p-subgroups (of fnte abelan groups) follows as a bonus We need a lemma 86 Lemma: Let A be a fnte abelan group and let q be a prme number If q dvdes A, then A has an element of order q Proof: Let A = n and let a,a,,a n be the n elements of A We wrte m = o(a ) for =,,,n We lst all products a k a k a n k n where each k runs through 0,,,m Our lst has thus m n entres Every element of A appears n our lst Two entres a k a k a n k n and a s a s a n s n are equal f and only f the entry a r a r a n r n, where r 306

6 s such that 0 r m and k s r (mod m ), s equal to the dentty element of A Thus any element of A appears n our lst as many tmes as does, say t tmes The number of entres s therefore m n = nt Snce q dvdes n, we see q m n and q dvdes one of the numbers,,,m n (Lemma 56), say q Let us put = qh, h By Lemma 9(), a h has order o(a h ) = o(a )/(o(a ),h) = /(,h) = qh/(qh,h) = qh/h = q 87 Theorem: Let G be a fnte abelan group and let G = p a p a p s a s be the canoncal decomposton of G nto prme numbers (a 0) () For n, we put G[n] := {g G: g n = } Then G[n] G for any n () Let G = G[p a ] for =,,,s Then G = G G G s (3) G = p a (and G s called a Sylow p -subgroup of G) (4) Let H be an abelan group wth H = G and H = H[p a ] ( =,,,s) Then G H f and only f G H for all =,,,s Proof: () Let n Fro n =, we get G[n], so G[n] We use our subgroup crteron () If x,y G[n], then x n = = y n and (xy) n = x n y n = = and so xy G[n] () If x G[n] then x n = and (x ) n = (x n ) = = and so x G[n] Thus G[n] G () We must show that G = G G G s and G G j G j = for all j =,,s a (Theore) We put G/p = m ( =,,,s) Here the ntegers,,,m s are relatvely prme and there are ntegers u,u,,u s such that u + u + + u s m s = We now show G = G G G s If g G, then g = g u g u g u sm s, wth g u m G snce (g u m ) p a = g u G = ( =,,,s) Thus G G G G s and G = G G G s Secondly, let j {,,s} and g G G j G j Then g = g j, where p a p a = = g j- j- j =, therefore g p a p a j- j- a = and o(g) p a p j- j On the other hand, g G j, so g p j a j a = and o(g) p j j Thus o(g) = and g = Thus G G j G j and G G j G j = Ths proves G = G G G s 307

7 a (3) By the very defnton of G = G[p ], the order of any element n G s a a dvsor of p Then, by Lemma 86, G s not dvsble by any prme b number q dstnct from p Thus G = p for some b, 0 b a From b p b p b p s a s = G G G s = G G G s = p a p a p s s, we get b p a = G = p for all =,,,s (4) Let : G H be an somorphsm For any g G, we have g p a =, so (g ) p a = (g p a ) = = Thus g H and G H Also, f h H, then h = g for some g G and (g p a ) = (g ) p a = h p a = Thus g p a Ker =, so g p a =, so g G and h = g G Hence H G We obtan G = H Consequently, G : G H s an somorphsm and G H for all =,,,s Conversely, assume G = H and G H for all =,,,s From part (), we get G = G G G s and H = H H H s and Lemma 6 gves G H Accordng to Theore87, the structure of a fnte abelan group s completely determned by the structure of ts Sylow subgroups Consequently, we focus our attenton on fnte abelan p-groups After two prepatory lemmas, the structure of fnte abelan p-groups wll be descrbed n Theore80 88 Lemma: Let G be an abelan group and elements of G Let n We wrte G n = {g n : g G} () G n G () If G =, then G n = n n n (3) If G =, then G n = n G/G n / n / n / g r n n n and (4) Let H be an abelan group If G H, then G n H n and G/G n H/H n Proof: () and () Snce (ab) n = a n b n for all a,b : G G n a a n G, the mappng 308

8 s a homomorphsm onto G n So G n = Im G by Theore06 Also, f G =, then G n = = g n,g n,,g n r by Lemma 84(3) (3) If G =, then G = and { } s ndependent (Lemma 84(7)) Then G n = n n n by part () Moreover, {g n,g n,,g n r } s ndependent, for f,,,m r are ntegers and (g n ) (g n ) (g n r ) m r nm =, then nm nm r =, so (g n ) m nm = g = because { } s ndependent From Lemma 84(7), we obtan that G n n = Lemma 7 n n The second asserton follows from (4) Assume : G H s an somorphsm For any g G, g n = (g ) n H n, and therefore G n H n Also, f h H n, then h = h n for some h H and h = g for some g G, so h = h n = (g ) n = g n G n and thus H n G n Hence H n = G n and G n: G n H n s an somorphsm: G n H n By Theorem (7), we have also G/G n G /G n = H/H n 89 Lemma: Let p be a prme number and G a fnte abelan p-group Let G be such that o( ) o(a) for all a G and put B = If Bx G/B and o(bx) = p m, then Bx = Bg for some g G satsfyng o(g) = p m Proof: Let o( ) = p s, o(bx) = p m and o(x) = p u Snce (Bx) pu = Bx pu = B = B, we have p m p u by Lemma 6 Also, Bx pm = (Bx) pm = B, thus x pm B = and x pm = n for some n wth n p s We wrte n = p k t, where k and t are ntegers, k 0 and (p,t) = Then p k p k t = n p s and, by Lemma 9, p u m = p u /p m = p u /(p u,p m ) = o(x)/(o(x),p m ) = o(x pm ) = o( n ) = o( p kt ) = o( )/(o( ),p k t) = p s /(p s,tp k ) = p s /p k = p s k So p s+m k = p u = o(x) o( ) = p s by hypothess and m k tp We put z = g k- and g = z x Then z = B and Bg = Bx (Lemma 0(5)) From x pm = g n tp = g k tp = (g k- ) pm = z pm, g pm = (z x) pm = (z pm ) x pm =, we obtan o(g) p m Also p m = o(bx) = o(bg) o(g) Thus o(g) = p m Ths completes the proof 309

9 We can now descrbe fnte abelan groups 80 Theorem: () Let p be a prme number and let G be a nontrval fnte abelan p-group Then G has a bass, that s, there are elements n G\ such that G = () The number of elements n a bass of G, as well as the orders of the elements n a bass of G, are unquely determned by G More precsely, let { } and {h,h,,h s } be bases of G, let o(g ) = p m ( =,,,r) and o(h j ) = p n j (j =,,,s), and suppose the notaton s so chosen that m r 0 and n n n s 0 Then r = s and the r-tuple (p,p,,p m r ) s equal to the s-tuple (p n,p n,,p n s ) The r- tuple (p,p,,p m r ) s called the type of G (3) Let H be a nontrval fnte abelan p-group Then G H f and only f G and H have the same type Proof: () We make nducton on u, where G = p u If u =, then G = p, so G s cyclc (Theore3) and the clam s true Assume now G s a fnte abelan p-group, G p and assume that, whenever G s a fnte abelan p-group wth G G, then G s a drect product of certan nontrval cyclc subgroups We choose an element of G such that o( ) o(a) for all a G and put = B Snce G, we have B If G = B =, the clam s establshed, so we suppose B G Then G/B s a fnte abelan p-group wth G/B G By nducton, there are elements Bx,,Bx r of G/B, dstnct from B, such that G/B = Bx Bx r Let us put o(bx ) = p m for =,,r Usng Lemma 89, we fnd g G such that Bx = Bg and o(g ) = p m ( =,,r) Let us wrte o( ) = p Then G/B = B Bg r and, by Lemma 85(), G =, where are dstnct fro snce B,,Bg r are dstnct from B and s dstnct fro snce o( ) o(a) for all a G and G Ths completes the proof of part () 30

10 () and (3) For convenence, a t-tuple (p a,p a,,p a t ) wll be called a type of a nontrval fnte abelan p-group f a a a s 0 and f A has a bass {f,f,,f t } wth o(f k ) = p a k (k =,,,t) We cannot say the type of A, for part () s not proved yet The clam n part () s that all types of a nontrval fnte abelan p-group (arsng from dfferent beses) are equal Let G and H be nontrval fnte abelan p-groups, let (p,p,,p m r ) be a type of G, arsng from a bass { } of G and let (p n,p n,,p n s ) be a type of H, arsng from a bass {h,h,,h s } of H If r = s and (p,p,,p m r ) = (p n,p n,,p n s ), then g m h for =,,,r and G = h h h r = H (Lemma 6) Ths proves the "f" part of (3) Now the "only f" part of (3), whch ncludes () as a partcular case (when G = H): we wll prove that G H mples r = s and (p,p,,p m r ) = (p n,p n,,p n s ) Suppose G H We make nducton on u, where G = p u If u =, then G = p = H, so G and H are both cyclc, hence G = and H = h Thus r = = s and p = o( ) = p = o(h ) = p n The clam s therefore establshed when u = Now suppose G p and suppose nductvely that, f G and H are somorphc fnte abelan p-groups wth G G, and f (p a,p a,,p a r ) s a type of G and (p b,p b,,p b s ) s a type of H, then r = s and (p a,p a,,p a r ) = (p b,p b,,p b s ) We dstngush two cases: the case when G p = and the case G p In case G p =, we have g p = for all g G, n partcular p m = o(g ) = p for all =,,,r Also H p = (Lemma 88(4)) and p n j = o(h j ) = p for all j =,,,s Hence p r = = G = H = h h h r = p s, so r = s and (p,p,,p m r ) = (p,p,,p) = (p n,p n,,p n s ), as clamed Suppose now G p Then H p Thus there are elements n G and H of order p, so p p and p n p Assume k s the greatest ndex n {,,,r} wth p m k p, so that (when k r) p m k+ = = p m r = p Let the ndex l {,,,s} have a smlar meanng for the group H Then (p,p,,p m r ) = ((p,,p m k,p,,p) ( ) r k tmes 3

11 (p n,p n,,p n r ) = (p n,,p n l,p,,p), ( ) s l tmes t beng understood that the entres p should be deleted when k = r or s = l By Lemma 88(3), G p = p wth o(g p ) = p m = p = p p g k p g k p p r k tmes for =,,k Hence { p,,g k p } s a bass and (p,,p m k ) s a type of G p In the same way, (p n,,p n l ) s a type of H p Here G p s an abelan p-group wth G p = p ( )+ +(m k ) p + + +m s = G Snce G p H p by Lemma 88(4), our nductve hypothess gves k = l and (p,,p m k ) = (p n,,p n l ) Then p m = p n for =,,k From p + +m k p r k = G = H = p n + +n l p s l = p + +m k p s l we get r k = s l = s k Thus r = s and a glance at ( ),( ) shows (p,p,,p m r ) = (p n,p n,,p n r ) Ths completes the proof 8 Examples: (a) We fnd all abelan groups of order p 5, where p s a prme number An abelan group A of order p 5 s determned by ts type (p,,p m r ), where of course p + +m r = A = p 5 Snce m + + m r = 5, the only possble types are (p 5 ), (p 4,p), (p 3,p ),(p 3,p,p), (p,p,p), (p,p,p,p), (p,p,p,p,p) and any abelan group of order p 5 s somorphc to one of 5, 4, 3, 3,,, 0 and In partcular, there are exactly seven nonsomorphc abelan groups of order p 5 (b) The number of nonsomorphc abelan groups of order p n (p prme) can be found by the same argument Ths number s clearly the number of ways of wrtng n as a sum of postve ntegers,,m r If n, an equaton of the form n = + + m r, where,,,m r are natural numbers and m r 0, s called a partton of n Thus 3

12 the number of nonsomorphc abelan groups of order p n s the number of parttons of n Notce that ths number depends only on n, not on p The parttons of 6 are 6, 5+, 4+, 4++, 3+3, 3++, ++, +++, ++++, and an abelan group of order p 6 s somorphc to one of 6, 5, 4, 4, 3 3, 3,,, (c) Let us fnd all abelan groups of order = (to wthn somorphsm) An abelan group A of ths order s the drect product A A 3 A 5, where A p denotes the Sylow p-subgroup of A (p =,3,5) Here A has order 5 and s somorphc to one of the seven groups of type ( 5 ), ( 4,), ( 3, ),( 3,,), (,,), (,,,), (,,,,) Lkewse there are fve possbltes for A 3 : and three possbltes for A 5 : (3 4 ), (3 3,3), (3,3 ),(3,3,3), (3,3,3,3) (5 3 ), (5,5), (5,5,5) The varous drect products A A 3 A 5 gves us a complete lst of nonsomorphc abelan groups of order Now that we obtaned a complete classfcaton of fnte abelan groups, we turn our attenton to torson-free ones 8 Lemma: Let G be an abelan group, B a subgroup of G and assume that G/B s a drect product of k nfnte cyclc groups (k ), say G/B = By By By k (y,y,,y k G) Then y, y,, y k are nfnte cyclc groups and G = B y y y k 33

13 Proof: Let Y := y,y,,y k G Then G/B = By,By,,By k and, from Lemma 84(5), we obtan G = BY We wll show that G = B Y and Y = y y y k To establsh G = B Y, we need only prove B Y = Let g B Y Then g = y a y a y k a k for some ntegers a,a,,a k (Lemma 84()) and B = Bg = (By ) a (By ) a (By k ) a k Snce {By,By,,By k } s an ndependent subset of G/B (Lemma 84(7)), we get (By ) a = (By ) a = = (By k ) a k = B But o(by ) = o(by ) = = o(by k ) = by hypothess, so a = a = = a k = 0 and thus g = y 0 y 0 y k 0 = Ths proves B Y = Hence G = B Y We now prove Y = y y y k In vew of Lemma 84(7), we must only show that {y,y,,y k } s an ndependent subset of Y Suppose,,,m k are ntegers wth y y y k m k = Then (By ) (By ) (By k ) m k = B, so = = = m k = 0 and y = y = = y k m k = Hence {y,y,,y k } s ndependent and Y = y y y k Fnally, snce By has nfnte order, we see that y has also nfnte order and y s an nfnte cyclc group ( =,,,k) 83 Theorem: Let G be a fntely generated nontrval torson-free abelan group () G has a bass, that s, there are elements n G\ such that G = () The number of elements n a bass of G s unquely determned by G More precsely, f { } and {h,h,,h s } are bases of G,then r = s The number of elements n a bass of G s called the rank of G (3) Let H be a fntely generated nontrval torson-free abelan group Then G H f and only f G and H have the same rank Proof: () Let G be a nontrval torson-free abelan group and assume that G = u,u,,u n We prove the clam by nducton on n If n =, 34

14 then G = u s a nontrval cyclc group and the clam s true (wth r =, = u ) Suppose now n and suppose nductvely: f G s a nontrval torsonfree abelan group generated by a set of m elements, where m n, then G s a drect product of a fntely many cyclc subgroups of G If u =, then G = u,u,,u n = u,,u n s generated by a set of n elements and, by nducton, G has a bass Let us assume therefore u Then o(u ) = We put B/ u := T(G/ u ) For any b B, the element u b of B/ u has fnte order, thus there s a natural number n wth b n u Consequently, for any b B, there s an n and m such that b n = u m We defne a mappng : B by declarng b = m/n for any b B, where n, m are such that b n = u m Ths mappng s well defned, for f n and m are also such that b n = u m, then u m n n m = (u m ) n [(u m ) n ] = b n n (b nn ) =, so m n n m = 0 (because o(u ) = ) and m/n = m /n s n fact a homomorphsm To see ths, let b,c B and b = m/n, c = m /n (where n,n, m,m ) Then b n = u m and c n = u m, so (bc) nn = (b n ) n (c n ) n = u mn u m n = u mn +m n and (bc) = (mn + m n)/nn = m/n + m /n = b + c Thus s a homomorphsm Snce Ker = {b B: b = 0/} = {b B: b = u 0 } =, the homomorphsm s one-to-one and : B Im s an somorphsm: B Im Clam: f B s fntely generated, then B s cyclc To prove ths, assume B = b,b,,b t and let b = m /n ( =,,,t) Usng Lemma 84(3), we see that Im = b,b,,b t = /n, /n,,m t /n t s a subgroup of the addtve cyclc group /n n n t Hence Im and B s cyclc s cyclc If B = G, then B s fntely generated by hypothess, so B = G s cyclc and () s proved We assume therefore B G Then G/B = Bu,Bu,,Bu n = Bu,,Bu n 35

15 (see Lemma 84(4)) s a nontrval abelan group, generated by n elements Moreover, G/B = G/ u / B/ u = G/ u / T(G/ u ) s torson-free by Lemma 8() So, by nducton, G/B = B Bg r wth sutable g G, where Bg s dstnct from B ( =,,r) Thus o(bg ) =, and ths forces o(g ) = ( =,,r) Lemma 8 yelds G = B We put = A Then G = B A and B G/A s fntely generated by Theore7(), Lemma 84(4) Hence, by the clam above, B s cyclc, say B = Snce u, we have o( ), so o( ) = and G = Ths completes the proof of () () and (3) For convenence, a natural number r wll be called a rank of a fntely generated nontrval torson-free abelan group A f A has a bass of r elements We cannot say the rank of A, for part () s not proved yet The clam n part () s that all ranks of a fntely generated nontrval torson-free abelan group (arsng from dfferent bases) are equal Let G and H be fntely generated nontrval torson-free abelan groups, let r be a rank of G and s be a rank of H, say G = and H = h h h s If r = s, then g h for =,,,r and G = h h h r H by Lemma 6 Ths proves the "f" part of (3) Now the "only f" part of (3), whch ncludes () as a partcular case (when G = H): we wll prove that G H mples r = s Ths s easy Now G/G / / / g r C C C s a fnte group of order r by Lemma 88(3) Also H/H h / h h / h h s / h s C C C 36

16 s a fnte group of order s If G H, then G/G H/H (Lemma 88(4)), so r = G/G = H/H = s Hence r = s 84 Remark: Theore83 states essentally that a drect sum of r copes of cannot be somorphc to a drect sum s := of s copes of unless r = s Ths s not obvous: there are many one-to-one mappngs from r onto s, and there s no a pror reason why one of these mappngs should not be an somorphsm The proof of r s r = s does not and cannot consst n cancellng one at a tme from both sdes of r s In general, t does not follow from A B A C that B C As a matter of fact, there are abelan groups G such that G G G G but G G G! r := 85 Theorem: Let G be a fntely generated abelan group Then T(G) s a fnte group and there s a subgroup I of G such that G = T(G) I Proof: G/T(G) s a fntely generated abelan group (Lemma 84(4)), and s torson-free (Lemma 8()) Thus ether G/T(G) ; or G/T(G) T(G) T(G) T(G)g r wth sutable G (Theorem 83()) and therefore G = T(G) (Lemma 8) Puttng I = n the frst case and I = n the second case, we obtan G = T(G) I Then T(G) G/I by Theore7() Snce G s fntely generated, so s G/I (Lemma 84(4)) and T(G) s also fntely generated From Lemma 84(), t follows that T(G) s a fnte group The subgroup I n Theore85 s not unquely determned by G However, ts rank r(i), whch s the rank of G/T(G) s completely determned by G when G/T(G) Let us defne the rank of the trval group to be 0 and let us call a bass of Then the rank of any fntely generated torson-free abelan group s the number of elements n a bass of that group, and r(i) s completely determned by G, also n case G/T(G) 37

17 As G = T(G) I, the fntely generated abelan group G s determned unquely to wthn somorphsm by T(G) and I Now I s determned unquely to wthn somorphsm by the nteger r(i) (Theore83(3) and the defnton r() = 0); and T(G), beng a fnte abelan group (Theore85), s determned unquely to wthn somorphsm by ts Sylow subgroups (Theore87(4)) Let s be the number of dstnct prme dvsors of T(G) (so s = 0 when T(G) ) Each one of the s Sylow subgroups (correspondng to the s dstnct prme dvsors) s determned unquely to wthn somorphsm by ts type (Theore80(3)) Thus the fntely generated abelan group G gves rse to the followng system of nonnegatve ntegers () A nonnegatve nteger r, namely the rank of G/T(G) Here r = 0 means that G s a fnte group If r 0, then T(G) I, where I s a drect product of r cyclc groups of nfnte order The subgroup I s not, but ts somorphsm type s unquely determned by G () A nonnegatve nteger s, namely the number of dstnct prme dvsors of T(G) Here s = 0 means that T(G) group () In case s and G s a torson-free 0, a system p,p,,p s of prme numbers, namely the dstnct prme dvsors of T(G) ; and for each =,,,s, a postve nteger t and t postve ntegers m,m,,m t, so that (p m,p m,,p m t ) s the type of the Sylow p -subgroup of T(G) Wth ths nformaton, G s a drect product of r + t + t + + t s cyclc subgroups r of them are nfnte cyclc; and (n case s 0) t of them have orders equal to a prme number p, more specfcally, t of them m have orders p m,p m,,p t Furthermore, two fntely generated abelan groups are somorphc f and only f they gve rse to the same system of ntegers Exercses Let G be an abelan group and H G Prove that (a) T(H) = T(G) H, 38

18 (b) T(G)/T(H) HT(G)/H T(G/H) and that HT(G)/H need not be equal to T(G/H) Let G be an abelan group Show that \{} (a) f G s fnte, then G/G n G[n] for all n ; (b) f G s nfnte, then G/G n G[n] need not hold for any n 3 Let G be a fnte abelan group The exponent of G s defned to be the largest number n {o(a): a elements n G Show that (a) the exponent of G dvdes G ; G}, e, the largest possble order of the (b) for any g G, o(g) dvdes the exponent of G; (c) the exponent of G s the least common multple of the order of the elements n G; (d) G s cyclc f and only f the exponent of G s G 4 Let G be a fnte abelan group and H G Let K G such that H K = and H L for any L G satsfyng K L Let g G (a) Assume g p K for some prme number p Prove that, f g K, then there are h H, k K and an nteger r relatvely prme to p such that h = kg r Conclude that g (b) Prove that G = H HK K f and only f, for any prme number p and elements g G, h H, k K such that g p = hk, there s an element h H satsfyng h = (h ) p 5 Let G be a fnte abelan group of exponent e and let g G be of order e, so that o(g) = e Put H = g Show that G = H K for some K G (Hnt: Use Ex 4 Consder the cases p e and p e separately 6 Let G be a nontrval fnte abelan group Usng Ex 5, prove by nducton on G that there are non-trval elements n G such that G = and (n case r ) o(g ) dvdes o(g + ) for =,,,r 7 Keep the notaton of Ex 6 Prove that the ntegers o( ), o( ),,o(g r ) determne the types of the Sylow p-subgroups of G unquely, and conversely the types of the Sylow p-subgroups of G completely determne the ntegers o( ), o( ),,o(g r ) (The ntegers o( ), o( ),,o(g r ) are 39

19 called the nvarant factors of G Two fnte abelan groups are thus somorphc f and only f they have the same nvarant factors) 8 Fnd the nvarant factors of the fnte abelan groups C 6 C 9, C 6 C 8 C 5 C 30, C 4 C 6 C 5 C 0 30