IJAPTEH ffiTopics in Group Tzeory

trI nis chapter takes a deeper look at various aspects of the classification problem for finitegr:ups, which was introduced in Section 7.5. After the necessary preliminaries are deveioped insecticn 8'1, allfinite abelian groups are classified up to isomorphism in section 8.2 (the Fundamen-tal Theorem ol Finite Abelian Groups). Nonabelian groups are considerably more compiicated. Thebasic tools or analyzing them (the sylow Theorems and conjugacy) are presented n Sections 8.3and B'4' Appiications of these results and several oiher facls about the structure of finite grgups areconsidered in seciton B.s, where groups of smail order are crassified.

sections B'3 and B'4 are independent of sections 8.1 and 8.2 and may be read first if desired.Sections 8.1-8.4 are prerequisites for Section 8.5.

s

If G and -Er are srouDs, then their cartesian product G x rr is arso a group, withthe operatio" Jun r"-.oorAiort"*iu" (iluor"* 7.4). Inthis secton we exbendthis notion to more than two groop* 'ii"o we examine the conditions underrvhich a group is (isomorphic to) , air".t product of cerbain ;il.ffi!,f,;:.

13." these subgroups are of a parbicurarly simpre kind, then the structure ofthe group can be completely determinecl, aswijl b" d"moirtrute in Section g.2.Throughout the general rr."."ro", i ;;il;H il;r-" ;ilfi phcativeiy, butspecific examples of famiiiar additive o"* are written uaitirr*ty as usuar.-

rr trt, Ge, . . . , G, are_groups, we d.efi.ne a coord.inat"*ir" opurution onthe Cariesian prod.uct G'r x d * .'. 1 *-C" as follows:-H

DIHECT PROOUCT

24s8.J Direct Prod,ucts

Il:,"".:l::,"^liy.th,"r c, ,x q+ . - . X G,is a group under this operarion:If e, is the identity element ofa, then (er, er,- v p ee usr!ry trrtruertt, oI u;, Inen (e1, gr, , . , n) is the identity ele_(at, az,

\ur2^'..r.riz and (ot-., a2-r,. . ran_r) is the inverse of, o,). Tlris group is cailed ihe direct iro :r,et. t e- -G_ rr *|roduet of Gr,Gr, . ,G-.*

E X A M P L E Recall t[at U, ls -ttr9

multiplicative group of units n Z^ andthat Un: {1, B} a,,d U. : {1, 5}. The direct product []n x Uu X Z, consistsofthe 12 tripies(1, 1, 0), (1, 1, 1), (1,7,2), (1, 5, 0), (1, 5, 1), (7, 5,2),(3, L, 0), (3, L, 1), (3,L,2), (3, 5, 0); tr, u, ii, (3,5, 2).

Note that Lrn has order 2, u. has order 2, zrhas ord.er 3, and the directproduct U4x (J6 X Z, has orde4 2.2.3:1_2. Similarly, in the generalCASE,

ilt:,-9:r : :

., g" "::_-:i:q:=1jBo

GtxGzx'''xG,hasorder lcrl .lcrl ... lG-I.

our viewpointthe component

0:0+0 1:g+4 2=A*Z3=3*0 4:0*4 5=3*2.

In the preceding "T?*pi" it is important to note that theloups f\n, (Jr,Td^r;iil,i"""jrT:.d plh* dirgct g1"a"91 u_ ;-,";;,":For insrance, 5 isan element_of Ur, byt 5_ is not n UnxTJ6XZ, because ifr"ii"*""i."1?Un x Uu X Ze are triples.In general, fo.*1 = i - ,

of the direet produet G, x G, x . . .This situation is not entirely satisfacto.y, t,rt fy drurrgirrgslightly we can develop

" ,,otioo of direct product in whichgroups may be considered as subgroups.

EXAMPLE Itiseasytoverify ihatM: {0, g}.ard}i :10,2,4} arenor_mal subgroups of zu (do it!). observe that every elem*i "r

zu can bewritten as a sum of an elementinM and an eremnt in/in one and onryone way:I

x .when each G is an additive {_erig group, ihe direct product of Gr. . . , G,is sometimes ca[edthe dieet sum and denoted G, @ , @ . . . O "*" It iEEIEEiE-Eaffi is a subgroup of G, x G,x . .. X G_ (seeExercise 12). s esvuuY ur el I

248 8 Tapcs in Group Theory

Y"rit that, when the elements of zuare written as sums in this v;ay, thenthe addition table for Zr looks like this:0+0 3+4 0+2 3+0 0+4 3+2

0+00+23+00+43+2

0+03+4a+23+00+4'J -r ^

3+40+23+00+4Qro0+0

0+23+00+4Qro0+0o'1-+

3+00+4Qao

0+03+40+2

0+43+20+0a+23+0

Qaq

0+03+40+23+0a+4

CompareMxN:

the Z, table with the operation table for the direct product

(0, 0) (3,4) (0 (3, o) (0,4) (3, 2)(3, 4)(0, 2)(3, 0)(0, 4)(3,2)(0, 0)

(0, 0)(3, 4)(0, 2)(3, 0)(0, 4)(3,27 I

(0, o)(3, 4)(0, 2)(3, 0)(0,4)(3, 2)

(0, 2)(3, 0)(0,4)(3, 2)(0, 0)(3,4)

(3, 0)(0, 4)(3,2)(0, 0)(3, 4)(0, 2)

(0,4)(3,2)(0, 0)(3,4)(0, 2)(3, 0)

(3, 2)(0, 0)(3, 4)(0,2)(3, 0)(0,4)

consequently, we can'express z, as a drect product in a purely irternalfashion, without looking at the set M x N, which is external to-Zr: write eachelement uniquel, as a sum a * b, with a eM and b eN. we now develop thissame idea in the general case, with multiplicalive notation in piace of aditionia Zu.

('fnmnfm A.t LeiNr,N, . . ,Nrbenormalsubgroupsof agroupGsuchthat1 euery e-lemerut in G can be written uniquely in the form ara2. ak, with a, eL

n, * Then G is isomorphic to the direct product N, X ff, x-. I . x ir.

The only difference in these two tables is that elements are written a * bin the flrst and (c, b) in the second. Among other things, the tables showthat the direct product M x N is isomorphic to zrunder the isomorphismthat assigns eachpair (a, b) e M x Nto the sum ofits coordinates a * b ev

The proofdepends on this useful facr:

*Uniquenssmeansthatifat&2.'.ar=brbr...rx-itheachc,,,e-Nr,thena,=b,foreveryi

g.l Direct prod,ucts' LEMMA E.Z Let M,:rd,ry be_normal subgroups of a group G such thati_M N : (e). If a e M and, b, , il""'Lu : or.

Proaf consider a-.lb-.iab. -since -tr1is ogr1l: b-_r?b e M byTheorem 7.34. C70-sure in' ^4f shows that a-tb-t"u :

":iu:ri.rv. sr*r]y, the normarity of./ implies t},at a-tb-.,1 5{-na] r,}r,".,'r_,;_;r;:7l'.,u_ro)uetr/. Thus;;3;#i {^r}{:=u;:] Td''l*l*'i""' *iJ", orJ:'u-\zu: e o, the rerr by'",:{"{ry:rii;1,,,?l;'*"jJJrn'*";ff *precedingthetheorem(butusing

f:ffrxr,X ".xi/*G by f(a1,ar,. . ,eh.)=el&z...d,h.since every erementof G can be written in the fory arar . . a h(with a, err) byhypothesis, /is sur'ectir; J;;;""),,_ , ou) = f(bt. b,. .. , ), rhen7ii';;!;r.i;!A;;,'":r' Bv the o.iq""'"'= Lvpir,=i=. o, - , rr each

?47

.(at, az, , @n) : (bt, br,

.

and /is injective..,b)inNrxNrx...xAL,

In order to prove that /is a homo30,rfh.T we must first show that the-A/'s are muruailv air:"i"r-"*r""oi'!nrt i., /, n,j: ir, *hen i * j. f a e4 n /j, then o ca, be writtenL "'prtar.t ;;i;#;1

"rii" tr/,s in two differ_ent ways: " "i: i, .: o: i' ' "i' , 1 Nj , /, , N^.The uniqueness hypothesis impries thai-the components r,, a4 *,r"t be equai:u : e. rherefore, A,, n T =

248 8 Topics in Group Theory

ft@,,

Therefore, /is homomorphism and., hence, an isomorphism. +

{^ whenever G is a group and. }/,, . . . , AIo are subgroups satisfying the'\ hypotheses of rheorem 8.1 we shall say that-G is the- direet pooa.rlt *r\ /r, . . . ,iy'u and write G:A/, x ...xN. EachN, is said to be a direet

I , factor of G. Depending on the context, we can think of G as L]e external direct

" product of the ?{ (each element a k-tuple (ar, . ,a*) }tr1 x...xIr)or as an internal direct produet (each element written uniquely in the forat&2, .oteG),

The next theorem is often easier to use than ?heorem 8.i. to prove that agroup is the direct product of certain of its subgroups. The statement of thetheorem uses rhe foilowing notation. rf M and a.e s,rbgroups of a group G,( th.r, /yf} denores the set of ali produ cts mn,with m e M and n e N.

\, Trr!!!Y *! I M and. N are norntal subgroups of a group G such that G : MN',a.nd Iu[ l-]:\': (e). then G : M X N.

For the case of more than two subgroups, see Exercise 25.

Proof oi rhearem 8.3 By hypothesis every element of G is of the form nzn, withmeM, rueN. suppose that an element had two such representations, sayffLn : trt,nr,with m_,m, e M andn, *, e N. Then multiplng on the left by nzr-land on the right by -r shows that mr-Lrn: nln:r.But mr-ameM i"afttfl-7 eI/and MnN =(e). Thus Lr-LL:e and h:*rrsimilariy, fl:fl1.Therefore, every element of G can be written uniquely in the form mn (m e M,n e N), and., hence, G : M x N by Theorem g.1.

EXAM PLE Themultiplicativegroupof units nZruis Ur, : |.7,2,4,7,8,11, 13, 14). The gxoups M : lL,11J andN: {1, Z-,-4, glL**.,.or*"i sub-groups whose ini'ersecticn is (1). E'ery element of -l[ is in JlfN (for in-stance, 2: 7:2), and similariy for M. Since 11 .2:7,11 .B : 18, and11 .4: 14. weseethat Lirs:Mll.Therefore, [.Iru : M xl/by?heorem8.3. Since.\,-is cyclic of order 2 andM cyclic of ord i +tZis a generaior), weconciude that Ur. is isomorphc to Z, x Z+ (see Exercise 10 and Theorem7.1E i.

@ EXEHCIES

N0TE: t/nles s stated otherwise. G], , G-, are groups.i.. Fin

8.1 Direct products

2. Srhat is the order of the group Us x [Ja * ,, * r,3. (a) List all subgroup s of Z2 x Zz. (Thereare mor than two)(b) Do the same for Z, x Zz x 22.4, If G and JI are glroups, prove that G x H

= H x G.

5' Give an exampre to shor, that the direct product of cyclic groups need not becyclic.6. (a) Write Z* as a direct sum of two of its subgroups.(b) Do the same for Zru.

(s) wriie Zss in three different ways as a direct sum of tivo or more of itssubgroups lHint: ?heorem g.B.l'' l.,,i"'#b,;; *;,?;,1" sroups.

Prove that G, x . . . x c, is abetian if and only if8. Let i be an integer with 1

= i

= n.prove that the function

r,:GrxGrx...xG,-G,given by

",(ar, ar, as, . . , d,) = o, is a surjective homomorphism of groups.9. Ts Z, isomorphic to Zn x Zr?1,0. (a) If f : G^r] Fr lod g : G, -+ rare isomorphisms of groups, prove that themap 0: G, x Gz- Htx }I, given by 0(a', : ( ib\:;.))," uo isomor_phism.

(b) If Gr:,tlrfor i:1,2, . . . ,ft,prove thatGrX"'XG,=IJrx xH,

11. Let H_:_K, M,i/ be groups such that K =

M x }y'. prove that12' Letibe anintegerwith 1

=i=n.LetG,*bethesubsetof G, x...X G,consisting of those elements whose l th cordinar. * .", ;i;ent of G, andwhose other coordin ares are "r.t ;h ;l;;;;; "i "* "rff, in'ili.,Grr' : {(er, . ., ri_t,air+t, . r ., er)laeG}.

Prove that(a) Q* is a normai subgroup of G, x . . . x G,.(b) G,*

= Gi.(c) G, x . . . x G, is the (internaU direct prod.uct of its subgroups Gr*, .

,G,*. LHint: Show that """.y "tu*",.i 9i C, * . . . * ;;L be wri*enuniquely in the form ara,I . . on,*itf. ,,'. Cr-, ,orhfiorem s.r.]

249

B.

250 8 Topics in Group Theory .

13, Let G be a group and let D : l(a, a, a)l a e GI.(a) ProvethatD is asubgroup ofG x G x G.(b) Prove thatD is normal in G x G x G if and onlyif Gis abelian.

14. lf Gt, . , Gn are finite gloups, prove that the order of (ar, d2, - , o.) in

la.l.15. Let ir,ir, . . . ,inbe apermutation of the integers 1,2, . . , n. Prove that

Gi,xGax..xG;is isomorphic to

GrxGrxlExercise 4 is the case n = 2.]

16. If/,Kare subgroups ofagroup Gsuchthat G: x Kand Misanormalsubgroup of N, prove that M is a normal subgroup of G. [Cornpare this inithExercise 14 i Section 7.6.1

17. Let Q* be the multipiicative group of nonzero rational numbers, @** the sub-group ofpositive rationals, and 11 the subgroup {1,

- 1}, Prove that Q* :

e*+ x.H.18. Let C* be the multipiicative group of nonzero complex numbers and R** be

the multiplicative group of positive real numbers. Prove that C* =

R*+ x R/2, where R is the additive group of real numbers.19. LetGbe a group andfr;G*Gr, fs:G--', Gr, . ., fniG*G,homomor-

phisms. For i = L,2, , n,let z', be the homomorphism of Exercise 8. Letf*:G--- G, X' -.x G,bethemap definedbyf*(a) = (f(a), fz@), . ., f,(a^)).(a) Prove that fr' is a homomorphism such that T " f* = f for each i.(b) Prove that f* is the unique homomorphism from G to G, x ' . . y. G, such

thaL r " f* : f for every .

-

2. Let Nr, , N* be subgroups of an abelian group G. Assume that every ele-ment of G can be written in the form o, 1 . an (with o N) and that when-eyer o,taz - ' - an: e, then a : for every i. Prove thatG:NrxN2x.'.x're.

21-,. LeL G be an additive abelian group rviih subgroups .Ey' and K. Prove that G :H x Kf and only if there are homomorphisms

such that 6r(rrr(r)) +Tt" 6z: 0, and r,, o

6zGz@D: for every-t6r : 0, where r, is the

x G,'

e G and f;r o 61 : L7, 7i2o32 = ry,identity map on X, and 0 is the

oe

24.

22.

.r 4,

8.f Direct Products 251

nrap tha'u sends every element onio the zero (identity) element. lHint: Let nbe as in Exercise 8.1Let G and -EI be fi.nite cyclic groups. Prove that G x -II is cyclic if and oniy if(lcl, lHl) = 1.(a) Show by example that Lemma 8.2 may be false if Itr is not normal(b) Do the same for Theorem 6.3.Let -l/, K be subgroups of a group G, with N normal in G. If N and ll are abe-iian groups and G : NK, is G the direet product of N ad K?Let ilr, , Nr be normal subgroups of a group G. Let NrN, ' ' ' flr, denotethe set of a-11 elements of the form arar' ' a.bwith a, e Nr. Assume that G =NrN, . . '.r\* and that

^, nN1' .' N'_rffr*, . . . tr/ : (e)

floreach i i1 s i < n). ProvethatG : Ni x N; x "' x AI.Lei N,, , 4'u be normal subgroups of a finite group G' If G :A'rNr' " r\ tnotation as ir Exercise 25) and lGl : lNrl ' lirl ' ' ' lul ,prove that G : Nr x N2 x ' ' ' x A.Let A-, .F/ be subgroups of a group G. G is called the semid.irect produot of Nand fI if is normal iri G, G : NH, and N f\' H : (e). Show that each of thefollowing groups is the semidireci product of two of its subgroups:(a) S. (b) D4 (c) SoA group G is said to be indeccmposable if it is not the direct product of twoof iis proper normal subgroups. Prove that each of these groups is indecom-posable:(a) Su &) ,n @) ZIfp is prime and is a positive integer, prove that Zr" is indecomposable.Prove that Q is an indecomposable g:'oup.Siiorv b5, example that a homomorphic image of an inriecomposabie groupneed not be i.ndecomposable.

Prove tha a g:roup G is indecomposable i ano or:.1, lf u,!srever i7 anci K arenorrnal subgroups such that G : H x.K, then H : \e) orK : (e).Let i be the set of positive integers and assume that for each i e 1, G, is agroup.* The infinite d"rect product of the G, is denoted Il. G, and consists

* .\ny in-nite index set 1 may be used here, but the restrjction to the positive integers simpli.fies thenoraiion.

2,6.

6)1

z8.

oo

30.

31,

DO

252 8 Topics n GrouP Theory

of ali sequences (or, &2, ' -' ') with a; e G'Prove that II G' is a group under

the coord.inatewise oPeration

(a.1, az, .)(bt, bz, .) : (-arbr, a2b2, .)-

S.2 FINITE AtsELIAN GRUFS

{. g4. With the noiation as in Exercise 33, let ) G, denote the subset of II G, con-sisting of all sequences (cr, c2'

-' ') such that there are at most a fiite num-

ber of .oordirrJt="l;ii .r: , *rr.:" e, is the identity_ element of Gr. Proveil;;tc, i. ,;;;;i;;ilsr"p of fl G,. r, c, is called the innnite directtel - ielt1sum of the G,'

Ss.LetGbeagroupandassumethatforeachpositiveinteger,,,':arormalsubgroup

"i c. r

"""r:, element of G can be wr-itten uniouelr, in the form

fti,'ni"' 'tui,,with i1 < i2< i-I"l'f f,'-p'""i tr'ut c = ) N' {seeExercise 34).* l*nt: Adapt the proof of fheorem 8'1 by definingVf'o"r',-;.r,. .' .l i" ["

'i'" ptoauct of those a that are not the identity

elernent.l

36. If (tn, n) = L,prove that U*oz (J* x (Jn'

37. Let /J be a group and rr:H ' G'' r2:\7 G'' " ' ' T^':H - Gn homomor-phism with this propei"y: WhenevJr G is a group and g' :G

* G''gz:G'Gr,'.--l l-, i^ti'9^are homomo'phi"*t' then there exists a

unique homomorphism g* : G * I/ such thai r o g* : g' for every i' Prove thatn = Crx G2 x''' x G,' lseeExereise 19'l

AII fi.nite abelian groups will now be classified. we shall prove that ever-v finiteabelian grorlp G is a Jir""t sum of cyclic subgroups an that the orders ofthesecyclic subgroop.

^r"',,""ique1v deteimined UV C' fe only prerequisites for ihe

proof other than Sectior, .f r" basic numbr theory (Section 1'2) and elemen-i"rv ""p theory (Chapter ?, through Section 7'8)'

Following tfr* o..r"f ",r.io* with

abelian groups, aII groups are writtenin ad.ditive notation i" titit section. 1'he following d'ictionary may be helpfu1for translating from multipiicative to add"itive notation:

x Uniqueness means that if a,then,b = f and for,r.= 1. 2, '

,*

i

MULTIPLICATIVE NOTATION

8.2 Finite Abelian Groups

ADDITIVE NOTATION

253

,abe

&h

^k_^u -.e

MN: {tnnlrn eM, n e-Atr}direct product M. x N

direct factor M

a*b0

rcu

ka:0M + N: lm * nlm eM,n eNlt

directsumM@Nrect summand. M

Here is a restatement in additive notation of several earlier results that rvi beused frequentiy here:

THE1BEM 7.8 Let G be an ad.ditiue group and,let a e G.(3) If a has order n, then lza : 0 if and onty if n I k.(4) If a has order td, with d > 0, then ta has order d-

THEIREM 8.1 If Nr,. , , N are normal subgroups of an ad.d.itiue groupG such that euery eletnent of G can be written uniquely in. the format* &z+ ... * awithareNr,thenG

=1 ONrO 1"" @i/u.THEAfrEM 8.3 If M and N are nornal subgroups.of u.n ad.ditiue group G su.chthot G : M + N andM A N : l0), then G : M @ N.

..

Finally we note that Exercise 11 of Section 8.1 wili be used without explicitmention at several points.

If @ and p is a prime, then G(p) denotes the set ofelements in G whose order is som QI.

-

Ii is eas;, to verify that G(p) is closed under addition and thai the inverse of anyelemeni in G(p) is also in G(p) (Exercise 1). Therefore., G(.p) is a subgroup of G.

EXAMPLE If G: Zrr,l}'en G(2) is the set of elements having orders20, 2', 22, elc.Verifu that G(2) is the subgroup {0, 3, 6, 9i ; similarly, G(3) :{0, 4, 8i. If G : Zs @ Zi, then G(3) : G since ever}, nnzero element in Ghas order 3.

The first step in proving that a nite abelian group G is the direct sum ofcyclic subgroups is to show that G is the rect sum of its subgroups G(p), one

1\))tJ

258, 8 Topics in Group Theory

for each of the:distinct primes viding the order of G. In order to do this, weneed

LEMMA g.4 Let G be an abelian group artd. a e G an element of finite'ord.er. T]rcnd: ali ar* * o,pt with areG(p), wltere Pt, , ph o.re the distinctpositiue prirnes that diuide the order of a.

Prool The proof is by induction on the number of distinct primes that dividethe order of a. If la I is d.ivisible oniy by the single primepr, then the order of a isa power of p, and, hence, a e G(p).so the lemma is true in this case. Assumeinductivel'that the lemma is true for all elements whose order is divisibie by atmost A

- I distinct primes and that lal is divisible by the distinct primes

pt, . ,pp. Then lal :PL"'''Ph", wiiheachr,)0'Lef,m=Pz"' 'pk"urrdr, : pr',, sothat lal : mn. Then(m, n): l andbyTheorem l-'3 there areintegers u, u such that i : nltl * nu. ConsequentlY,

But maa e G(pr) because a has order mn. and-, hencq pr'i (mua) : (nm)ua :u(mtza): u0:-0. Simiiarly, m(nua):0 so that by Theorem 7.8 the order ofzuo dideS n'1, arLinteger with only k - l distinct prime divisors. Therefore, bythe indrrction assumptiorLnua: az* as * ' * ou, with a eGtp)'Leta,=mua;t}rer'a.: n'Luo, + rla.: dt* az*'.''* a*,with ure G(p)' 6

G:G(p)@G(p) o"'@G(p,), '|"

where pt, . , pt are the distirtct positiue primes that d.iuide tlte order of G.Proal I{ a eG, then its order divides lGl bv coroilary 7.27. lenc, a:or * . . * o, N.ith a e G(pJ by Lemma 8.4 (where ai: 0 if the prime p,does not divide ll). To prove that this expression is unique, supposear* ar+''' + ar: bt + bz+''' * bfiwitho,, b, eG{ni).sinceGisabelian

at- bt: (bz -or) + (b, - o.) +' " + {bt - a,):

For each i, b, -

a, e G(p)and, hence, has order a power ofp,. say pi'. If m :pz', ." p.',, then m(br

- a) : 0 for i = 2, so that

nt(ar- b) : in(b2- ar) +''' : m(b, -o,) = 0 + "''i 0 : 0.Consequentiy, the order of a, - br- nust d-iYide m by Theorem 7'B' Butat

- bt e G(pr), so its order is a porver of pr. The oniy power of p, that divides

m =

pz"' 'Pr'' ispro: 1. Therefora, at- br: 0 and ar: br' Similar argu-ments for I : 2, . , f show that a, : b f'or every i. Therefore, every elementof G can be written uniquely in the form a, +''' * c, with o, eG(p,) and,hence, G : G(pr) @''' O G(p,)byTheorem 8.1. +'

/

51

L

IB.Z Finite Abelian Groups 285

lf+u-tryEe, tlela 4@nt has order a pow er of p,:SI:d

"{-sr:}g. Each_of the G(p,) io Theodeii mtron. E e.,ement o of ap-group B is es e.n elgment sf maximal ord.erif li = lotf@at p"b: pn*i(pib): 0 for every b eB. Not thatelements of ma-iimal order always exist n-afinitep_group.

The next step in elassifying finite abelirr. g*op" is to prove that everyfinite abelianp-group has a cyclic direct.o**rrrd., after which we will be ableto prove that every finite abeiianp-group is a direct sum of cyclic g:oups.

|WW Let G be a finiie abelian p-group and, q. an erement'of maximar ord.er\- in G. Tlten there is a subgraup K of G such that G : (a) @ K.L-

The following proof is more intricate than most of the proofs earlier in thebook. I{evertheless, it uses only elementary group theory, so if you read itcarefully, you shouldn't have trouble foliowing th" rgr**rrtProof at Lemma 8.6 consider those subgroups .Er of G such that (o) n rr : (0).There is at ]east one (rl: (0)), and since G is finite, there must be a largest'subgroup K with this property. Then (a) n E: (0), ,rr v Theorem B.B weneed only shorv that G : (a) + K. If this is ro the ease, then there is a nonzero such that I @) + K. Let be the smallest positive integer such that p e(a) + K(there must be one sice G is ap-group and., hencelp;6 : 0 : o i o e(a) + K for some positivej). Then(1)

(2)andpc : phb is in (o) + K, say

If o has order p", then p"r : 0Consequently, by (Z)

c -

pn-r6 is no in (a) + K

pc:ta + k (t ez,k eK).for all r e G because a has maximai order.

Pn-7ta * pn-tp - pn-l(ta + l?): p"-t(pc): p,c:0.Therefore, p"-7ta:

-pn-tk e(a) n K: (0) and p"-tta:0. Theorem 7.gshows thatp" (the order of o) divides p"-tt, and t follo*s thatp | /. ?herefore,pc:ta*k:pma +& for some fr, and, consequently, i:Or_p":pk

- ma).Let

(3) d.:c-ma.Then pC : p\c. nw).= h__uK,but d.'gK (since c

- ma: k, eKwould implvthat c : ma + k' e (a) + K, contradiciing (r)). use Theorem T.ri to ,.rirv tf,uiH: V + zdlxeK,z eZIsasubgroupofC*itnf e fi. Since d,: A + iA ,

?ld d f K, H is larger than r{. Bui K is ttre largest sroop such that (a) K =(0,1, so '*'e must hai,e (o) n H + (0). If u is a nonzero element of (c) rr, then

,L-.

256 8 Topics in GiouP Theory

W:SA:k1+rd (k., e K; r, s eZ).We claim thaLp/r; for if r:py, then sincepd eK,0* w: sa:kt+ ypde(a) f,K, a contradiction. Consequenti-v, (p,r):1, and by Theorem 1.3 thereare integers /, u withpu. * ru : 1' Then

c:'c: (pu * ru)c::if:;ri'i',oro + ma)) iby (2) and (B)l

: u(ta + h) + u(rd + rnta)= u(ta + h) + u(sa - k, + rma) tby (a)l: (ut * us * rm)a + (uk

- ukr) e (a) + K.

This contrad.icts (1). Therefore, G:(a)+ K, and, hence, G:(al@KbyTheorem 8.3. &

THEOREM.-L,7 (THE FUNDAMENTAI THEOREM OF F]NITE ABELIAN.ffi,finiteabeliarugroupGisthedrectsumafcyc!icgroups,eachof prime power order.

Praof By Theorem 8.5, G is the direct sum of its subgroups G(p), one for eachprime p irrut divides lGl . Each G(p) is ap-group. So to complete the proof, werreed oniy show that every fi.nite abelian p-group fl is a direct sum of cyclicgroups, each of ordel a powel ofp. We pr.ove this by induction on the order of'Ff'he ssertion is true wenf/has order 2 by Theorem 7.28. Assume inductivelythat it is true for all gloups w-hose order is less than lI{ | and }et a be an e}ementof maximal orderp" in fl. Then.Ff : () e Kby Lemma 8.6. By induction, Kis adirect sum of cyclic g:'ollps, each u-ith order a power ofp' Therefore, the same istrue of .Fl : (a) O K. +

EXAM PLE The number 36 can be written as aproduct of prigrepowersin just four ways: 36:2'2'3'3:2'2'32:l2z;' 3'3 = 22-)' 31' Con-se{uentl,,, by Theorem B.Z every abe}ian group oT'order 36 must be iso-morphic to one of the following groups:

z, @ z;@ zsoz,", z2@ z2@ zs, 24@zs @zs' z^ @ zs.Y;; ;^" ;rsiiy veri' that no two of these groups are isomorphic (thenunrber of elements of order 2 or 3 is differeni for each group). Thus rvehave a complete classification of all abelian groups of order 36 up to iso-morphism.

You probably noticed that a familiar gToup of order 36, nameiy zr., doesn'tappear explicitiy on the lisi in the preceding example. However, it is isomorphicLo Z, @ Zq, vs ln e now Prove.

(4)

8.2 Finite Abelian Groups

LEMMA 8.8 If (m, k) : 7, then Z^ @ Zu = Z*r.Prasf The order of ( 1, 1) in z* @ z is t]ne smallest positive integer such that(0, 0) : r(1, 1) : (t,t).Thus = 0 (mod. m) and = (mod), so that mlt and.& l' But (m,k): l implies t]oatrnl l by Exercise 1T in section 1.2. Hence,mk < t. Stnce mk(l,1) : (mk, mk) : (0, 0), we must have mh,: : I(1, 1)l .Therefore, z* @ zh (a group of order mlz) is the cyclic group generated. by (1, 1)and, hence, is isomorphictoZ^pby Theorem 7.18. aTHEAREM 8.9 If n: p{p2"". . .p{,, with pr,zn=zo,,,o''.@2o,,,.Praof The theorem is true for groups of order 2. Assume inductively thai it istrue for groups of order less than rz. Appiy Lemma 8.8 with m: p{, and :p2n..-'pr",. Then Zn=Zo,^@Zp, ar.Ld. ti:e induction hSrpothesis shows thatZB=2p,": O. ' ' AZo,",.

258 8 Topics in GrouP Theory .

we pause briefly here to present an interesting corollary that wiil be usedin Chapter 10. It ln'as proved earlier as Theorem 7'15'

chfrlLLABY 8.11 If G is a finite subgroup of the multiplicatiue group of nonzeroelements of a fietd F, then G is cyclc'

Praof Sinee G is a finite abelian BrouP, Theorem 8.10. implies that G-V- @ ' ' '@ Z--,wheteeaehm divides m,'Everyelement b^12^,9'.

".

@(*';#i;.. iZ' twhv?). Conslquently, every elementg of the multiplieative*orn G ust satiJff i^' : Tritt'ut

-is, must be a root of the poly-nomial

i-,' 1";. Since Gtrs orde. *r*r''' ffitandf-' - -lfh'u-d mostrn' distinct

,oot, io f by Corollary 4'l-6, we must have i : 1 and G = Z*,' S

If G is a finite abelian grouP, then the integers rrllt ' ' 'ttt'irtTheorem8.10 are called the il;;i""t facrs of G. \lhe G is written as a direct sum ofcS,clic groups of prime power orders, as in Theorem 8.7, the prime powers arecailed the element".y i"i"ors of G. Theoi'ems 8.7 and 8.10 show thattheorder of G is the product of its elementary diyisors and also the product of itsinvariant factors'

E X A M P L E All abelian gToups of order 36 can be classified up to isomor-phism in terms of their elementary divisors (as in the example precengL"**u 8.8) or in terms of their invariant factors:

ELEMENTARY INVARIANT ISOMOHPHICprvls0Rs

-

FACT0RS GRpUPGROUPzz@z-\@z.ozs 2'2,3,3 6,6 z6@26

Zz@Zz@Z-s 2,2,32 2'78 Z2@Z$'i zn@zr@Z, 22,3,3 3,12 z3@zL- z4 a zs 2', 3' 36 zru \

\The Fundamentai Theorem B.? can be used to obtain a iist of al} possible

abelian gToups of a given order. To complete the ciassification of such gloups,rve must shor, that no t*o groups on the lisi are isomorphic, that is, that theelementary divisors of a group are uniquell- determined'*

THEREM 5.12 Let G and.I{ be finite abelian groups. Then G s isom.orphic to Hif and. onty if G and. H haue the same elemeruta4i diuisors'

* The remainder of this section is optionai. Thsorem E.12 is often considered to be part of theFundamental Theorem of Finite Abelian Groups'

r:

(:..t.i'li.,

-R"t-

A.Z Finite Abelian Groups

It is also true that G =

H if and only if G and -H have the same invariant

Proaf of Thearem 8.12 If G and.tlhave the same elementary disors, thenbothG and H are isomorphic to the same dtect sum of cycli. grrrp. and, hence, areisomorphic to each other. converseiy,rf f:G--->H is an isomorphism, then aa,:d f(a) have the same order for each a e G.It follorvs that for Lach prime p,f{G(p)) : H(p) and, hence, G(p)

= H(p). The elementary divisors of G that aie

poyers of the prime p are precisely the elementary divisors of G(p), and simi-larly for ll. so we need orrly prove that isomorphic p-groups have the sameelementary visors. ln other words, we need to prov this }-aif of the theoremonly when G and

-El arep-groups.Assume G and H a:.e isomorphicp-groups. we use induction on the order

of G to prove that G and fr have the same elementary divisors. Al1 groups oforder 2 obviously ha'e the same eiementary divisor, 2, by Theor"* z.e. soassume that the sratement is true for all groups of order less than lGl . Supposethat the elementary dir-isors of G are

PN" P", , Pn', p, p, . , p with n, > n2>r copies

>nrll

and that the eiementary divisors of -Ef are

P*', P^', ,P^r,p,p, ,p with ffitzTTtzz---_-V-

s copies

> m.r) L.

-,-1 -"-1P tP' ) , p''-'A similar argument shows that the elementary divisors of pIf are

nr-! m2- i.!- )1, t , p^'-'

Verifv thatpG : lp*lx eG) is a subgroup of G (Exercise Z).If Gis the rectsum of groups c, verify thatpG is the direct sum of the groupspc, (Exercise 4).rf c, is cyclic with generator a of orderp", thenpc, is the cycfic group generatedby pa. sincepo has orrierp"-I by part (4) of rheorem T.g,pc,is cy"lic of orderp"- 1. Note that when : 1 (that is, when c, is cyclic oro.e.p), thlnpc, : (0).Consequentiy, the elementary dir.isors of pG aie

259

c-'? .

I

\ t-

r*.

-

lf f :G -

I/ is arr isomorphism, verify that f(pG): p.Ef so that pG : pE.Furthermore,pG =

G (Exercise g), so that lpci < G. Hence pc r,,iprthave the same elementari; fliyi=ors by the induction hypothesis; tat r, t ! iand

{iEi:

6riEi'ffr:i1E-,ffi;[ffiil

pn'-} : F^,-', so that n..- !: rrli- l for j : L,2, . ,t.

I

260 8 Topics in Group TheorY

Therefore, rL: trLfor each l. So the only possible difference in elementarydivisors of G and fl is the number of copies ofp that appear on each list. SincelGl is the product of its' elementary divisors, and simir1y for lill ; and sinceG

= H, we have

pn,pn,. . . po,p, : IGI : lHl : b*,p*,. . . p^,pr.Since rn, : n;for each a, we must havep' : ps rrd hence, r : s. Thus G andHhave the same elementary divisors. +

& rxrRclsrsN0TE: Ail groups are written add,itiuely, and p always denotes a positiue prime,unless noted otherwise..

A. \ 1. If G is an abelian Broup, prove that G(p) is a subgroup.I!.2. lf G is an abelian Broup, prove thatpg = {p* lr e G}is a subgroup of G,.4tlS. l-irt ail abelian groups (up to isomorphism) of the g:iven order:

(a) 12 (b) 15 (e) 30 @) 72(e) 90 (f) 744 (g) 600 (h) 1160

- /' 4. ffG and G, (1 < I < n) are abelian groups such that G : Gr@''' @ G.,v show llnatpG = pGt@'' '@pG".

5. Find the eiementary divisors of the given g'noup:

, (a) Zruo (b) z6 @ zD @ 218

, (c) ZLo @ Z2o @ ZBo @ Z4o (&) ZL2 @ Zso @ Zaoo @ Zz4a .;

Il

:

I

)ii

l

,l:

IVs. 7./

,,"/ a. Find the invariant factors of each of the groups in Exercise 5.Find the elementary divisors and the invariant factors of the given group.Note that the group operatioo is multiplication in the first three and additionin the last.(a) Ue (b) Li1? (c) Urs @) M(Zz)If G is the adtive group @lZ,wihat are the elements of the subgroup G(2)?Of G(p) for any posiiive prime p?(a) If G is a flnite abelian p-group, prove that pG * G.(b) Shor, that part (a) may be false if G is infinite. lint: Consider the group

G(2) in Exercise 8.1If G is an abeiianp-group and (n, p) : 1 prove that the map f :G -- G givenby fb): na is an isomorphism.

8.

t," g.

10.

:1

::ii'