The Derived Series of a p-Group Citation Hobby, Charles Ray (1960) The Derived Series of a p-Group. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/QY7G-Q706. https://resolver.caltech.edu/CaltechETD:etd-06152006-083527 Abstract NOTE: Text or symbols not renderable in plain ASCII are indicated by [...]. Abstract is included in .pdf document. Olga Taussky (see W. Magnus, Math. Ann. vol. 111 (1935)) posed the problem of determining whether there is an infinite chain of p-groups G1, G2,..., such that G1 is abelian, [...], and [...] where [...] is the nth derived group of [...]. N. Ito (Nagoya Math. J., vol. 1, (1950)) constructed such a chain for p > 2 and G1 of type (p,p,p). It is shown (by an explicit construction) that if p > 2 there is a chain of the required kind for G1 any non-cyclic abelian p-group. If p = 2 there is a chain of the required kind if G1 contains a subgroup of type [...], of type [...], of type [...], or of type (2,2,2,2,2). As a consequence, for p > 2 it is impossible to estimate the length of the derived series of a non-abelian p-group G from the type of [...]. This gives a negative answer (for p > 2) to a question posed by O. Taussky (Research Problem 9, Bull. Amer. Math. Soc. vol. 64 (1958) pp. 124). Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: (Mathematics) Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Taussky-Todd, Olga (advisor) Zassenhaus, Hans (advisor) Thesis Committee: Unknown, Unknown Defense Date: 1 January 1960 Record Number: CaltechETD:etd-06152006-083527 Persistent URL: https://resolver.caltech.edu/CaltechETD:etd-06152006-083527 DOI: 10.7907/QY7G-Q706 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 2608 Collection: CaltechTHESIS Deposited By: Imported from ETD-db Deposited On: 23 Jun 2006 Last Modified: 07 Nov 2023 18:04 Thesis Files Preview PDF (Hobby_c_1960.pdf) - Final Version See Usage Policy. 2MB Repository Staff Only: item control page

THE DERIVED SERIES OF A p-GROUP Thesis by Charles Ray Hobby In Partial Fulfillment of the Requirements For the Degree of Doctor of Philosophy California Institute of Technology Pesadena, California 1960 ACKNOWLEDGMENTS It wish to thank Doctor Olga Taussky Todd and Professor Hans Zassenhaus for many helpful suggestions, The research presented in this thesis was performed while the author held a National Science Foundation Summer Fellowship. My thanks are extended to Miss Evangeline Gibson for her patience and efficiency in typing the manuscript. ABSTRACT Olga Taussky (see W. Magnus, Math. Ann. vol. 111 (1935)) posed the problem of determining whether there is an infinite chain of p-groups Gy, Gp» «ee » such that G, is abelian, G, = a, ,/ate), and gin) #1 where gin) is the nth derived group of Gay? N. 146 (Nagoya Math. J., vol. 1, (1950)) constructed such a chain for p > 2 and G, of type (pspep). It is show (by an explicit construction) that if p> 2 there is a chain of the required kind for Gy any non-cyclic abelian p-group. If p=2 there is a chain of the required kind if G, contains a subgroup of type (27427), of type (27,27 ,27), of type (292527527), or of type (252525252). AS a con~ sequence, for p> 2 it is impossible to estimate the length of th 1) 6 f derived series of 4 non-abelian p-group G from the type of G/c This gives a negative answer (for p > 2) to a question posed by \ 0. Taussky (Research Problem 9, Bull. Amer. Math. Soe. vol. G4 (1958) pp. 124). Chapter 7 Olga Taussky [9] posed the following problem on finite p~ groups (i.e6, groups of prime power order): Is it possible to estimate the length of the derived series of a p-group G from the type of a(t). (a) a «rad 2 Ths « G/G\"’, where G denotes the nth derived group of G? This ques- tion is known to have an affirmative answer in certain cases. For example, it follows from the Burnside Basis Theorem [11, page 111] that (1) if G/G is cyclic, then G is cyclic, and consequently gVacts, 9. Taussky [10] has shown that for 2-groups G, with o/h? of type f (252), the derived group gil) is eyclie and hence g(?? = <j>, This result was rediscovered by Blackburn [1] who showed that if G is a (1) 2~group with G/G of type (2,2), then G contains a cyclic sub- group of index 2. The following related result was obtained by Scholz and Taussky [8] for a large class of 3<groups. Let uy v, w be elements of a group G, and define symbolic powers by the rules: a = vo u V3 veiw vow . ‘ . . u =u ue This enables us to consider polynomials as symbolic exponents, even though addition is not necessarily commutative. Let G be a 3-group having generators a, b and denote by H the subgroup 4 4 ~ @ ; a b , ol Pea ; of G generated by e and 6 where ¢ =a b ab. Denote by M the ideal (3: (a-i *, (amt) (b-1), (b=1 2) in the ring of all 3 symbolic powers of elements of G. Let G be the group generated by the third powers of the elements of G. Then the result of Scholz and Taussky can be stated as follows: If co EG for every mc M, fry 2 3 gh , and if #H a) 2G then g(?) = <{>, A stronger version of this result can be obtained from a recent theorem of Blackburn [2] onDee 2 which states that if G is a 3egroup with g! ) # <1>, then the class of G is greater than four. The stronger version of the result of ma (2) 2 , Scholz and Taussky is then: If e¢ e€ G* for avery meM, then 2) g! = <]>, bo tae - as tee . ver, palmil) 26 The obvious conjecture, "If G is a p=group with G/G Ox , a . ; oa. type (psp), then G* ) = <1>," is known to be false. Burnside [3] 6 constructed (for every prime p> 3) ap-group G of order p such that G g(t) has type (pyp) and g(?? #<1>,. However, as was noted by Blackburn [2], the construction of Burnside actually gives a group G of order 3! if p=3, where again c/o? has type (p sp) and (2) # <I>, W. Magnus [6] first noted that the length of the derived series of a p-group cannot always be estimated from the type of ofa), He constructed, for every positive integer n, a 3=group G with c/o") of type (3,3,3) and o(™) ¢ <tp, The result of Magnus was generalized by N. Ité who considered a problem which arose in the work of Scholz and Taussky 18] on class field towers. This problem can be stated as follows (see Magnus [6], [7}): Determine whether there is an infinite chain of p<groups Gy» Gos se. » such that G, is abelian, G, = aan), af} # <i>, It is easy to see that, in such a chain, a Joh"? = G, and ee e n for every n. (See the proof of Theorem 1.) Thus, since gi) # <1>, ah it follows from the existence of such a series that the first mentioned question of 0. Taussky has a negative answer for any p=group G with G/G = Gye It6 [5] constructed a chain of the kind deseribed above, for every prime p #2, with Gy of type (psp5p)0 = H, Zassenhaus, in an unpublished work, constructed a large number of such chains, for every prime p #2, with G, of type (p,psp). It follows that such a chain exists for G, any abelian p-group (p # 2) having at least three generators. (See Theorem 1, Chapter 2.) As noted earlier, the derived series of a pegroup G must terminate if c/o) is either cyclic or of type (2,2). Thus no ehain of the type described above can exist for G, eyclic or of type (2,2). It will be shown that such a chain exists in the following cases: p #2 and G, has at least 2-generators (Theorem 3, Chapter 3); p=2 and G, eontains a subgroup having one of the types (27,27), (27,2727), (292527927), amd (2,2,2,252) (Theorems 4 and 6). It remains an open question whether an infinite chain exists with Gy a non-cyclic 2-group which is neither of type (2,2) nor contains a subgroup having one of the types listed above, Chapter 2 contains preliminary lemmas which are valid for general groups. Two theorems are established which reduce the con- struction of an infinite chain of the kind described above to the construction of a single infinite group having factor groups of a specified form. Infinite chains with G, having two generators are constructed in Chacter 3, The chapter ends with a refinement of an inequality for p-groups of P, Hall (Theorem 5), Infinite chains with G, a 2=group having 3, 4, and 5 generators are constructed in Chapter 4. exfex Chapter 2 This chapter consists of notation, preliminary lemmas, and theorems which are valid for general groups. Theorems 1 and 2 give two reductions of the problem of constructing an infinite chain of p-groups Gye Gop eee 9 with G, = att [oe : gin} #<1>, and G, an abelian Mn, Ry "ke group of type (p ‘9 P “9 «ee 9 P Theorem 1 shows that it is sufficient to construct such a chain for Gy of type my in, m (p “9 Ps eee 9 P 8) where m Sn, and s<k. Theorem 2 further reduces the problem to the construction of a group containing an infinite chain of normal subgroups satisfying certain conditions. The remaining chapters of this thesis are devoted to the construction of groups satisfying the conditions of Theorem 2. The following notation will be used: (M,N) is the group i =] : : n for m in M and n in Neg < X9¥oeee9% > yt) is the net derived generated by all mm is the group generated by X» Yo -ce » 23 group of the group H; R is the ring consisting of all expressions uty Jp For us, v integers and p a fixed prime; P is the ideal of R generated by /p3 I, and 0, are, respectively, the 2 x 2 identity and zero matrices; X, is the set of all 2% 2 matrices having elements in X, where X is an arbitrary ring. Frequent use will be made of the following weak form of the Burnside Basis Theorem [11, page 111]. Burnside's Basis Theorem. Let G be any p-group and let X4 9Xq gece sk, be coset representatives of a minimal basis of the abelian eroun c/a'?, Then G = < X49 Xo 9 eee 5 x Fe wa me Theorem 1. Let Gay Go» ees bean infinite chain of p=-groups such that 1? "2 (a) (n) %, a S = a a . G, = Gi _/G Gr? Spe] ~ <i>, and G, is abelian of type (o ‘sp “seses m eS te 4 e es ° 4 p ). Then an infinite chain of p~groups Hys Hos eee Can be con- ay Te me structed with H, abelian of type (p ‘sp “sg eee 9 P Dy =H ae and rhe #<1> whenever n, >m, and k 2s. Proof. Suppose that k= s and let Hy be an abelian group of type ca (p 9 P *, ese p DP) Such that H, G = <i> and (Hy » G) = <{> for every n. Then Hy =< hy re SS hy > X eo8 KK he > where each hy is an element of order Ny If G is an arbitrary group and N is a normal subgroup of G, then (G/N) yf) = oy Ay. Thus it follows from @, = ain/ thas ol!) Sg) /e(™), ana rence o,/o{!? = cn AG af) g()), n Go ott n+] CoH Therefore G. W/o) = = Guay /S all}, and G, = Gy jell) ss a /ah!?, Thus , by the Burnside Basis Theorem, each Gy has k independent generators, a om 3 1 and these generators have orders p 9 P 5 «#s 9 p modulo af 1) Suppose Ge =< X49 Xqe coe 9 XH, >» where the x, are mM. elements of order p * modulo ofl), Define Hy to be the subgroup of Go x Hy generated by X4hy» Kobe » eos 9 Xb That is, (1). gf! Ho =< %4hy » Koko one 9 hy >. It is clear that Hy since each h,. commutes with every element of G.,. Since n. >m, the ah ¥ 2 2. - ‘ order of h, is not less than the order of x, modulo afl), Thus cohe ele . (i) « SV cee . «ws (1) Hy / Hy =<h,~ > X <b, > X X< ho > =H Alsoe Hs # <> / (1) Ga since was assumed to be different from <i>. be fei [If Hae Hig coe ¢ A have been constructed by letting bs oo =a < . > er G = < 4 a@ > Hy hyZs49 soe 9 I Ke tk > where G, = Zs 4? Bens 2 8 Ber 5 and if H, =H HA), g@) ¥ a> gor i<net, then # i ~ if H, = Lag / Beige Beas or ign-t, then H., is con structed as follows. Let G = < 249 2409 eos 8 2 >, and let + (n) apr nfl : s » Ia gy TBs be the coset of G nt Ga corresponding to Z,¢ in the . . (n) (n) . . isomorphisn Geiae Then G_.,/G BO z Bs isomorphism = G4 / a+ nen nti / nti 490 npserr3e nk ® Let Vs be a coset representative of T Bye an Gay? Then, by the sy £ Sex Pe 0 25, YF <= € 9p a PE an ’ Burnside Basis Theorem, Gat < ¥49 Vos see 9 Vy > Define He to be the subgroup of G. 4 x Hy generated by hyyys hoYos eee 9 bi T ye , = ch , (1) 2 That is g Hae] ~ 44 9 boYos eee 9 LI} ® Clearly He +4 = Gy since each hs commutes with every element of G ntd Also, ) Hot Boe Me " ain) _ (n) Haag / Sn < bay yGnay? see 9 BT One of 0 ale), Thus n . ca 8 . Baa / Hen) =< hye O24 9hne TB o9e00 shy eo 2) > which is isomorphic , (n) = to < hhyB,.4 bnZ ose ee sh fy > 6 That LB g Hatt H nt En @ Alse, Hw) = gl) ¥ ep, Hay = ae The existence of the required chain Hys He 9 eee now follows by induction. If k>s, let B be an abelian group of type Doig sti st+2 "ky (p 9 P e «ee g D Construct a chain Kis Kos eee With Ky La n n, of type (p . D "9 eee 9 PB *) by the methed described above, and let Hs = K, X Be. The chain Hy» Hos e-e CGlearly has the required be properties. aja Theorem 2. Let H be a group having an infinite ehain of normal sub- groups H=H,> Hj, D.+. « Suppose f(n) is a monotonic increasing, integer valued, positive function defined on the positive intevers. Let He(n) = py for every n; k 21 and define G, = wf). x n) then G, = Gig / 0 for every n. - ae +] Proof. It suffices to show that gn} = HH y/ ae MH eat)? for Cai / cave = (Hf oH) He (n4+1 yt \ a Me (n Vf i ee Hy (n+1 )) = uA) Hein) = Sys = 8) then Note that 1°) Hpi) for all positive inte- Ho (se . thoes (s) gers s and t. For, by hypothesis, H He(sit) = He(s) and hence uis) g a = gS) (g(8) 9 (s) Bs > ®@ £(st+t) e(stt)? ai He(s) The reverse inequality is trivial since Hog +t) S He(g y° The first derived group of G, +4 is given by (1) 2 w(t), (n+1), (n+) - a(t +1) Gi, = HCH ° He(ntt //# ° He(ntt) ~ at ‘Be caat fa HeGa+)° Thus att = Hy) / HO a) since at )y = (Uy n+] f(nt1) ~ ° £(1)° The proof will follow by induction if it is shown that k) — a(k) +1), . . k+1) _ (c+ +1) One =a! "e ce) / ue "Be (ntt) duplies Oe ) = att ‘He cea // u® eet) whenever 1<k <n. But, if g(t) has the above form, then n+ (kH1) 2 (ply \()y (+t) / (a1) . Saet B Beacy) FB e(nay)/ BO "Ee(gy)? and it only remains to show that 7 (x) ay)? (n+1) ag (k+1) (x Hp H Hornet) = } Hetciq)° £ 4\ Note that Hott!) > pint) since k<n. Clearly ( OW) \) a yleHy i), (k) { > ; Cc e A(t c “(ety since H is normal in H. These inequal- He(k) = £(n+1) f(n+1) ° ° ities, and the fact that all groups considered are normal in H and henee commute, give (aC Key | (Uy (ott ly > yet ul) alt Hay He tic) ) e(nt1) = Het) = ylktty = qed, f£(n+1) £(k+1)° . . («) = yk) On the other hand, it follows from H Het) =H He (44) that (x) (1) 2 fx) (1) (+1) (a Hee) ) = (H Het) ) se Get) where the last inequality holds since He(n+4) is normal in H. But | . (k+1) (n+1) k <n implies that H 2H and He(K+4) = Hons)? hence c lk, (n+4) 5 (k+1 Hoey) / (1 ng Paley e(ntt) c(t) ® Bengt) = 4 te (ct) and the proof is complete. The following lemma is due to He Zassenhaus [12]. Lemma i: Let U and V be ideals of K, a commutative ring with identity. Define D, Lor every ideal. X of K as the set of all 2x 2 matrices (955) for o;,; in K guch that (a, yi 7 ty = 0, modulo X5- Then (DysDy) Dp» where (DysDy) is generated by the a i | , : set of all xyx y for x in D, abd y in Dye & Ot 8! ao ze ace. mh My ~Qm Proof: It suffices to show that yxy! = i, is in (OV), if x is in Dy and y is in Dye But -— . =| xyx y 1 I, = (xy - yx) x ‘y 4 = H ~j if (x=) G-b)--t) &-4)) xy Now x = i) and y= a belong to U, and ao respectively, hence The ring R is commutative, hence (UV), = (WU) 5 Since oo, is an ideal in the ving of all 2%* 2 matrices with elements in K, it follows from the above that x-LJyv-bh)-G-Le-L)e W&M); and henee, [= In) = I) -(y- Ip) (x -L )] x ly wig (UY). Lemmg 2: Suppose the group G contains a descending chain of normal Subgroups, G = G, > GoD cee . Let H be a subgroup of G such that, for i1<i<n, H contains a set of elements X; which maps on a complete set of generators of G, ,/%; i+ in the homomorphism G7 G/G, 44° Then HG, = G Proof: The relations < X,_, > G/G,, 2G, 4/6, and X,. GH imply nak? «Then < Kinny 7 Ca! Ge 2 Gn G, and X47 GH imply that HG... 2G, 44° Therefore HG, = H(HG,) DEG, 2G, 445 and the proof follows by induction. The next lemma is due to N. Blackburn. «x4 Ooo Lemma 3. [2, Theorem i.1]. Let G bea group generated by a set xX communis 5 ‘ ° \ ates . of elements. Define subgroups Ys (G@) recursively by the rules W4(G) =G, and Y3,,@) = (G.y,(@)) for 121. If Y isa set ef elements which together with Vga4(@ generate YWq(G)s then V5 44 (G) is generated by “Ys 45(G) together with all commutators A =| Le os xyx y i where X, y run through X, Y respectively. This is true for 4 = 1s 2s ee @ ® wi1< Chanter 3 The following theorems are provéd in this chapter. Theorem 3. If p> 2 and G, is an arbitrary non-cyclic abelian pegroup, then there exists an infinite chain of p-groups G, G5 aeeg such that G, = G so (03 (n) and gin) # <i>. Theorem 4. If G, is an arbitrary abelian 2=group which contains a 3 subgroup of type (2,2 ), then there exists an infinite chain of 2-groups G, 8 Go § ©8@ such that G. =¢ Gay ie, (n ) and G. (n (n) 7? <j > This chapter ends with a refinement of an inequality for p-groups of P, Hall (Theorem 5). The notation of Chapter 2 is used. 1 vp Definition 1. The group generated by the matrices a = ( 1 90 and b = is denoted by A, Jo 1 Definition 2. The subgroup of A, consisting of all matrices X in A such that X- 1, modulo Poe is denoted by A. Ons Cutline of the proof of Theorems 3 and 4: The subgroups AeA, > Ae PAs —.e0. are geen to form an infinite descending chain of normal subgroups of A, A monotonic increasing, integer valued, positive function f(m) is defined on the positive integers, and it is shown that Ken) Clg, for every n and k. (The definition of f(n) depends upon whether p is even or odd.) Groups G_ are n+#1° ete ns = n/a '2) _ . (n) defined by 6, A/A Aen) then by Theorem 2, G, = G4,/6 It is seen that gin) f<i>, If p> 2, then G, is abelian of type =] Dm (osp), and Theorem 3 follows from Theorem 1. If p= 2, then Gy is 4 2 2 a) i on ee. abelian of type (2°92°), and Theorem 4 follows from Theorem 1. The following lemmas establish some elementary properties of the group A = Age Lemna 4: (A, 24) S An Proof: This is an immediate consequence of Lemma 1 and the definition of An? if the ring K is chosen as R and the ideals U, V are taken as P’ and P". Lemma 5: A, is a normal subgroup of A. Proof: This follows from Lemma 4 since (AsA,) = (Ayobn) S Anay S Age ita, 4 /1+s, t\ Lemma 6: Let X = » Y= be elements ar 1+ \ U tiv / ef A such that X-Y=0,, modulo Poe Then X¥"' is an element gf Aye fi Proof: It suffices to show that XY ' = I, = 0, modulo Po. By (2 ‘ : ‘ : 2 hypothesis, there exist S49 bas Urs Vi in R such that s = aip’ “8,5 & n/2 $2 G+ pt u= Yr om, and v= &§ +p Vie When these expressions are substituted in ¥, it Pollows that ce f Fak & ee al YY a ww (2 om - XY i. (Itat 5 +o Bx) +p (v, tov, Bus)s Pp (As, cb %4) | ary c 3 nfa c \ BAL 60)9 (Hat S40 5-8 Y) +p? (6,4 $5, Ye,)/. Since X belongs to A, the determinant of X must be 1. That is, wn] Jomo teat St+taSe-2Y¥ =1, and the lemma follows. Lemma 7. Let n be a positive inteser, Then (1) A /i- 4 is eyclic of order vp, and re as ~ eee n (2) af aP --> a, bP —> b in the homomorphism A ——> A/Bonio? then Ao a/donte =<a,b >, where <a,b > is a non-cyclic “y group of order ps [ Ita, Vas \ , \ sag a2 mae Proof of (1): If X= | _ | ds in the set difference \ WY, its / wn c & a. an & ~y P Aon Aone 9 then a and o are integers, © and are integers multiplied by p> (The generators of A have this property, and it is clearly preserved under multiplication.) Thus a and 5 can be written as oe and q “+ S=p PS, pT Sa tee O<|§,1 <p. Not both of a, and oy can be zero, for this would imply that X belonged to Aona4? Since X is in A, the determinant of X mst be 17. That is, debX=1tat5+af- &V=1. The coefficient of p in det X is ct + by: hence $8, = 0 modulo (p). Since not both ay =O and Oy = QO ean hold, it follows that Gy 7 0 and O, # 06 ~{4~ Tt is clear that a, can be assumed to be positive: for, if \ at ay <0, then a can be weitten as a= (p + a,)p" + (a, = 1)pnt $ see, where now p + a, > 0. / Ita! 9 £3 \ Let Y= | be an arbitrary element of ‘. ‘y! 3 T+ ) Aon = Monty such that n , atl, af = pal +p nas Ftee yg O< lajl <p. It will be shown that if af =a,, then xy! belongs to Ap, yse This will imply that the order of A, nf dont is at most p. Suppose now that ad = Ose Then ~{ (We BS \ i+ S', « St \ (Fto+ § tea Gta By, * \ \ Es 145) / \- Yiig ita! \ % : HStatatsmaty | - where it is clear that the off-diagonal elanents belong to P pat Since this is already true for the off-diagonal elements of X and 4 ef ~ \ Yo. Now a, & » at, and SS! belong to Pp a. 4 By fy Vso and ¥! belong to pent | Therefore modulo pent | il Oo a ft- At and atS - 2tY 20 modulo peo*t, Thus it is only necessary to show that 1ta+ S'21 moduto pm, and 1+ 5 +a'=1 moauio po, j wo] 5 in order to verify that x¥"' is an element of Aniaqe But S +a! can be written as Stat = (S, tat) p+ (G+ ag) pel + eve ands since af = Oy the coefficient of > is 4 + 54° Since 4 + by =O module p, it follows that 1+ S+al BY module pent The other congruence is proved similarly. - oj aya LOW es © The order of Aon/bonas has been shown to be at most p This is a pegroup, which is not the identity group since pil tt a bak b =] g belongs to Ay, = Aonaq? Therefore bon/sonst has order p, and hence must be eyclic. Proof of (2). Note that a® = Pox | n 1; ve n / 4 : 0\ and b =| | Os 1 belong to Mons - Annan? Also, _| 15 so” Ve \/ 1, 0\ _ [trate sp" fp \ 2") pny (a (b = | , \O, 1 | ef, 1/ tp” Jp 5 1 is in Aonto if, and only if, s=t=0O modulo p. By Lemma 4, (bons Monty? S Aunt SAonigs hence Ao. 1/ko.i is abelian. Thus, ~16~ n n if a? > a, and bY ==> b in the homomorphism A <=> A/bonso9 then < 9D > is a none-eyclic subgroup of order p” of Bong4/bontn® Ita, 2 \ Now let x = . | be an arbitrary element of Ys I+ S/ . a Adapt (2n+1)/2 Mona ~ Aonige Since a and & are integers divisible by p 9 of it follows that they must be divisible by p” , Alsoy 2 = up” /p , n : and y= vp Jp for some integers u and v. Therefore 4 1+ B ex) 1 » O \ 9 1+ { ()( Voc! ps 1 / ~ x ant ef \ = b-up" Jot Yaw", 1- Yup" fet 6 / clearly belongs to Aonaoe Thus the homomorphism A > A/ka 19 maps ena : =i, & oA = < > x x where xa b i That iss dont /bonio Beh 5 and the proof is complete. The following three equations will be used to determine the derived series of the group A. 8 pe’ -p® b Let h(s,t) =a bP a? bP. then 4 + petty + prstett2 | pesite Jp (I) h(s t) = pataty Js ; a petn : ALSO, (II) a” n(syt) a?” n(e,t)7! = ( "17 12 \ 9242 S99 / where 124 4344943, +4429 sh O44 = 1+ potetiqt + p*8 3b+g43 + ps Abt2gth + ps Attgtg | =] Jeo : 42Gb btock 28 agpetitatt Jp . ® 2t+g42 Jp (p4 + p* toatl 4. 35 + 2p" +b+1 4+2b+q+ 426+ + ps abtbtqt2 + ps b 2), “42 ~ Listtot 2) and 4A > oes Gyo =x fo peretiate - prstattat3 . ps rabrqty, The corresponding identity for b is given by / (IIT) wP n(syt) bP h(s,t)7! = wt Boys where a 28+t+q42 3424043 A, 21 + petate _ jfetziiats | 42b-+q+ ,. = ps 2t+g43 hes 12 — 9,Stttqt astitqt2 t +btg+ Boy = 2p SUH fF ~ p2stbra & (+ ~ pi + patttatt ), and : Boy = 1 = prortiat? 4 yJet2ttatd , pAstabteat Lemma 8 Suppose p #2. Then (1) the homomorphism A ~ A/ ho 10443 maps h(s,t) ona gener~ ator of the cyclic group bog s0t40/ Moston for every BS, t > O03 f ‘ po =p =| (2) the homomorphism A > Mba 51 0%42q+h, maps a* h(s,t)a* h(s,t) q q we and vb h(s,t)b ” h(s,t) ( on a complete set of generators Of Ao s19t 10943 Bastot+2qth, Sy ts q 2 O01 Proof of (1). This is an immediate eonsequence of equation I and statement (1) of Lemma 7. =| Se Proof of (2). Equation II and Lemma 6 imply that g gq 4 stttott Pp “p \eb = ep , a h(s,st) a h(s,t) a modulo Mo stotdoo-td® Similarly, using IIT, g q . stttq+i wP n(sst) bd? h(syt)7! =P MOdULO Ao 5 1o442q-h," The proof now follows from statement (2) of Lemma 7, since p # 2. Lemma 9. Suppose p # 2, Define f(n) for every positive integer n as follows: f(1) = 2; if f(n) is_even, then f(nt1) = 2f(n) + 1; (n) hy if f(m) ig.odd, then f(n+1) = 2f(n). Then Mem) SA for every k> 1. Proof: Clearly h(0,t), ah(Ost) a” h(O,t)”', and bh(0,t)b™'n(o,t)7! (1) (1) belong to A for every t>0O. Take s=q=0 in Lemma & Then, for m>2, A is seen to contain elements mapped by the homo=- morphism A > A/A ik on a complete set, of generators of A! 44° Lt \ follows from Lemma 2 that Ay © a A, Yor every k> 1. That is, k Let n be chosen such that Ae(n) cS fa) Ay for every k > 1. The lemma will follow by induction if it is shown that A call), cor every k> 1. But, if A S 3) A,» ‘then f(nt1) = k TY me fe ee Rn) S ko . 4 1 ast. c (aa, ) (1) eat), since A, is normal in A. Thus See S cas ‘As henee it will suffice to show that Aj < ak, for ev k >] “e(nt1) = “£(n)"k oy ets in-tc mt Pp Tf f(n) is even, say f(n) = 2m, then aP 4b? , and =] 9= h(O,m—1) belong to etn) =A, whenever k 20. Thus nim sm) s mH rote 4 mh tree p ~1 , uP ~1 a’ hlOym=t)a? n(Oym-1)7', and bP h(Osm-7)b-P h(o,m-1) 1 : o belong to aft) = = aft) for every k > O. Therefore, whenever w>d4m+i, it follows f on a complete set of eenerators A, Por every k > 13 3 putk _ tk Tf f(n) dis odd, say f(n) = 2m +1, then a bs, and h(0,m) belong to Ae(n) for every k>0O. Thus h(mtkym), \ hy mk otk | mle mtk ~4 a® oh(OsmjaP h(Oym)7', and bP h(Oym)b™ —n(O,m)7! belong to as). Therefore, whenever w> 4m +2, it follows from Lemma & that ag) contains elements mapped by the homomorphism A > A/a on a complete set of generators of ASA, 44° Thus, by Lemma 2, A c ai) A, for every k > 1; that is, A S ahd A 4mt2 = “t(n) “k cosa ? *“e(nt1) = “£(n) “k° This completes the proof. Proof of Theorem 3: Let G, as i/a™ ap Aen) where f(n) is the function defined in Lemma 9. It follows from Lemma 9 and Theorem 2 that Oe \ G, = Gi. [enh Clearly G, = A/hy has type (psp). If it is show Kh * 2 oe a3 that gin ) #<1> the proof will follow from Theorem 7. ) Let k be the smallest integer such that A, /) 4 contains an element, say x, which does not belong to Area ‘a ,@), . {n-+4) Then gf) SA.» and ae) S (Ay sAy de By Lemma 4» (Ay 2A,) S Ao, (n+1) Thus A S Aor S Ayg 9 and alot } Ay. a4 = Ayaye Therefore, since xe Apias X g fat) gtd), > (ot) “+ Ae(nit)* It follows from =20= ; (n) _ ,{) (n+1) ee the proof of Theorem 2 that Ga = Ae(n) A Aetna) Thus (n) b ort ds gin) #<1> since xcA : This n+] Ae(n) and x f£(n+1)° completes the proof. An analogue of Lemma 8, for the case p= 2, is needed in the proof of Theorem 4. Lemme 10. Suppose p= 2. Then (1) the homomorphism A > A/ bog t0643 maps h(s,t) on_a gener- ator of Moga otio/tag40b+3 whenever s 3 t 2 QO; (2) the homomorphism A > M/bossott2q46 maps q q - q | - aP n(syt)a? h(syt)”' and bP h(t.s)b? h(t,s)”' one complete set of generators of Ao gsotoqs5/Aastottoqt6 whenever t>i, 8, q20. Proof of (1): This is an immediate consequence of equation I and statement (1) of Lemma 7. Proof of (2): Equation II and Lemma 6 imply that nAStttot2 1 = me * q wan ws a h(sotia © hlsyt) a modulo A 28t2ti2q7+6. Similarly, by III and Lemma 6, stttgt2 q q bP h(s,t)b > h(s,yt) b MOdUlo bog tot +2046, The proof now follows from stetement (2) of Lemma 7. The next lemma gives a sharper version of Lemma 4 for the special case p = 2. Lemma 11. If p= 2, then (Aj show) S domaine =2{= Proof: Ib follows from Lemm, 7 and equation I that Ao =<h(O gm=1 )PAomige By Lenma 4, (A, shomt4) SAsnige Since A, =<asb > it follows from equations IT and ITT that (A, 9<h(O,m—1 )>) S domo ® The proof now follows from the commutator identity (xeyz) = Grey) (kez) (las) sy) Lemma 12: Let p=2. Define f(n) for every positive integer n as follows: f(1) = 6, and f(nti) = 2f(n) +2 for n>. Then As Ae(n) GAA, Lor every k 2 1. 1) Proof: If m> 6 then, by Lemma 10, af contains elements mapped on a complete set of generators of bind Amn +4 by the homomorphism Ao A/Baaae Thus, by Lemma 2, Ae eng for every k > 1. Let n be chosen such that Aen ) cS AO )y for every k > 1. The proof will follow by induction if it is shown that (nt) (n) he(niq) GA Ay for every k> 1. But, if Ae(n) GA As then Aste) vor 7 since A, is normal in A. Thus race Sr S (aa i) = garth), K Af(n) k? (1) and it will suffice to show that Ae(n H)S f(n)*« for every k > 1. By definition, f(n) is an even number, say 2m. Therefore, pitk mitk by I, af 4 DP 4 h(m1,0), and h(O,m-1) belong to Ap(n) =A pittk apie 4 for every k>0O. Thus h(mtkym), a h(O,m-t)a ~ nO .m-1)”', and mtk whe 4 C4 \ 1 « 4 La .~ ~ bP oh(m-1,0)b ” _h(m=1,0)7! belong to Aaisy = = all) for every Belly s k > 0. Therefore, whenever w> Am +2, it follows from Lemma 10 one ge gel se that hein) contains elements mapped by the homomorphism A ~ A/Moa4 on a complete set of generators of A, /Bavae Thus, by Lemma 29 acl? for every k > 1. This is just the statement that Aum S Ag(y) 4 , ail) 4 af “6 Ae(niy) S Aen) K? and the proof is complete. Lemna 13. Let p= 2. Then Vee is an abelian group of type ) (27,27). Proof: It is clear that Vee is an abelian 2=eroup. The order of A/he is, by Lemma 7, equal to 2°, The order of a) (1) Ae/Be will be shown to be 2, Thus the order of A/A s Ag is 2, Since A =< a,b >, and aes b (1), both belong to Aes no element of the abelian group A/A 4 can have order greater than 8. Thus this group, having order 5, must have 2 generators. The lemma will foklow, since an abelian Z2-sroup having two generators, order 2°, and no elements of order greater than 2”, must have type (27,27). Let yy (A) be the subgroup of A defined in Lemma 3. Since A=< a,b >, it follows from Lemma 3 that Vola) = < by ¥3(A) >s al al 3 bo, and n where h = h(0,0) = aba v; (a) =< ahah !, pan, il Y A a By Lemna, Leg Yo (A) (Ay pA) Cc Ao} henes , by Lemma Tt. ¥5(A) = (ay ¥y(A)) S yedg) SA, and if ¥, (a) (As ¥3(A)) S (Ayal, SA 23m Thus Yn (A) = <h, ana th”, pho n™! 1 Piel | iol § YW, (A) > ds ineluded in <h, aha bh, bib hy Ag >. Note that Yn (A) = AM), hence 11 h', and php”! \ ete | aa Me is generated by images of h, aha bh in the homomorphism A > A/Age By equation I, ( ) (1+2 4% v2 ) h = h(0,0) = ON a 1-2! herice 2 145427, =302* /2 \ x | €# - A 9 \ 32% 2, 1-22 | 5 6 4 . , ; . _ and h’ €A,. Since, by Lemma 4, (Ay As ) SAg, the group <h > Ay 2 is normal in A. Clearly <h” > Ag/e has order 2. It remains to show that ay h* > Ag has order A. It is easy to see that the square of 4-1 T+ Alt, 4919 2 \ aha bh = i 8f/2 5 1-497 / is in Ase Also, h and aha th”! map on independent generators of and A, D< he > Age But ,_[ 174 82 fay 1+ 45/4 call Piel | ( T= 8, 12 /2 \ a? (ob 'n7!) (ahaa?) = | | xh \ 212 /2; Al / oe . ied Pied | ied Mikel . That is, aha 4h and bhb ‘h map onto the same element of 2h, athea as hence (aha7'n7!) h A < nfs he > Age AlSG, (aha! hv! ) h= aha . ; 2 GQ) e . has order 2 modulo < h° > Aes Thus A A,/< ho > Ag has two inde< pendent generators of order 2 whose product has order 2, hence the group has order 4. This completes the proof. \ Proof of Theoren 4: Let G, = a/a™ aa. where p=2 and f(n) is Q (a; the function defined in Lemma 12. 4 follows from Lemma 12 and Theorem \ 4 : = aL + 2 ad 2 that G, = a fo). By Lemma 13, G, has type (2° ,2°}. The 48 he § » AD cad zy m " argument used in the proof of Theorem 3 shows that g(t) # <1>. Theorem 4, is now an immediate consequence of Theorem 1. Remark, The full strength of Lemma 13 is not needed for the proof of f fheorem 4. It would suffice to know that VE has type (27,2 f for n<2, m<3. The exact type of aaa was determined in order to show that Theorem 4 is the strongest result obtainable from the group A. Refinement of an Inequality of P. Hall. The groups A, can be used to obtain a refinement of an inequality of P. Hall. This inequality is contained in the following theorem [4, Theorem 2.57]: If p #2 and G is a p-group of minimal, “ ° hh rid a * ® order for which rs ) #<1>, then |G] (the order of G) satisfies crewed n nw i; j 2 tn f] & 4 p <igisp" ° ° n - 3.9 The upper bound of this inequality was refined by N. 1t6 [4] to p ° An additional refinement is given by the next theorem. 25 mx Theorem 5: If p #2 and G isa p-growo of minimal order for which a™) 4 <1>, then |o| satisfies rat antl _ pe Me lah cp? 7) nt] Remark. It is interesting to note that > \ is precisely the upper bound found by Hall in the special case p = 2. Proof of Theorem 5: Suppose p #2 and let f(n) be the function defined in Lemma 9. It is clear that, for any fixed n> 1, a normal subgroup H of Ae(n) ean be found such that p(y Hl =p and “_ 2 * f read ¢ » HD Apty)aqe Since (Ay eH) ¢ (A, sho(n)) SAs(n)ay SH» it follows that H is normal in A. The required refinement of Hall's inequality nti =p? ot ana (af) # <i>. will be obtained by showing that |a/a Theorem 5 will then follew from the theorem of Hall. By Lemma 9, Ae(m) S a), for every m, k2> 1. Gonse- quently » sn = aur 2 WR: since eq) S a aca) and H2AMpraiq)+ Suppose (afn) ©) 3 OO a Ale Then (any) > (ie), ste) du/a. But . Dy > ert MD a > a (ett) (A e(k) Aecty)? hence (fa) HT) 3g Ger Beta yq)/He Therefore, by induction, (any 2A a Ais which is not the identity group since Ae (y )/Hl = De It follows from the definition of f(n) that 3 23 _ 2+] _ Qtk'-2 +P tere +2 Say for kk 2 0, f£(2k+1) = 2+ 2 = 2bn and : 2k#2 x can iC oe a oe ae It is easy to see from Lemma 7 that 36 mor jafa,| =p ~ af m ds odd, and BR I [A/A._| =p if om is even. Thus, if n is even, f(n) is odd, and nt2 2 (Beene =|) antl 8 3 2 2 |A/berny! =P =p ; If n is odd, then f(n) is even, and 2 le A 2 5 y @ n+ That is, [A/ber,)| =p” 2 tor n= 15 25 oe » But [An \/H] =p, (n)} gatt hence |[A/H] = p » This completes the proof. Ghapter 4 Let Gy be an arbitrary abelian 2~group which contains a 2 42, 2 2 subgroup having one of the types (2° 9292") (252.2 127)s or (252,2,2,2). An infinite chain of 2-groups Gy,» Gos «++ will be ~ (n) (n) constructed such that G, = G4 /Gn and Go # <1>. The method of construction will be similar to that used in Chapter 3. However, instead of considering the group A, three new groups are introduced. The notation of Chapter 2 is used. The prime p used in the definition of R and P is assumed to be 2. 14+ 2; fe) 1, 2\ Definition 3. Let ¢ = 9 Us 9 \ 0 5 142-2 0, 1 a {i, 0 { 3 and v= \ ) « Groups B, CG, and D are defined as follows: ee 1 B=< bet >, ©3 il A uo e, = v 6 and D i A C2 a < Vv ® Definition 4. The subgroup of D consisting of all matrices x e€ D such that = n x = I, = 0, modulo Py» n>, is denoted by D,. Subgroups BY and C, of B and G, respec- tively, are defined similarly. That is, B, = 8B a D3 O, =6 fe) Do ae Lemma, 14: (By»B.) SBS GisCn) Sones Oy D,,) S Diane Proof: This is an immediate consequence of Lemma 1 and Definition 4. =D Sew Ee : Is co fete S normal jn Bs; C. is normal in C; D, is normal Proofs This follows from Lemma 74. The next lemma will be used to determine generators of Cag? and D ne for n= 1, 25 eos « Lemma 15: Leb H bea sroup of 2x2 matrices with elements in Re 72 4 Let H,» for n= 15 25 «oe 5 be the subgroup of H consisting of all. = n matrices X ¢ H such that det X= 1 and X - 5 = 05 modulo Pos Suppose B contains elements X, Y, 4 such that - at] X- 6 = 05 modulo Po 3 n+] Y- B= 0 modulo P,''s and Z-\Y = 0, modulo Py s 5 where 8 = | |, @= \ > and \O5 1 / (¥2)", 1) 1+ (/2)"s 0 \ oO 4, t+ (ay 8. « Then H=< Xylo% gH >, and. HS nt] = Proof: It is easy to see that x vy zt. oF i ye = 0, modulo eel for any integers r, Ss, and t. Let W be an arbitrary element of fa om 1 vy Hy Bot ® Then =2Q— 1 +t as 2 Wo ] Ys,y itd/ where a = ay ( 2)” modulo ( fan", B= B, (/2) modulo ( /2)""", Y= \, ( /2)" modulo ( faye*t, and b= A, ( /2)" modulo ( faye, for O49 Ors vy9 by each either O or 1. Since We He? det W=1+ (at 5) +06 -BY =1. Therefore a, + 54 * O modulo 2. yar “ ue + A simple computation shows that 6& ‘gf | v y. W= = 0, modulo Poe L Therefore By 4 x iy 'z!-w2o, modulo Patt, and it follows that H, = < XsYs%si.4 > a © ¢ anne _ apap ge By Lemma 1, (H,sH,) GH,» hence H =< XYZ stay? where r, 8 and t= range over the integers. It is easy to see that nt] z S yt _ = 4 e gf Wy I, = 0, modulo P, if, and only if, 2 divides each of ry s, t. It Follows that LoS ob Xv ¥" 2 eg Lan +4 if, and only if, 2 divides each of r,s, and t. Therefore Ho: nity = S, and the proof is complete. Lemma 17: De Define £ f(n), for every positive integer n, as follows: (nti) = 3f(n) +4 for n>1. Then, for al) positive ne (n) G3 Dery) EDM 'D,. integers n and kg aay 2, Com) So! 2 -a-1_/{! 4 2 Proof: Note that a “cac a =| =U» and \o 4 be beb =| | =v, belong to B. Thus, for t > 0, the \ L 1 y following three matrices belong to B é atti Ly gb (i, (2272 ght? \ eu eu = | } \0, 1 )? 4 gbtl jit 1 , 0) cy ev. = ae (2+2 J2)2"" a 1/ ; atta tte, (1. + bt 4 2PttO) _p2tt5 5 \ and aX ba” pis! Taking these matrices as the elements X, Y, and Z of Lemma 16, with H=B and A = Be it follows from Lemma 16 that, for every +t > 0, gil) contains elements mapped by the homomorphism B > B/Bos a7 on a complete set of generators of Bou 16) Bogan The three matrices ott 4 aot [is (212/328 2 G a c a = { \ O, 1 9 tH tH 1. 5 0) =| .2 2 _ | cb eb = \ te] \ (242 f2)ae", 1] ? ‘ ‘ (7 ¢ tt2 FB 4 peti | apttt3 2° 2 =2" =2 and a va v = | \ atts , te ott fa)? ‘ also belong to git) for every t 2 0. Taking these three matrices as X, Y, and Z in Lemma 16, and letting H =B, H = Bw? it follows (1) that 8B contains elements mapped by the homomorphism B => B/Bos 4 oth | com on a complete set of generators of Bot 45/ Bona whenever t > 0. Combining these results, and using Lemma 2, it is seen that BM) 2B, for every k 2 Te The proof (for B) will follow by induction if it is shown . (n) as (n+1), (n) that Be(n) S B BY implies Bein) S B Bye rf Ben) cS B Bis then, since Be. is normal in B, £ Bet ) x (ps 1) E platt). «Therefore (1) (nt+1), \. — o(ntt) Be(n)Be S (BY BIB, = BY "Bye and it will suffice to show that Be(ni1) S hla. By definition, f(n) is odd, say f(n) = 2m +1 where m> 2. It is easy to see that a 5» Db » & 9 u » and v belong to Bont whenever +t > 0. Since m> 2, a calculation shows that | Smt 3-H (2 om gmt (om an\"" gt ( 1, 2 \ , & 90 Uu \a 9b” j wu | modulo Bemtoey? \ f vA \ Os | and (am ey at ee amt - ( Too» O\ modulo B \ eh gr / \ 3 / \ gomttt3 4 ‘. ie] 6mnt2t+7 e \ 9 7 Also, by equation I, ( pem —pmt2\ 4 4 gamed omtat 6 pdm 2643 | \a- 9b je \ Aantte+5 fo , i- nit t+3 Thy es whe ole ‘7 ) = ah! ) %, > - hese elements belong to Betn) = Bont whenever t > 0. Hence, by Lemna 16 (with the above matrices as X, Y, 2 and H=B, H, = Bs pil) Betn) a complete set of generators of Bemtogrel Bem L247" contains elements mapped by the homomorphism B > B/Bem 42 Db On It follows from Lemma 6 and equations II and IIT that ; Bmnt-+3 aitt+t aot H ~| — 1 9 apt rr? f2 \ a h(mym) a h(mym) ° = 0, , / modulo Be oregs and. met tt _oatttl = ( 1 os 0 b h(mgm) b n(mgm)™* \ p3mtt+3 LBs 1 | modulo Bem ete? / Another computation gives By E Far . 1 t, L&é \ ( gemrbte of \ [ 1+27 att+3 J2 + yom? wT ngomte +6 le \ émitts 5 , i- nanrt+3 J} \ whenever t > 0. Hence, by Lemma 16 (1) These elements belong to Bom 4 (with the above matrices as X,Y, Z and H=B, H =B), bd 2 2 cet ¥ pil) (4) = 8B ontains elements mapp he homomorphism Bein ) = Borne contains elements mapped by the homomorphism > B/Bg, éneoteg OF 4 complete seb of generators of Bemogay BE 6mt2548° Thus BM (n) contains elements mapped by the homomorphism B > B/B., +4 a on a complete set of generators of B,/B 4 whenever w> 6m +6. It ) for every k > 4 [x n)"« 2 > B, émtG tor every k be dn oo oak ( follows from Lemma 2 that Be particular s. Bhi, 2 a = Bae(n 44 = Bem) This completes the _ (n) proof of the statement Be(n) CB Bie The above proof remains valid if B is replaced, throughout, by either © or D. The lemma follows. ‘ty &. °F ay he ry Theorem 6G: Let Gy 2) subsroup of one of the types (2° 92792"), (24252”, 9 OF (242525252). Then there exists an infinite chain of 2-groups Gas G59 ese Such that \ n = a ,,/a{n) and a uy 2 <P. Proof: let H, = a/R) i= coos and L, = D0 bo) Tt follows from Lemma 17 and Theorem 2 that H, = H, WA, (n) nt eet The argument used in the proof K, = Kae, and ~ =, of Theorem 3 shows that Ae # <1>, xn) #<1>, and 1,04 # <1>. Theorem 5 will follow from Theorem 1 if it is shown that: Hy can be generated by three elements of order less than or equal to 43 Ky can be generated by four elements, two of which have order 2 while the remaining two generators have order less than or equal to 4; Ly can be generated by five elements of order 2. Since B= < a,b,c >, where a’, bo”, and oft belong to Bs = Borys the group Hy = B/a\e 4) can be generated by three elements of order less than or equal to 4. \ Since © =< asbycgu >, the group Ky = of", can be 4 A generated by four elements. It is easy to see that b” and e¢ belong to Gre But Ts 4, f2 + 8 u*(e,u) = EC, S o(Mo,, 0, 1 and a*(esa) = (su), hence uw and, a” belong to oo, Thus two of these generators have order 2, while the remaining two generators have order less than =3h= or equal to 4. Since D =< a,b,cgu,v >, the group Ly = oop, can be generated by five elements. Clearly ole, eo, hence a and f/f 4 5 0\ a” belong to Dp, Also, ve 1 aw) = | | £ De c op, 4 f2 +8, 1 / and, b°(e7! yb) = (ov! yw). Therefore b” and v" belong to py ° y) j ’ 2 {1 4 2.[ 1 0\ Note that ui =| ; P v= and \.0 2 21, 8 /2\ (a,b) = ( | « Thus, by Lemma 16, \ 42, 1-4/ Dy =< u” sv" (a”sb) Dg Ae Similarly , Ds =< a” sb s(asv) D, an Since a” yb" yu", and v belong to pp, it follows that op, 2Dz. Tt is easy to see that (2+ 2, 62-8 \ 642-8, 3428) Thus Pa & pp, and hence Ly ean be generated by five elements of order 2. This completes the proof of Theorem 6. Remark. An argument similar to that used in the proof of Lemma 13 shows that B/a'p, has type (2792 52°), ofoN¢, has type (25252792), and oo, has type (252,252,2). Thus Theorem 6 is the strongest result obtainable from the groups B, GC, and D. Whether Theorem 6 is the best possible result, for Gy a 2-group with at least three generators, remains an open question. [1] [2] [4] [5] [6] (7] [8] [9] [10] [11] [12] =35— REFERENCES N. Blackburn, On prime power groups with two generators, Proc. Camb. Phil. Soc. Vol. 54 (1958), pp. 327-337. N. Blackburn, On prime power groups in which the derived group has two generators, Proc, Camb. Phil. Soc. Vol. 53 (1957), PPe 19-27, W. Burnside, Some properties of groups whose orders are powers of primes, Proc. Lond. Math. Soc, (2), Vol. 11 (1912), PP. 225-245, P, Hell, A contribution to the theory of groups of prime power order, Proc, Lond, Math. Soc. (2), Vol. 36 (1933), pp. 29-95. N. It6, Note on p-groups, Nagoya Math. J. Vol. 1 (1950), pp. 113-116, W. Magnus, Beziehungen gwischen Gruppen und Tdealen in einem speziellen Ring, Math. Ann, Vol. 111 (1935), pp. 259-280, W. Magnus, Neuere Ergebnisse uber auflésbare Gruppen, Jber. Deutsch. Math, Verein, Vol. 47 (1937), pp. 69~79, A, Scholz and 0. Taussky, Die Hauptideale der kubischen Klassenkérper imagingrquadratischer Zahlkorper, Journal fiir Math, Vol. 171 (193L), po. 19=416 O. Taussky, Research Problem 9, Bull. Amer, Math, Soe. Vol. 64 (1958), pp. 124. 0, Taussky, A remark on the class field tower, J. Lond. Math, Soc. Vol. 12 (1937), po. 82-85. H, Zassenhaus, Theory of Groups, (trans.) New York, Chelsea, (1949). H. Z2assenhaus, Unpublished.