An Exposition on The Laws of Finite Pointed Groups

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1 An Exposton on The Laws of Fnte Ponted Groups Justn Laverdure August 21, 2017 We outlne the detals of The Laws of Fnte Ponted Groups [1], wheren a fnte ponted group s constructed wth no fnte bass for ts laws. 1 Introducton 1.1 Defntons Defne a word on the countably nfnte alphabet {x = x 0, x 1, x 2,... } to be a strng of sad symbols, along wth ther formal nverses. That s, such a word s an element of the free group generated by {x 0, x 1, x 2,..., }. Gven a group G, say that a word w s a unversal law (or more often smply a law ) of G f, gven any assgnment of group elements g 0, g 1, g 2,... to the symbols x 0, x 1, x 2,..., the correspondng element of G s trval. For example, G s an abelan group f and only f x 1 x 2 x 1 1 x 1 2 s a law of G. We say that a ponted group (G, g) s a group G along wth a dstngushed element 1 g G. A homomorphsm of ponted groups φ : (G, g) (H, h) s a homomorphsm of groups φ : G H such that φ(g) = h. That s, homomorphsms must take dstngushed elements to dstngushed elements. A sub-ponted-group (H, h) of a ponted group (G, g) s a subgroup H G, whch addtonally s a ponted group such that h = g. That s, sub-ponted-groups are just subgroups contanng the dstngushed element. Thus, the sub-ponted-group generated by a subset S G of cardnalty S = ν s the subgroup generated by S {g}; ths sub-ponted-group s sad to be ν-generated (as a ponted group). It s a fact that the somorphsm theorems for groups hold equally well for ponted groups [4]. These ponted groups too have a noton of a law; the only dfferences wth the pror defnton 1 Sometmes, a ponted group wll be such wthout havng ts dstngushed element be explctly mentoned, f t can be nferred from context (especally f g has been mentoned already). Ths s about as dangerous as mentonng a rng wthout explctly mentonng ts addtve dentty. 1

2 are that we add a letter y to the alphabet from whch our words are constructed, and that we requre y to be nterpreted as g. For example, (G, g) has g Z(G) f and only f xyx 1 y 1 s a law of (G, g). Note how, wth ths addton to our language, we can talk about certan propertes held by g alone. Certan laws mply others; for example, f squares commute (.e. the law x 2 1 x2 2 x 2 1 x 2 2 ), then fourth powers commute too. Thus, we mght ask: gven a group G, what are the sets of laws of G from whch all laws of G may be deduced? Such a set of laws s a bass for the laws of G. Certanly the set of all laws s a bass for tself; we mght ask nstead, when can we fnd a smaller, more tractable bass? When can we fnd a fnte bass? A class of groups G s a varety f t s defned by some set of laws; that s, f there exsts some set W of words such that G G f and only f, for every w W, w s a law of G. It turns out that we may talk about the varety generated by a class of groups H: the smallest varety contanng every H H. Call ths varety Var(H). Fact: Var(H) s closed under takng homomorphc mages, subgroups, and (arbtrary) cartesan powers. So, f a group H s obtaned from a group H H by takng mages, subgroups, and powers (not necessarly n that order), then H Var(H). Analogous facts hold for varetes of ponted groups. General facts about the theory of varetes may be found n [4]. 1.2 Outlne When G s fnte, a theorem of Oates and Powell [3] tells us that a fnte bass always exsts. We now ask an analogous queston for ponted groups: does every fnte ponted group have a fnte bass for ts laws? Ths tme the answer s no, a result of Roger Bryant [1]. We follow the counter-example of Roger Bryant n constructng a fnte ponted group (P, p) wth no fnte bass for ts laws. Ths ponted group (P, p) has a one-parameter famly of words w n = w n (x 1,..., x n, y) as laws 2 ; we frst construct a one-parameter famly of ponted groups {(Q n, q n )} n N, such that each (Q n, q n ) does not have w 2 n as a law. Therefore, n partcular, no (Q n, q n ) s n the varety generated by (P, p). Each (Q n, q n ) wll turn out to addtonally have the property that every (n 1)-generator sub-ponted-group s n the varety generated by (P, p). Therefore, (P, p) wll not have have a fnte bass for ts laws; f t dd, then the number of 2 Among others, of course. The w n are not an exhaustve lst. 2

3 varables appearng across ths fnte bass would be fnte (say, there would be m of these varables), and hence Q m+1 (beng outsde of Var(P, p)) would have some a 1,..., a m Q m+1 whch fal some word w B. Yet, the m-generator sub-ponted-group a 1,..., a m, q k Q m+1 lands n the varety generated by (P, p), and hence would satsfy w, a contradcton. For a general reference on fnte groups, see [2]. 2 Constructng the (Q n, q n ) 2.1 Constructng the group A Let s see the (Q n, q n ). Fx a postve nteger n, and defne A = A n to be a (multplcatve) elementary abelan 2-group generated by bass elements e 1,..., e n. That s, A := e 1,..., e n e 2 = 1, [e, e j ] = 1. So, A has order 2 n, and each element a of A s wrtten unquely as n a = e ɛ, =1 wth each ɛ ether 0 or 1. It wll later be useful to now observe that each a, actng on A by leftor rght-multplcaton, s a permutaton of order 2. Beng elementary abelan, A s automatcally a vector space over the feld wth two elements F 2 = Z/2Z. Thus, we may speak of lnearly ndependent elements of A, of a bass for A, and so on, whch we shall do over the course of ths paper. It may also be helpful to realze that A has a rather large automorphsm group: GL n (F 2 ) Lemma 1. Aut(A) acts doubly transtvely on the non-trval elements of A. Proof. Say that we have non-trval elements a 1 a 2, a 1 a 2, and that we wsh to send a 1 to a 1 and a 2 to a 2. Because a 1 and a 2 are dstnct, non-trval elements of A, they are lnearly ndependent (because the only non-dentty multple of any a A s a 2 = 1 A). Thus, we may extend {a 1, a 2 } to a bass for A. Smlarly, {a 1, a 2 } s an ndependent set, and so may be extended to a bass. Thus, the desred automorphsm s gven by mappng the former bass to the latter. Morally, ths means that the e wll have no specal role to play, except n a countng argument 3

4 later. Secondly, ths means that we may, n an argument, wthout loss of generalty, pck any non-trval a A, and once we have done ths, agan pck, wthout loss of generalty, a non-trval, dstnct a A. We ll use ths repeatedly. 2.2 Constructng the group B Now, let β be a multplcatve cyclc group of order 3, and let B be the (multplcatve) elementary abelan 3-group generated by bass elements β a. At the moment, we regard these β a as formal symbols. That s, B := {β a } a A (β a ) 3 = 1, [β a, β a ] = 1. So, we may also vew B as β A, the set of β -valued functons on A, wth pontwse multplcaton. Ether way, ths s a group of order 3 2n, and each element b of B s wrtten unquely as b = a A(β a ) ka, where each k a Z/3Z = F 3. When we nterpret elements of B as functons, we have b(a) = k a. Let us vew β as beng an element of B by dentfyng β = β 1, where 1 A. Lke A, B s elementary abelan, and so s automatcally a vector space (n ths case over F 3 ). Therefore we shall freely use lnear algebra arguments also when dscussng B. 2.3 Constructng the acton of A on B by Now, let A act (on the rght) on the set of bass elements {β a } a A, permutng the bass elements β a a 0 := β aa 0. Observe that ths notaton makes sense, n the sense that we may vew β a as ether the prmtve symbol β a, whch generated B n the frst place, or as β acted on by a. Now, let us extend ths multplcatvely (and therefore lnearly) to the whole of B, defnng b a 0 = b a 0 by ( ) (β a ) ka a 0 := a A(β aa 0 ) ka. a A Thus, the acton of A on B s fathfully gven by a set of permutatons of the bass elements β a. Thus, unlke n A, the bass elements of B are dstngushed from other elements of B, specfcally by the acton of A on B. 4

5 2.4 A tangental observaton about B We may also see that (B, ) s somorphc, as a group, to the group rng (F 3 A, +), va a A(β a ) ka k a a, a A and that we may regard each element of B unquely as β a A kaa := a A(β a ) ka, off-loadng all of the nformaton about an element b = β r B nto the exponent r, and so that the above somorphsm becomes β r r. Accordngly, the rght-acton of A on B extends to an acton of the group algebra F 3 A on B, gven by β a0 k a a = (β a0a ) ka, a A a A and extendng multplcatvely. Thus, A acts by automorphsms of B, and generally, elements of F 3 A act by endomorphsms. Under the somorphsm (B, ) (F 3 A, +), ths looks lke k a a a 0 = k a aa 0. a A a A That s, the acton of A on B s just (rght-)multplcaton n F 3 A. 2.5 Constructng the group W Now, consder the semdrect product W := B A = BA (wth respect to the aforementoned acton of A on B), whch s also the wreath product β A. So, W s a group of order 2 n 3 2n. As wth any semdrect product, every element of ths group s wrtten unquely as ba, for some b B and a A. Further, as wth any semdrect product, we have the equaton ab = b a 1 a, whch n our case (where a 1 = a) s more smply stated as ab = b a a. So, we may further nterpret b a as b a = aba 1 = a 1 ba B W. 5

6 Ths s a specfc case of the prncple: whenever a group G acts (on the rght) on another group H, we shall wrte h g for the mage of h under the acton of g. Ths wll always be conjugaton wthn a semdrect product HG = H G. 2.6 Countng maxmal subgroups Now, A, as an elementary abelan 2-group, s automatcally a F 2 -vector space; maxmal subgroups are then just subspaces of dmenson n 1. Further, by defnng an nner product on A by e, e j := δ j and extendng blnearly, where δ j s the Kronecker delta, we see that, for all k, subspaces S A of dmenson k are n bjecton wth subspaces of dmenson n k by S S, where S := {a A : a, S = 0}. So, f we would lke to count the number of maxmal subgroups of A, t s suffcent to count the number of 1-dmensonal subspaces {1, a} A. But, these are n bjecton wth the non-trval elements a A, of whch there are 2 n 1. So, suppose that the maxmal subgroups M of A are gven by M 0, M 1,..., M 2r, where r := 2 n 1 1. One mght note that the automorphsm group of A also acts doubly transtvely on these maxmal subgroups; that s, the M are also hghly nterchangeable. Specfcally, we have the same stuaton as that for non-trval elements a A, as establshed n Lemma 1: n an argument, we may wthout loss of generalty pck any M, and then wthout loss of generalty pck any dstnct M. 2.7 Certan subgroups of commutators of W Gven a maxmal subgroup M of A, consder now the commutator subgroup [B, M ] = b 1 m 1 bm : b B, m M. Now, we have b 1 m 1 bm = b 1 b m, and b = a A (βa ) ka. The β a commute, so that [B, M ] (β a ) 1 β am : a A, m M, 6

7 because each b s a product of the commutng β a s. commutator [β a, m] [B, M ], so we have Conversely, each (β a ) 1 β am s n fact the [B, M ] = (β a ) 1 β am : a A, m M. Now, for each m M whch s not 1 (of whch there are 2 n 1 1), consder the element β m β 1. Smlarly, fx some a 0 A \ M, and consder the 2 n 1 1 elements β ma 0 (β a 0 ) 1, where m ranges agan over the non-trval elements of M. We ll see n a moment that ths set S := {β m β 1, β ma 0 (β a 0 ) 1 : m M \ {1}} of sze 2 n 2 s a bass for [B, M ], so that each [B, M ] s a group of order 3 2n 2 and ndex 3 2 n B. () That S s ndependent may be seen by consderng the matrx , where each block has 2 n 1 rows and 2 n 1 1 columns. It s easy to see that ths matrx has full rank: 2 n 2. Now, we may nterpret the columns of ths matrx as the vectors n S, represented wth respect to the ordered bass ((β m ) m M \{1}, (β a0m ) m M \{1}), and n partcular the frst row corresponds to β, and row number 2 n corresponds to β a 0. Therefore, [B, M ] has dmenson at least 2 n 2. () Secondly, S spans because [B, M ] s n the kernel of the lnear mappng Ψ : B F 3 F 3, 7

8 where Ψ(b) = (Ψ 1 (b), Ψ 2 (b)) and ( ) Ψ 1 (β a ) ka := k a a A a M ( ) Ψ 2 (β a ) ka := k a. a A a A\M Ths follows from the fact that the generators (β a ) 1 β am of [B, M ] are n the kernel of Ψ. It s easy to see that ths mappng has mage of dmenson two, and therefore that ker f has dmenson 2 n 2, so that [B, M ] ker f has dmenson no more than 2 n 2. Note the general form of ths argument. We break up A nto a partton X = {aa : a A} gven by the cosets of some subgroup A A, exhbt lnearly ndependent vectors n B that, f we nterpret them as functons b : A β, have support contaned exclusvely n one element of ths partton, and fnd lnear functonals f whch only depend on components of the vector correspondng to some element of the partton, that s, functonals f of the form f(b) = f (b aa ), where aa s some coset of A and f s some affne functonal defned on aa. Arguments of ths form wll be used n Secton 4 and Certan quotents W of W correspondng (loosely) to sad subgroups Now, each [B, M ]M s a subgroup of W, as, gven an arbtrary generator b 1 b m [B, M ] and an arbtrary n M, (b 1 b m ) n = (b n ) 1 (b n ) m [B, M ]M, because b n B. Thus, [B, M ]M = M [B, M ]; that s, M acts on [B, M ] by conjugaton, and [B, M ]M = [B, M ] M s n fact a semdrect product tself. Thus, each [B, M ]M s a subgroup of order 2 n 1 3 2n 2 and ndex n W. In fact, each s a normal subgroup, as, gven an arbtrary generator b 1 b m n [B, M ]M and an arbtrary b 0 a 0 BA, we have (b 1 b m n) b 0a 0 = [ (b a 0 ) 1 (b a 0 ) m][ (b a 0 ) 1 (b a 0 ) n] n [B, M ]M where we ve used the fact that B and A act on themselves trvally by conjugaton (.e. B and A are abelan), and that a b = [b, a]a for any a A and b B. Thus, we may consder the quotent W = W/[B, M ]M 8

9 along wth the natural homomorphsm φ : W W. Now, n the mage of φ, the subgroup M collapses down to 1, the bass elements {β m : m M } are all dentfed, as are the elements {β a0m : m M }. Thus, W s somorphc to W n the n = 1 case,.e. W (Z/3Z Z/3Z) Z/2Z, where Z/2Z acts by nterchangng (1, 0) and (0, 1). Say that β := φ (β) and β := φ (β a 0 ) correspond to these elements, and further let us defne ξ = φ (a 0 ). Thus, W = β, β, ξ. 2.9 Auxlary groups Z Z and actons on them by the W For each, consder two copes of the quaternon group (of order 2 3 ), Z = ζ, η, θ, ι, ζ 2 = 1, η 2 = θ 2 = ι 2 = η θ ι = ζ and Z generated smlarly by elements ζ, η, θ, ι, wth smlar relatons. These ζs, ηs, θs, and ιs are, of course, analogues of 1,, j, k Q 8 C. Let each W act on the correspondng Z Z (on the rght), where: a) β acts by permutng η, θ, ι Z and fxes Z pont-wse, b) conversely, β permutes η, θ, ι Z and fxes Z pont-wse, and c) ξ permutes Z and Z,.e. (a, b) (b, a). Observe that the actons of β and β are commutng, exponent-3 bjectons, whch are conjugate by the (nvolutve) acton of ξ. These are exactly the relatons of W ; thus, ths acton s well-defned (and fathful) Constructng the groups C and the actons on them by the W Now, consder the central, non-trval elements ζ Z and ζ Z, along wth ζ ζ Z Z. Ths latter element s agan central, and so we can consder the quotent C := Z Z ζ ζ, a group of order 2 5. We wll abuse notaton, and refer to elements of C as tuples. Let γ C (somewhat arbtrarly) be the mage of η Z Z, so that γ has order 4. Also, let δ := γ 2, so that δ s the mage of both ζ and ζ n C. So, computng n C, we have that δ = (ζ, 1) = (1, ζ ) C 9

10 s a central element of order 2, and s addtonally a commutator, as j = j Q 8, so, agan computng n C, δ = [(η, 1), (θ, 1)] C. In fact, δ alone s the centre and commutator subgroup of C, whch may be seen explctly: 1 Q 8 s the centre and commutator subgroup of Q 8, so that (1, 1), ( 1, 1) s the centre and commutator subgroup of Q 8 Q 8, and the quotent of ths group by ( 1, 1) s exactly analogous to our case. Now, because the aforementoned acton of W on Z Z fxes ζ ζ, ths acton yelds up a well-defned quotent acton on C Facts and computatons We wll later need to observe that γ β = (1 β, η β ) = (1, η ) = γ C, that [γ, γ β ] = [(1, η ), (1 β β, η )] = [(1, η ), (1, θ )] = (1, ζ ) = δ C, and that, smlarly, whch all together mply that [γ, γ β 1 ] = [(1, η ), (1 β β, η 1 )] = [(1, η ), (1, ι )] = (1, ζ ) = δ C, [γ, γ β β ] = [γ, γ (β β ) 1 ] = δ C. Further, we should know that δ ba = (δ φ (b) ) φ(a) = δ because both β and β fx δ, as does ξ. Thus, each δ s fxed by the acton of W. At ths pont, we may mostly strke from our mnds the ntty-grtty of the quaternon groups Constructng the groups C and D and the actons on C and C/D by W Let 2r C := =0 C 10

11 be the drect product of the groups C (and therefore a group of order 32 2r+1 = 2 5 2n 5 ), and let us consder the subgroup D C generated by the elements δ δ j, for 0, j 2r. That s, we defne D := δ δ j : 0, j 2r. Note that the group D := δ : 0 2r C s generated by (2 n 1)-many commutng nvolutons, and s therefore elementary abelan of order 2 2n 1. Vewng the δ as an F 2 -bass of D, we have the lnear functonal f : D F 2 gven by δ 1 for every 0 2r and extendng lnearly; the mage of an element d D we may vew as ts party. In partcular, we also have the restrcton f D of f to D, and we may talk about the party of these elements. In fact, lookng agan at the generatng set of D, every element of D has even party; better yet, D s exactly the elements of D wth even party. Thus, D = f 1 ({0}), and so D has order 2 2n 2 and ndex 2 2n+2 3 n C. We wll later use the fact that δ 0 δ 2r D, as there are 2r + 1 such ndces (an odd number). Now, let W act on each C by pullng back each φ, that s, let us defne, for c C, w c := φ (w) c. Now, let us extend ths component-wse to an acton on all of C, that s, for c = (c ) 2r =0 C, w (c ) 2r =0 := (φ (w) c ) 2r = Constructng the ponted group (Q n, q n ) So, we may consder the semdrect product CW = C W, a group of order 2 5 2n +n 5 3 2n. Now, as we ve seen, each δ s fxed by W, so that all of D s, so that W nherts a quotent acton on C/D. Now we get to the frst punchlne: let Q n := (C/D)W, and gven the element γ := γ 0 γ 2r = (γ ) 2r =0 C, defne q n := (γ 1 D)β (C/D)W. Thus, we have our one-parameter famly of ponted groups (Q n, q n ), each respectvely of order 2 2n+2 +n 3 3 2n. 11

12 We now have three thngs remanng to show: that each word w 2 n s not a law of the correspondng (Q n, q n ), to construct the ponted group (P, p), and that every (n 1)-generator sub-ponted-group of (Q n, q n ) s n the varety generated by (P, p). Let s see these n ths order. 3 Falure of the law w 2 n n (Q n, q n ) We defne the words v n (x 1,..., x n, y) := y x1 y xn w n (x 1,..., x n, y) := [y 3, (y 3 ) vn ]. Now, let the elements of A be enumerated as α 1,..., α 2 n, and let ρ := (γ 1 β) α1 (γ 1 β) α 2 n = v 2 n(α 1,..., α 2 n, γ 1 β) CW σ := [(γ 1 β) 3, ((γ 1 β) 3 ) ρ ] = w 2 n(α 1,..., α 2 n, γ 1 β) CW. These elements ρ and σ are thus mages of the words v n and w n n CW. In partcular, t follows that w 2 n(α 1,..., α 2 n, (γ 1 D)β) = σd (C/D)W too s a value of w 2 n, whch s non-trval f and only f σ D, so t wll suffce to show the latter. Frst, we observed that γ β = γ, whch just means that γ and β commute. Further, ths mples that γ β = γ, by the coordnate-wse defnton of the acton of W on C. Thus, γ and β commute. But ths means that (γ 1 β) 3 = γ 3 β 3 = γ 3 = γ. In fact, we may generalze ths fact a lttle bt: gven any α A, we wll see that γ α and β α commute. Remember that C s a drect product of the C ; let us fx any such, and consder the th component of (γ α ) βα, whch s (γ φ (α ) ) φ (β α), agan usng the defnton of the acton of W on C. If α M, then φ (α) = 1 and φ (β α ) = β, so that (γ φ (α ) ) φ (β α) = (γ 1 ) β = γ. In the other case, when α M, we have φ (α) = ξ and φ (β α ) = β, so that (γ φ (α ) ) φ (β α) = (γ ξ )β = γ ξ. because γ ξ has trval frst co-ordnate. But, n ether case, the acton of φ (β α ) s always trval 12

13 on γ φ (α ). Thus, (γ φ (α ) ) φ (β α) = γ φ (α ) whenever 0 2r. Thus, (γ α ) βα = γ α,.e. γ α and β α commute. So, ρ = (γ 1 β) α1 (γ 1 β) α 2 n = (γ 1 ) α1 (γ 1 ) α 2 n β α1 β α 2 n. It s also true that γ commutes wth γ α, for any α (whch may be seen by agan lookng at each co-ordnate). Thus, γ (γ 1 ) α 1 (γ 1 ) α 2 n Ths mmedately mples that [γ, γ ρ ] = [γ, γ τ ]. = γ, so that γ ρ = γ τ, where τ := β α1 β α 2 n. Now, breakng down nto components, we may wrte [γ, γ τ ] = [γ 0, γ φ 0(τ) 0 ] [γ 2r, γ φ 2r(τ) 2r ]. And then, we observe that 2 k 1 (mod 3) f and only f k s even, and otherwse 2 k 2 (mod 3). Ths, along wth the fact that exactly half,.e. 2 n 1, of the elements of A are elements of M, shows that φ (τ) = φ (β α1 β α 2 n ) = (β β ) 2n 1, whch s β β or (β β ) 1 f n s odd or even, respectvely. Ether way, a fact we ve prevously observed. So, fnally But, as we ve observed prevously, δ 0 δ 2r D. [γ, γ φ (τ) ] = δ, σ = [γ, γ ρ ] = [γ, γ τ ] = δ 0 δ 2r. 4 A few prelmnares We d lke to move on to constructng (P, p) now, but we need two thngs frst: to outlne the strategy (whch s to see that every (n 1)-generator sub-ponted-group R Q n s n the varety generated by (P, p)), and a lemma (whch s the fact that the subgroups [B, M ]M W have trval ntersecton). We wll see that 3 R s a subgroup of a group (C/D)BM, whch s n turn a homomorphc mage of a group (C/E)BM. Ths latter group wll be somorphc to a subgroup of a group r k=0 Ω k, 3 In ths dscusson, we gnore the dstngushed elements n our ponted groups for notatonal clarty. 13

14 whch wll tself be a quotent of a power of P. Hence, we wll see that (R, q n ) HSP((P, p)) va a chan of Hs, Ss, and P s. Of course, because HSP Var, ths wll show the desred fact. Lemma 2. The subgroups [B, M ]M W have trval ntersecton. Proof. Because the M have trval ntersecton, t s suffcent to see that the [B, M ] have trval ntersecton. Recall that each [B, M ] s assocated wth the two lnear functonals Ψ, Ψ : B F 3, gven by ( ) Ψ (β a ) ka := k a, a A a M ( ) Ψ (β a ) ka := k a. a A a A\M As we ve observed prevously, [B, M ] = ker Ψ ker Ψ. For each coset of a subgroup αs A, we have the lnear functonal (β a ) ka k a. a A a αs If S has ndex k, say that ths map s type k. Suppose that b = (β a ) ka [B, M ]. a A 0 2r We wll argue that, f b s n the kernel of every lnear functonal of type k + 1, then t s n the kernel of every lnear functonal of type k, for 0 k n 2. Ths, along wth the fact that k a = k a + k a = 0 a A a M a A\M for some (and every) (because b s n every [B, M ]) wll mply that k a = k a0 = 0, a {a 0 } for every a 0 A. But, ths s the fact that b = 1. Suppose so, that s, that b s n the kernel of every lnear functonal of type k + 1. Let αs be an arbtrary coset of a subgroup S of ndex k. A/S s elementary abelan of order 2 n k, whch s at least 4. Thus, A/S contans a subgroup of order 4 contanng αs, necessarly somorphc to the Klen four-group. Suppose that ths subgroup s gven by {αs = α 1 S, α 2 S, α 3, α 4 S}. 14

15 Defne Σ p F 3 to be a α k ps a, for 1 p 4. Then, because the sx quanttes k a = Σ p + Σ q a α ps α qs are all 0 by our nductve hypothess, for 1 p < q 4, (and because 2 s nvertble modulo 3), ths mples by a lttle lnear algebra that each Σ p = 0; n partcular, Σ 1 = k a = 0. But, ths s the concluson of the nducton. a αs Thus, k a0 = 0, so that b = 1, so that 0 2r [B, M ] = {1}. 4.1 Step 1 n the HSP chan Now, let s construct the ponted group (P, p), keepng n mnd also our goal: to see that every (n 1)-generator sub-ponted-group of (Q n, q n ) s n the varety generated by (P, p). The dea s ths: supposng that (R, q n ) s such a sub-ponted-group of (Q n, q n ), then note Q n = (C/D)BA, a semdrect product, and n Q n we know C/D to be a normal subgroup. Smlarly, B s a normal subgroup of BA, agan a semdrect product. Thus, we have the projectons (C/D)BA BA A, and we consder ther composton χ : Q n A. Because (R, q n ) (by assumpton) s an (n 1)- generator ponted group, (χ(r), χ(q n )) (A, χ(q n )) s also such. In partcular, though, χ(q n ) = χ(γ 1 β) = 1, so that χ(r) s really just an (n 1)-generator subgroup of A (.e. one whch s trvally ponted). Now, because A s not (n 1)-generated (whch we may see by a vector space dmenson argument, say), χ(r) s a proper subgroup of A, and s therefore contaned n some maxmal subgroup M A. In partcular, (R, q n ) s a sub-ponted-group of ((C/D)BM, q n ), a group of order 2 2n+2 +n 4 3 2n Thus, a choce of R gves us a partcular M, but as we ve observed prevously n Lemma 1, because all of the M are hghly nterchangeable, t doesn t matter whch M we choose, n the sense that the groups we ll construct are all somorphc. 5 Constructng the group (P, p) Now we ll construct (P, p) for real. 15

16 5.1 Parng the maxmal subgroups M of A Choose some maxmal subgroup M A. Now, gven any other maxmal subgroup S A, the ntersecton S M s always a subspace of co-dmenson two, so that A/(S M) s somorphc to the Klen four-group (.e. the unque F 2 -vector-space of dmenson 2). In the quotent, both M and S are sent to subgroups of order 2, and we can see that there s exactly one other such subgroup of the Klen four-group; call ts premage under the quotent map T. Thus, T S and T M = S M; set-theoretcally, we have T = S M (A \ (S M)). Thus, we have a parng S T of the maxmal subgroups of A not equal to M. So, let s suppose that the subgroups are enumerated as M 0,..., M 2r so that M = M 0, and M 2k 1 s pared wth M 2k, for 1 k r. We wll use the partcular fact that M 2k 1 M = M 2k M repeatedly. 5.2 Constructng the group E; step 2 n the HSP chan We defned earler the subgroup D of C generated by all products δ δ j ; let us now consder the subgroup E D generated only by consecutve products δ 2k 1 δ 2k, for 1 k r. That s, defne E := δ 2k 1 δ 2k : 1 k r, a group of order 2 r = 2 2n 1 1 and ndex 2 9 2n 1 4 n C. Thus, C/D s a quotent group of C/E, and n partcular ((C/D)BM, q n ) s a quotent pontedgroup of ((C/E)BM, (γ 1 E)β), the latter beng of order 2 9 2n 1 +n 5 3 2n. 5.3 Constructng the groups Ω k ; step 3 n the HSP chan Defne N 0 := [B, M 0 ]M 0 (whch we recall to be of order 2 n 1 3 2n 2 ), and for 1 k r, N k := [B, M 2k 1 ]M 2k 1 [B, M 2k ]M 2k. Then, because an ntersecton of normal subgroups s normal, all of the N k are normal subgroups of W. Further, because by smlar dmenson-countng arguments 4 to those n Secton 2.7, these 4 We can partton A nto the four cosets of M 2k 1 M 2k. Then, for each element X of ths partton, demonstrate 2 n 2 1 lnearly ndependent elements of B whose support s contaned n X, for a total of 2 n 4 such vectors. Further, there are four lnear functonals, of the form a A (βa ) ka a X ka the ntersecton of whose kernels contans [B, M 2k 1 ] [B, M 2k ]. 16

17 latter N k have order 2 n 2 3 2n 4. Proceedng smlarly, we defne the correspondng F 0 := C 0 (whch we recall to be order 2 5 ), and for 1 k r, F k = (C 2k 1 C 2k )/E k, where these E k are δ 2k 1 δ 2k, so that these F k are order 2 9. Because these δ 2k 1 δ 2k commute, we have E = E 1 E r, so that F 0 F 1 F r = C/E, or at least, these may be dentfed nnocuously. Yet another defnton: for 0 k r, let K k := (F 0 F 1 F k 1 F k+1 F r )N k (C/E)BM, so that K 0 s a group of order 2 9 2n 1 +n n 2 and ndex n (C/E)BM, and the other K k are order 2 9 2n 1 +n n 4 and ndex n (C/E)BM. The drect summand F 0 F 1 F k 1 F k+1 F r s a normal subgroup of the drect product C/E, as well as a normal subgroup of BM by the defnton of the acton of BA on C/E. Also, each N k s a normal subgroup of (C/E)BM by a smlar computaton to that n secton 2.8. So, each K k s a normal subgroup of (C/E)BM. Note how the F 0 F 1 F k 1 F k+1 F r have trval ntersecton; because the N k have trval ntersecton as well by Lemma 2, the K k themselves have trval ntersecton. For 0 k r, defne Ω k := ((C/E)BM)/K k as ponted groups. Let us take a moment to recall that, a pror, each Ω k depends on n as well; let us sgnfy ths by wrtng Ω k,n as needed. We have the map r : (C/E)BM k=0 Ω k (ce)bm (((ce)bm)k k ) r k=0 whose kernel s the ntersecton of the K k, whch we ve seen to be trval. somorphc to a sub-ponted-group of Ω 0 Ω 1 Ω r, va ths map. Thus, (C/E)BM s 17

18 5.4 Constructng the ponted group (P, p); the fnal step n the HSP chan Fnally, up to smorphsm, there s only one Ω 0,n, as well as only one Ω k,n across all k 1. Ths s because, n general, for a semdrect product G H and a normal subgroup G 1 H 1, we have where the acton on the rght s gven by G H G 1 H 1 G G 1 H H 1, (gg 1 ) hh 1 = g h G 1, f and only f [G, H 1 ] G 1. In our case, ths holds, as, gven any (q 1, q 2 ) C 2k 1, any a A, and any m, n M 2k 1, we have that (q 1, q 2 ) (βa ) 1 (β a ) mn = (q 1, q 2 ), because ether both a and am are n M 2k 1, or nether are; n ether case, (β a ) 1 (β a ) m and n both act trvally on C 2k 1. Thus, [(q 1, q 2 ), [β a, m]] = (q 1, q 2 ) 1 (q 1, q 2 ) (βa ) 1 (β a ) mn = (1, 1). The correspondng fact holds for C 2k. Thus, [C 2k 1 C 2k, [B, M 2k 1 ]M 2k 1 [B, M 2k ]M 2k ] = 1, and so [(C/E), N k ] F 0 F 1 F k 1 F k+1 F r. Thus, Ω 0 F 0 BM BM [B,M 0 ] and Ω k F k [B,M 2k 1 ] [B,M 2k ] for k 1. But, yet agan by elementary abelan-ness and dmenson-countng, these groups do not depend on n. So, because Ω k,n Ω k,n for non-zero k, k (and separately so for zero k), and because these Ω are ndependent of the choce of M, t s suffcent to take P := Ω 0 Ω 1, for an arbtrary, fxed n and M. So, P s a group of order Thus, r k=0 Ω k s a quotent of P r, a power of P. Thus, n summary, we have r R (C/D)BM (C/E)BM Ω k P r Var(P ), so that (R, q n ), the arbtrary (n 1)-generator sub-ponted-group of (Q n, q n ), les n the varety generated by (P, p). k=0 18

19 6 The laws w m of the ponted group (P, p) Now, we have just one task: show that, for every m N, the word w m s a law of (P, p). By the constructon of P, t s suffcent to show that w m s a law of each Ω k, and therefore to show that w m s a law of ((C/E)BM, (γ 1 E)β). Suppose we have arbtrary κ 1,..., κ m CBM, and consder now the elements g := v m (κ 1,..., κ m, γ 1 β) = (γ 1 β) κ1 (γ 1 β) κm, h := w m (κ 1,..., κ m, γ 1 β) = [(γ 1 β) 3, ((γ 1 β) 3 ) g ]. For reasons smlar to those mentoned n Secton 3, t s suffcent to observe that h E. that Because each κ j s an element of CBM = C (B M), we may let λ j CB and µ j M such κ j = λ j µ j for all 1 j m. Then, by a smlar calculaton as n Secton 3, we see that h = [γ, γ b ], where b = β µ1 β µm. In partcular, both γ and γ b are n C, so that h s too. Thus, because C = 2r =0 C s a drect product and by the defnton of the acton of b W on C, we may wrte and each [γ, γ φ (b) [γ, γ b ] = [γ 0, γ φ 0(b) 0 ] [γ 2r, γ φ 2r(b) 2r ], ] s n C. In fact, each [γ, γ φ (b) ] s ether δ or 1, by our observaton that the commutator subgroup of each C s δ. Now, φ (b) = φ (β µ1 β µm ) = β l β m l, where l s the number of ndces j for whch µ j M. For = 0, we have µ j M = M 0 for every 1 j m, so φ 0 (b) = β m 0, and so [γ 0, γ φ 0(b) 0 ] = [γ 0, γ βm 0 0 ] = [γ 0, γ 0 ] = 1. Otherwse, we recall the parng of M 2k 1 wth M 2k, for 1 k r. Because M 2k 1 M = M 2k M 19

20 and each µ j M (as we just observed), µ j M 2k 1 µ j M 2k, for any 1 k r. But ths means that, fxng some such k, for the same l. φ 2k 1 (b) = β l 2k 1 βm l 2k 1 φ 2k (b) = β l 2k βm l 2k Thus, [γ 2k 1, γ φ 2k 1(b) 2k 1 ] = δ 2k 1 f and only f [γ 2k, γ φ 2k(b) 2k ] = δ 2k. That s, h s a product of elements of the form δ 2k 1 δ 2k, for varous 1 k r, whch s exactly the fact that h E. 7 Appendx Below s a table detalng most of the groups whch appear n the constructon n the order of ther defnton, along wth ther orders. The bounds for ndces and k are omtted; ther values are the same as those n the man body of the paper. Group Order Group Order A 2 n (C/D)BM 2 2n+2 +n 4 3 2n M 2 n 1 E 2 2n 1 1 B 3 2n C/E 2 9 2n 1 4 W 2 n 3 2n (C/E)BM 2 9 2n 1 +n 5 3 2n [B, M ] 3 2n 2 N 0 2 n 1 3 2n 2 [B, M ]M 2 n 1 3 2n 2 N k 2 n 2 3 2n 4 W F C 2 5 F k 2 9 C 2 5 2n 5 K n 1 +n n 2 D 2 2n 1 K k 2 9 2n 1 +n n 4 D 2 2n 2 Ω C/D 2 2n+2 3 Ω k Q n 2 2n+2 +n 3 3 2n P

21 References [1] Roger M Bryant. The laws of fnte ponted groups. In: Bulletn of the London Mathematcal Socety 14.2 (1982), pp [2] Danel Gorensten. Fnte groups. Vol Amercan Mathematcal Soc., [3] Shela Oates and MB Powell. Identcal relatons n fnte groups. In: Journal of Algebra 1.1 (1964), pp [4] Hanamantagouda P Sankappanavar and Stanley Burrs. A course n unversal algebra. In: Graduate Texts Math 78 (1981). 21