BLAISE PASCAL
Khaled Jaber & Frank O. Wagner
Largeness and equational probability in groups Volume 27, no1 (2020), p. 1-17.
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Publication éditée par le laboratoire de mathématiques Blaise Pascal de l’université Clermont Auvergne, UMR 6620 du CNRS
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Largeness and equational probability in groups
Khaled Jaber Frank O. Wagner
Abstract
We definek-genericity andk-largeness for a subset of a group, and determine the value ofkfor which ak-large subset ofGnis already the whole ofGn, for various equationally defined subsets. We link this with the inner measure of the set of solutions of an equation in a group, leading to new results and/or proofs in equational probabilistic group theory.
1. Introduction
In probabilistic group theory we are interested in what proportion of (tuples of) elements of a group have a particular property; if this property is given by an equation, we talk aboutequational probability. In [9] a notion oflargenesswas introduced for a subset of a group, and it was shown that certain equational properties of a group hold everywhere as soon as they hold largely. In this paper, we shall introduce a quantitative version of largeness, and deduce some results in equational probabilistic group theory.
Throughout this paper,Gwill be a group andµa left-invariant probability measure on some algebra of subsets ofG.
Example 1.1.
(1) Gfinite,µthe counting measure.
(2) G1 a group,µ1a left-invariant measure onG1, andG=Gn1 with the product measureµ=µn1.
(3) More generally,G1a group,G≤Gn1 andµa left-invariant measure onG.
(4) Garbitrary and the measure algebra reduced to{∅,G}. While this set-up trivialises the probability statements, the largeness results remain meaningful.
IfXis a measurable subset ofGwe can interpretµ(X)as the probability that a random element ofGlies inX. IfHis another group, f :G→His a function andc∈Hsome constant, we putµ(f(x)=c)=µ({g∈G: f(g)=c}).
The second author was partially supported by the ANR-DFG project AAPG2019 GeoMod.
Keywords:probabilistic group theory, largeness.
2020Mathematics Subject Classification:20A15, 03C60, 20P99.
Example 1.2. LetG1be a group,G≤G1na subgroup, ¯g ∈G1mconstants, andw(x,¯ y)¯ a word in ¯xy¯and their inverses, with|x|¯ =nand|y|¯ =m. Thenw(x,¯ g)¯ induces a function fromGtoG1.
We shall now list some known results, starting with Frobenius in 1895.
Fact 1.3. LetGbe a finite group.
• Frobenius 1895 [5]Ifndivides|G|then the number of solutions ofxn=1is a multiple ofn. In particular,µ(xn=1) ≥ |Gn|.
• Miller 1907 [14]IfGis non-abelian, thenµ(x2=1) ≤ 34.
• Laffey 1976 [11]IfGis a3-group not of exponent3thenµ(x3 =1) ≤ 79.
• Laffey 1976 [12]If p is prime and divides|G|, butGis not ap-group, then µ(xp =1) ≤ p+1p .
• Laffey 1979 [13]IfGis not a2-group, thenµ(x4=1) ≤ 89.
• Iiyori, Yamaki 1991 [8]Ifndivides|G|andX ={g ∈G:gn=1}has cardinality n, thenXforms a subgroup ofG.
• Erdős, Turan, 1968 [3]Ifk(G)is the number of conjugacy classes inG, then µ([x,y]=1)=k(G)|G|.
• Joseph 1977 [10], Gustafson 1973 [6]IfGis non-abelian, thenµ([x,y]=1) ≤ 58.
• Neumann, 1989 [16]For any realr >0there aren1(r)andn2(r)such that if
µ([x,y]=1) ≥ rthenGcontains normal subgroupsH≤Ksuch thatK/His
abelian,|G:K| ≤n1(r)and|H| ≤n2(r).
• Barry, MacHale, Ní Shé, 2006 [1]Ifµ([x,y]=1)>13 thenGis supersoluble.
• Heffernan, MacHale, Ní Shé, 2014 [7]Ifµ([x,y]=1)> 247 thenGis metabelian.
Ifµ([x,y]=1)>67583 thenGis abelian-by-nilpotent.
In Section 2 we shall introduce largeness and prove the main connection between largeness and measure, Lemma 2.5, which will be used throughout the rest of the paper.
Section 3 deals with central elements, or more generally FC and BFC groups. We shall treat equations of the formxn=cfor arbitrarycin Section 4, recovering Miller’s result
for n = 2, and a weaker bound than Laffey for n = 3 (namely 67). In Section 5 we shall consider commutator equations; while our methods allow us to deal with more complicated commutators, they are too general to obtain the bounds from Fact 1.3.
Section 6 deals with nilpotent groups via linearisation, and the short Section 7 places Sherman’s autocommutativity degree in our context.
Notation. We shall writexy=y−1xy,x−y=(x−1)y=y−1x−1yand[x,y]=x−1y−1xy= y−xy=x−1xy.
2. Largeness and Probability
The following notion of largeness was introduced in [9].
Definition 2.1. IfX ⊆G, we say thatXisk-largeinGif the intersection of anykleft translates ofXis non-empty, andX isk-genericinGifkleft translates ofXcoverG. A subsetXislargeif it isk-large for allk; it isgenericif it isk-generic for somek.
Of course, analogous notions exist for right and two-sided genericity/largeness. Both genericity and largeness are notions of prominence, increasing withkfor largeness and decreasing withkfor genericity. Clearly, ifX ⊆GandXis (k-)large/generic, so is any left or right translate or superset ofX. Largeness and genericity are co-complementary:
Lemma 2.2. LetX ⊆G. ThenXis1-large if and only ifX ,∅, andXis1-generic if and only if X =G. More generally,X isk-large if and only ifG\X is notk-generic.
Finally,Xisk-generic/large if and only ifX∩Y ,∅for allk-large/genericY ⊆G.
Proof. We only show the last assertion. IfXis notk-generic/large, thenY :=G\Xis k-large/generic, andX∩Y =∅. Conversely, ifXisk-generic, sayG=Ð
i<kgiX, andY isk-large, then
∅, Ù
i<k
giY =G∩Ù
i<k
giY =Ø
i<k
giX∩Ù
i<k
giY
=Ø
i<k
giX∩Ù
i<k
giY
!
⊆Ø
i<k
(giX∩giY)=Ø
i<k
gi(X∩Y).
ThusX∩Y ,∅.
Remark 2.3. If φ: G → His an epimorphism and X ⊆ Gis (k-)large/generic, so is φ(X) ⊆H. Conversely, ifY ⊆His (k-)large/generic inH, so isφ−1[X]inG.
In particular, ifX ⊆G×His (k-)large/generic, so are the projections to each coordinate.
Conversely, ifX ⊆GandY ⊆Hare (k-)large, so isX×Y ⊆G×H; ifXisk-generic andY is`-generic,X×Y isk`-generic.
Lemma 2.4. SupposeX is k`-large inGandH ≤Gis a subgroup of indexk. Then X∩His`-large inH.
Proof. Let(gi :i<k)be coset representatives ofHinG, and consider(hj : j < `)in H. Byk`-largeness ofXinGthere isx∈Ñ
i<k,j<`gihjX. AsÐ
i<kgiH=G, there is i0 <kwithx∈gi0H. But then
g−i1
0 x∈H∩ Ù
i<k, j<`
gi−1
0 gihjX ⊆H∩Ù
j<`
hjX=Ù
j<`
hj(X∩H),
soX∩His`-large.
The link between largeness and probability is given by the following lemma, which will be used throughout the paper. Recall that theinner measureof an arbitrary subsetX of a measurable groupGis
µ∗(X)=sup{µ(Y):Y ⊆Xmeasurable}, and theouter measureis given by
µ∗(X)=inf{µ(Y):Y ⊇Xmeasurable}.
Clearly the inner measure is superadditive, the outer measure is subadditive, and µ∗(X)+ µ∗(G\X)=1.
Lemma 2.5. IfXisk-generic inG, thenµ∗(X) ≥ 1k. If µ∗(X)>1−k1 thenXisk-large inG.
Proof. IfXisk-generic there areg1, . . . ,gk inGwithG=Ð
i≤kgiX. Hence 1=µ∗(G)=µ∗ Ø
i≤k
giX
!
≤Õ
i≤k
µ∗(giX)=kµ∗(X) by left invariance, whenceµ∗(X) ≥ 1
k.
Now ifX is not k-large, its complement is k-generic, soµ∗(G\X) ≥ k1. But then
µ∗(X) ≤1−1k.
These bounds are strict, as we can takeXa subgroup of indexk(resp. its complement).
Remark 2.6. For any groupGthe set(G× {1}) ∪ ({1} ×G)is 2-large inG2; ifGis infinite it is of measure 0.
We shall now prove some results about finite groups, which owing to their non-linearity do not generalise easily to the measurable context.
Remark 2.7. LetGbe a finite group of ordern, andX⊆Ga non-empty proper subset of sizem. ThenXis(n−m+1)-generic and at mostm-large, since we can form the union of Xwithn−mtranslates ofX to cover all then−mpoints ofG\X, and we can intersect Xwithmtranslates ofXto remove allmpoints ofX.
Theorem 2.8. LetGbe a finite group of ordern, andX ⊆Ga non-empty proper subset of sizem. Ifm>n−12 −
q
n−34, thenXis2-generic. Hence ifm< 12+q
n−34 thenX is not2-large.
Proof. Ifm>n−12 − q
n−34, then n−3
4 >(n−m−1
2)2 =(n−m)(n−m−1)+1 4. PutZ ={xy−1 :x,y∈G\X}. Then
|Z| ≤ (n−m)(n−m−1)+1<n,
so there isg ∈G\Z. But ifh∈G\(X∪gX), thenh,g−1h∈G\X, andg=h(g−1h)−1 ∈Z, a contradiction. ThusG=X∪gXandXis 2-generic.
The second assertion follows by taking complements.
Theorem 2.9. LetGbe a finite group of ordern. If the exponent ofGdoes not divide` thenµ(x`=1) ≤1−√1
2n, whereµis the counting measure.
Proof. Put X ={g ∈ G: g` =1}, of sizem <n, and take anyg ∈ G\X. Note that X∩gX∩CG(g)is empty, as otherwise there would bey∈CG(g)withy`=1=(gy)`, whenceg` =1 andg∈ X.
Thus|CG(g)| ≤2|G\X|. MoreovergG∩X=∅, and
|G|/|CG(g)|=|gG| ≤ |G\X|.
Thusn=|G| ≤2|G\X|2andpn
2 ≤n−m, whence µ(x`=1)= m
n ≤ n−pn
2
n =1− 1
√
2n.
Definition 2.10. Let f :G → Hbe a function, andc ∈ H. The equation f(x)=cis k-largely satisfiedinGif{g ∈ G : f(g) =c}is k-large inG. By abuse of notation, if G =Gn1 and x =(x1, . . . ,xn), we shall also say that f(x1, . . . ,xn) =cis k-largely satisfied inG1.
3. FC-Groups
In this section we shall work in the set-up of Example 1.2:G1will be a group,G≤G1n, w(x,¯ y)¯ a word in ¯xy¯and their inverses withn=|x|¯ andm=|y|, ¯¯ g ∈Gm1 andc ∈G1
constants, and f(x)¯ =w(x,¯ g).¯
Recall that a group isFCif the centraliser of any element has finite index; it isBFCif the index is bounded independently of the element.
We shall first need a preparatory lemma. For two tuples ¯g = (gi : i < k) and
¯
g0=(gi0:i<k)inG1kwe shall put ¯g−1=(gi−1:i<k)and ¯g·g¯0=(gigi0:i<k).
Lemma 3.1. Supposeg,¯ g¯0 ∈ G1mandh,¯ h¯0 ∈ Gn1 are such that all elements fromg¯h¯ commute with all elements fromg¯0h¯0. Then
w(h¯·h¯0,g¯·g¯0)=w(h,¯ g)¯ w(h¯0,g¯0).
Proof. Obvious.
Theorem 3.2. LetG1be anFC-group. If the equationw(x,¯ g)¯ =cis largely satisfied in Gthen it is identically satisfied inG.
Proof. Consider ¯h ∈ G, and C = CG1(g,¯ h), a subgroup of finite index in¯ G1. Put H=Cn∩G, a subgroup of finite index inG, andX ={h¯0∈ G:w(h¯0,g)¯ =c}. Then X∩h¯−1X∩His large inH, whence non-empty. So there is ¯x∈Hwith
w(¯1,g)¯ w(x,¯ ¯1)=w(x,¯ g)¯ =c=w(h¯·x,¯ g)¯ =w(h,¯ g)¯ w(x,¯ ¯1).
Hencew(h,¯ g)¯ =w(¯1,g)¯ for all ¯h∈G, andw(¯1,g)¯ =w(x,¯ g)¯ =c.
For aBFC-group, we can bound the degree of largeness needed:
Theorem 3.3. Suppose every centraliser of a single element has index at mostkinG1. If the equationw(x,¯ g)¯ =cis2kn2+mn-largely satisfied inGthen it is identically satisfied inG.
Proof. In the notation of the previous proof,C=CG1(g,¯ h)¯ has index at mostkn+min G1, so
|G:H|=|G:G∩Cn| ≤ |G1n:Cn|=|G1 :C|n≤ (kn+m)n =kn2+mn. Now 2kn2+mn-largeness ofXinGimplieskn2+mn-largeness ofX∩h¯−1XinG, whence 1-largeness ofX∩h¯−1X∩HinH. So we can find the ¯xrequired to finish the proof.
Corollary 3.4. Suppose every centraliser of a single element has index at mostkinG1. Ifw(x,¯ g)¯ =cis not an identity onG, then
µ∗(w(x,¯ g)¯ =c) ≤1− 1 2kn2+mn.
Proof. Ifµ∗(w(x,¯ g)¯ =c)>1− 1
2kn2+mn, then{x¯∈G:w(x,¯ g)¯ =c}is 2kn2+mn-large in Gby Lemma 2.5, andw(x,¯ g)¯ =cis identically satisfied inGby Theorem 3.3.
Remark 3.5. This holds in particular for the equationx`=c, withn=1 andm=0.
If the group is central-by-finite, the largeness needed does not depend on the number of parameters.
Corollary 3.6. Suppose Z(G1)has indexkinG1. If the equationw(x,¯ g)¯ =cis2kn- largely satisfied inGthen it is identically satisfied inG.
Proof. H=G∩Z(G1)nhas index at mostkninG. We finish as in Theorem 3.3.
Corollary 3.7. If |G1 : Z(G1)| ≤ k and w(x,¯ g)¯ = c is not an identity in G, then µ∗(w(x,¯ g)¯ =1) ≤1− 1
2kn.
Of course, for an abelian groupG1we havek=1 in the above results.
Remark 3.8. Ifw(x,¯ g)¯ =cis 2-largely satisfied inGn, then it is identically satisfied in the abelian quotientG/G0. If moreoverGis aBFC-group, thenG0is finite by B.H.
Neumann’s Lemma [15], andGnsatisfies a finite disjunctionÔ
c0∈G0w(x,¯ g)¯ =cc0. We can also deduce results for central elements just from 2-largeness (although for infinite index|G1:Z(G1)|there is no reason that ifXis large inGthe intersectionX∩ Z(G1)nis still large inG∩Z(G1)n).
Theorem 3.9. Ifw(x,¯ g)¯ =cis2-largely satisfied inG, thenw(x,¯ ¯1)=1identically on G∩Z(G1)n.
Proof. Consider ¯h∈G∩Z(G1)n. PutX={h¯0∈G:w(h¯0,g)¯ =1}. ThenX∩h¯−1Xis non-empty, so there is ¯x∈Gwith
w(x,¯ g)¯ =c=w(h¯·x,¯ g)¯ =w(h,¯ ¯1)w(x,¯ g).¯
Hencew(h,¯ ¯1)=1.
Corollary 3.10. Ifx1k1. . .xnkn =cis2-largely satisfied inGnandk =gcd(k1, . . . ,kn), thenxk =1identically onZ(G).
Proof. We havex1k1. . .xnkn =1 onZ(G). Puttingxi =g∈ Z(G)andxj=1 forj ,iwe
havegki =1 for all 1≤i≤n. The result follows.
Corollary 3.11. If the exponent ofZ(G)does not dividegcd(k1, . . . ,kn), then µ∗(xk1
1 . . .xnkn =c) ≤ 1 2.
4. Burnside and Engel Equations
In Remark 3.5 we have already seen that if every centraliser of a single element has index at mostkinG, thenµ∗(xm=c) ≤1−2k1 unless the exponent ofGdividesm. In this case necessarilyc=xm=1.
We shall first prove Miller’s Theorem mentioned in the introduction.
Theorem 4.1. Letc ∈ G. If x2 = cis4-largely satisfied inG, thenGis abelian of exponent2, andc=1.
Proof. Fixg,h ∈ G. Then there isxwithc= x2 =(gx)2 =(hx)2 =(ghx)2. But this impliesx−1gx=g−1,x−1hx=h−1andx−1ghx=(gh)−1. On the other hand,
x−1ghx=x−1gx x−1hx=g−1h−1=(hg)−1.
Hence gh = hg andG is abelian. But nowc = x2 = (gx)2 = g2x2 = g2c, whence
g2=1.
IfGsatisfies 4-largelyxax=bfor somea,b∈G, then it satisfies 4-largely(ax)2=ab, whencex2=ab. HenceGis abelian of exponent 2, anda=b.
Corollary 4.2. IfGis not of exponent2ora,b, thenµ∗(xax=b) ≤ 34.
Recall that thenthEngel conditionis the condition[x,ny]=1, where[x,1y]=[x,y]
and[x,n+1y]=[[x,ny],y]. Note that
[x,y,y]=[y−xy,y]=y−1yxy−1y−xy y=[y−x,y]y.
Thus the 2-Engel condition[x,y,y]=1 is equivalent to[y−x,y]=1, that is all conjugacy classes being commutative.
Theorem 4.3. IfGsatisfies7-largelyx3=1thenGis2-Engel.
Proof. PutX={g∈G:g3 =1}. Forg,h∈Gconsider
x∈X∩g−1X∩h−1X∩gX∩ (gh)−1X∩gh−1X∩gh−1g−1X.
Then (yx)3 = 1 for y ∈ {1,g,h,g−1,gh,hg−1,ghg−1}, which means that xyx = y−1x−1y−1. We calculate the productxhx2gxin two ways:
xhx2gx=(xhx)(xgx)=h−1(x−1h−1g−1x−1)g−1
=h−1ghxghg−1 and
xhx2gx=xh(g−1x)−1x=xh(g−1x)2x=(xhg−1x)g−1x2
=gh−1(x−1gh−1g−1x−1)=gh−1ghg−1xghg−1.
Thush−1gh=gh−1ghg−1andghg=ggh. Ash ∈Gwas arbitrary, the conjugacy class ofgis commutative; asgwas arbitrary, all conjugacy classes are commutative.
Theorem 4.4. LetGbe2-Engel. IfGsatisfies2-largelyx3=1thenGhas exponent3.
Proof. For anyg∈Gthere isx∈Gwithx3=(gx)3 =1. AsxGis commutative, gxg−1gx−1 =x−1gxg−1xgx−1 =gx−gx xgx−1=gx−gxgx x−1=g.
SincegGis commutative, we have
g3 =g2gxg−1gx−1=g2g−1gx−1gx=(gx)3=1.
Corollary 4.5. IfGsatisfies7-largely x3 =1, thenGhas exponent3. IfGis not of exponent3thenµ∗(x3=1) ≤ 67. If moreoverGis2-Engel, thenµ∗(x3 =1) ≤ 12.
Note that the bound 67 is not as good as Laffey’s bound 79 cited in the introduction.
Problem 4.6. A group which satisfies 5-largelyx3 =1, is it 2-Engel? This would improve our bound to 45.
Corollary 4.7. If|G:Z(G)| ≤7andGsatisfies7-largelyx3 =cfor somec∈G, then c=1andGhas exponent3.
Proof. {x∈G:x3=c} ∩Z(G)is 1-large, whence non-empty, and contains an element z. But now there isx∈Gwithx3 =1=(zx)3 =z3x3 =cx3, whencec=1. We finish by
Corollary 4.5.
If|G:Z(G)|is prime, thenGis abelian, and 2-largeness is sufficient by Corollary 3.10.
5. Commutator Equations
Consider the equation[x,g]=cfor somec,g∈G. Since{x∈G:[x,g]=c}is a coset ofCG(g)or empty, and a coset of a proper subgroup cannot be 2-large, it follows that if Gsatisfies 2-largely[x,g]= ctheng ∈ Z(G)andc =1. The following argument generalises this result.
Theorem 5.1. Supposef :G→Hsatisfies f(x x0)= f(x)h f(x0)for someh∈Hwhich depends on x,x0 ∈ G. IfG0andG1 are groups, f0 :G0 → Hand f1 :G1 → H are functions such thatG0×G×G1satisfiesk-largely f0(x0)f(x)f1(x1)=cfor somek≥2, then f(G)=1andG0×G1satisfiesk-largely f0(x0)f1(x1)=c.
Proof. Fixg∈G. By 2-largeness there is(x0,x,x1) ∈G0×G×G1such that f0(x0)f(x)f(x1)=c= f0(x0)f(gx)f(x1).
Thus f(x) = f(gx) = f(g)h f(x) and f(g) = 1. It follows that f0(x0)f(x)f1(x1) = f0(x0)f1(x1)onG0×G×G1. The result follows.
Corollary 5.2. IfGsatisfies2-largelyÎ
i<n[xi,gi]=cfor somegi ∈G, thengi ∈Z(G) for alli <nandc=1. If not allgi are central orc,1thenµ∗(Î
i<n[xi,gi]=c) ≤ 12. Proof. We have[x x0,y]=[x,y]x0[x0,y]. Now use Theorem 5.1.
Remark 5.3. Theorem 5.1 also holds if f(x x0)= f(x0)f(x)h, with almost the same proof.
Hence Corollary 5.2 also holds if some factors are of the form[gi,xi].
Gustafson [6] has shown thatµ2([x,y]=1) ≤ 12(1+µ(Z(G)) ≤ 58 for a non-abelian compact topological groupG, whereµis the Haar measure onGand µ2 the product measure on G2. Pournaki and Sobhani [17] have generalised this to calculate that µ([x,y]=g)< 12 for anyg,1 in a finite group, using Rusin’s classification [18] of all finite groups with µ([x,y]=1)> 1132(see also [4]). We have only been able to establish results using 4-largeness, giving the bound of 34 in Corollary 5.7, so the following two problems remain open:
Problem 5.4.
(1) IfGsatisfies 2-largely[x,y]= 1, isG0 =C2 andG/Z(G)of exponent 2, or G0=C3andG/Z(G)=S3?
(2) IfGsatisfies 2-largely[x,y]=cfor somec∈G, isc=1?
Theorem 5.5. Ifw(x,¯ g)[x,¯ y]=cis satisfied4-largely inGn+1, wherex∈x¯andy<x,¯ thenGis abelian andw(x,¯ g)¯ =c.
Proof. For anyh∈Gthe set
{(x,¯ x,y):w(x,¯ g)[x,¯ y]=c=w(x,¯ g)[x,¯ hy]}
is 2-large in Gn+1. Hence {(x,y) ∈ G2 : [x,y] = [x,hy]} is 2-large in G2. Now [x,hy] = [x,y][x,h]y, so [x,h] = 1 is satisfied 2-largely inG, whence h ∈ Z(G). It follows thatGis abelian. But thenw(x,¯ g)¯ =cis satisfied 4-largely inGn, and must be an
identity inGby commutativity and Corollary 3.6.
Corollary 5.6. IfGis a group withµ∗(w(x,¯ g)[x,¯ y]=c)> 34, thenGis abelian satisfying w(x,¯ g)¯ =c.
Corollary 5.7. IfGsatisfies4-largely[x,y]=c, thenGis abelian andc=1. IfGis not abelian orc,1, thenµ∗([x,y]=c) ≤ 3
4.
Remark 5.8. The same holds for the equationxcy=yc0xwithc,c0: puttingx0=xc andy0=yc0, this is equivalent to[x0,y0]=c−1c0.
Theorem 5.9. Let g,h ∈ Gand k = min{|G : CG(g)|,|G : CG(h)|}. IfG satisfies k-largely[g,hx]=1, thengGandhGcommute.
Proof. If k = |G : CG(h)|, then {x ∈ G : [g,hx] =1} ∩CG(h) is 1-large, whence non-empty, and[g,h] =1. Now note that for anya ∈ Galso |G : CG(ha)| = kand [g,hax]=1 is satisfiedk-largely, whence[g,ha]=1 and[g,hG]=1.
Ifk=|G:CG(g)|, then{x∈G:[gx−1,h]=1} ∩CG(g)is 1-large (still on the left) and non-empty, whence[g,h]=1 and we finish as above.
Corollary 5.10. If[gG,hG]is non-trivial for someg,h ∈G, thenµ∗([g,hx]=1) ≤1−1
k, wherek=min{|G:CG(g)|,|G:CG(h)|}.
Theorem 5.11. If g,h,c ∈Gand[x,g,h]= cis2k-largely satisfied, wherek =|G: CG(h)|, then[G,g,h] = 1. Similarly, if[g,x,h] = c is2k-largely satisfied for some c∈Z(G), then[g,G,h]=1.
Proof. Choosea∈G. Then the setX ={x ∈G:[x,g,h]=c=[ax,g,h]}isk-large, and forx∈Xwe have
[x,g,h]=c=[ax,g,h]=[[a,g]x[x,g],h]=[[a,g]x,h][x,g][x,g,h], whence[[a,g]x,h]=1. By Theorem 5.9 we have[a,g,h]=1.
If[g,x,h]=cis 2k-largely satisfied withc∈Z(G), then fora∈Gwe obtain ak-large X ⊆Gsuch that forx∈ Xwe have
[g,x,h]=c=[g,ax,h]=[[g,x][g,a]x,h]=[g,x,h][g,a]x[[g,a]x,h],
whence[[g,a]x,h]=1, and[g,a,h]=1 by Theorem 5.9.
Corollary 5.12. Ifg,h∈Gandk=|G:CG(h)|, then[G,g,h],1impliesµ∗([x,g,h]=c) ≤ 1−2k1 for anyc∈G, and[g,G,h],cimpliesµ∗([g,x,h]=c) ≤1−2k1 for anyc∈Z(G).
We shall now generalise Corollary 5.7 to higher nilpotency classes. However, the proof requires an additional assumption.
Theorem 5.13. Supposes< ωis such that for alli <kthere is a setAiof size at mosts such thatZ(G/Zi(G))=CG/Zi(G)(Ai). IfGsatisfies2(s+1)k-largely[x0,x1, . . . ,xk]=c, thenc=1andGis nilpotent of class at mostk.
Proof. We use induction onk. Fork=1 note thats≥1 (otherwiseGis abelian and we are done), so the result follows from Corollary 5.7.
Now suppose the assertion is true fork, and
X={x¯∈Gk+2:[x0,x1, . . . ,xk+1]=c}
is 2(s+1)k+1-large inGk+2. If A0 = {ai : i < s} consider the projectionY of X ∩ Ñi<s(1, . . . ,1,a−1i )Xto the firstk+1 coordinates, and note that it is 2(s+1)k-large. Then for all(x0, . . . ,xk) ∈Y there isy∈Gsuch that
[x0, . . . ,xk,y]=c=[x0, . . . ,xk,aiy]=[x0, . . . ,xk,y] [x0, . . . ,xk,ai]y
for alli<s, whence[x0, . . . ,xk] ∈ Z(G). By inductive assumptionG/Z(G)is nilpotent
of class at mostk, and we are done.
Corollary 5.14. Letsbe as above. IfGis not nilpotent of class at mostkorc,1, then µ∗([x0,x1, . . . ,xk]=c) ≤1−12(s+1)−k.
Remark 5.15. Recall that anMc-group is a groupGsuch that for every subsetAthere is a finite subset A0 ⊆ A such that CG(A) = CG(A0). Equivalently, G satisfies the ascending (or the descending) chain condition on centralisers. Roger Bryant [2] has shown that in anMc-group, for every iterated centreZi(G)there is a finite setAi such thatZ(G/Zi(G))=CG/Zi(G)(Ai). So in anMc-group we can find somesas needed for Theorem 5.13 and Corollary 5.14.
Problem 5.16. To what extent do we need theMc-condition (or similar) in Theorem 5.13 and Corollary 5.13? It is not needed for nilpotency class 1 (Corollary 5.7). In general, assuming just 2k+1-largeness of[x0, . . . ,xk]=c, we obtain that{x¯∈Gk :[x0, . . . ,xk−1] ∈ CG(g)} is 2k-large in Gk for any g ∈ G. Does this imply γk(G) ≤ CG(g), or even γk(G) ≤Z(G)?
6. Nilpotent groups
We shall first introduce the notion of a supercommutator from [9].
Definition 6.1. Any variable and any constant fromGis asupercommutator; ifvandw are supercommutators, thenv−1and[v,w]are supercommutators.
Alternatively, we could have said that x,x−1 andgare supercommutators for any variablexand anyg∈G, and that ifvandware supercommutators, so is[v,w].
Definition 6.2. The set Var(v) of variables of a supercommutator v is defined by Var(x) = {x}, Var(g) = ∅, Var(v−1) =Var(v), and Var([v,w] = Var(v) ∪Var(w). We put var(v) = |Var(v)|, thevariable number ofv. If ¯x is a tuple of variables, we put Varx¯=Var(v) ∩x, Var¯ 0x¯(v)=Var(v) \x, var¯ x¯(v)=|Varx¯(v)|and var0x¯(v)=|Var0x¯(v)|.
Clearly var([v,v0]) ≥max{var(v),var(v0)}, and similarly for varx¯and var0x¯. Lemma 6.3. LetHEGandv(x,¯ z)¯ a supercommutator.
(1) vdefines a function fromH|x¯¯z|toγvar(v)(H).
(2) Ifvarx¯(v)>0andx,¯ y¯andz¯are pairwise disjoint, then v(y¯·x,¯ z)¯ =v(x,¯ z)¯ v(y,¯ z)¯ Φ(x,¯ y,¯ z),¯
whereΦis a product of supercommutators whose factorswsatisfy
(†) Varz¯(w)=Varz¯(v), and ifxi ∈ Varx¯(v)then xi ∈ Var(w)oryi ∈ Var(w), and both possibilities occur for at least onei.
(3) Ifv(x,¯ z)¯ is a product of supercommutators whose factorswsatisfyvarx¯(w)>0 andvar0x¯(w) ≥n, then
v(y¯·x,¯ z)¯ =v(x,¯ z)¯ v(y,¯ z)¯ Φ(x,¯ y,¯ z),¯
whereΦis a product of supercommutators whose factorswsatisfyvarx¯(w)>0 andvar0x¯(w)>n.
Proof. (1) is proved as in [9, Lemme 6 (1)] by induction, using thatγn(H)is characteristic in H, whence normal in G, and [γn(H), γm(H)] ≤ γn+m(H). We shall show (2) by induction on the construction ofv.
Ifv = x ∈ x¯ we havev(yx)= yx= xy[y,x] =v(x)v(y)[y,x]; ifv = x−1 we have v(yx)=x−1y−1=v(x)v(y). This leaves the casev=[v1,v2]for two supercommutators v1andv2. We shall assume varx¯(v1)>0 and varx¯(v2)>0 (the case varx¯(v1)varx¯(v2)=0 is analogous, but simpler).
By inductive hypothesis, there are Φi fori = 1,2, products of supercommutators satisfying (†) relative tovi, such that
vi(y¯·x,¯ z)¯ =vi(x,¯ z)¯ vi(y,¯ z)¯ Φi. Then
v(y¯·x,¯ z)¯ =[v1(y¯·x,¯ z),¯ v2(y¯·x,¯ z)]¯
=[v1(x,¯ z)¯ v1(y,¯ z)¯ Φ1,v2(x,¯ z)¯ v2(y,¯ z)¯ Φ2]
=[v1(x,¯ z),¯ v2(x,¯ z)] [v¯ 1(y,¯ z),¯ v2(y,¯ z)]¯ Φ=v(x,¯ z)¯ v(y,¯ z)¯ Φ, whereΦis a product of supercommutators[w,w0]
(i) wherew∈Φ1∪ {v1(x,¯ z),¯ v1(y,¯ z)}¯ andw0∈Φ2∪ {v2(x,¯ z),¯ v2(y,¯ z)}, except for¯ [v1(x,¯ z),¯ v2(x,¯ z)]¯ and[v1(y,¯ z),¯ v2(y,¯ z)]¯ ; it is clear that these must satisfy (†).
(ii) where one ofw,w0is from (i), so[w,w0]satisfies (†).
(iii) where one ofw,w0is equal tov(x,¯ z)¯ and the other contains at least oneyi, or one is equal tov(y,¯ z)¯ and the other contains at least onexi; again[w,w0]satisfies (†).
(iv) which are obtained iteratively from supercommutators from (ii) and (iii) by commutation with other supercommutators, thus satisfying (†).
Here (i) takes care of the commutators of various factors of the two products, while (ii)–(iv) takes care of the correct order. Note that the only factor without a variableyi isv(x,¯ z)¯ and the only factor without a variablexj isv(y,¯ z).¯
To show (3) note first that for a single supercommutatorvthe factorisation given in (2) satisfies the requirement. So for a product of supercommutators, we apply (2) to every factor, and then use commutators to get them into the right order. Note that we never have to commute aw(x,¯ z)¯ with aw0(x,¯ z), or a¯ w(y,¯ z)¯ with aw0(y,¯ z), as they already appear¯ in the correct order with respect to one another. It follows that all new commutators
satisfy (†), whence var0x¯ >n.
Theorem 6.4. IfGis nilpotent of classkandvis a product of supercommutatorswwith varx¯(w)>0andvar0x¯(w) ≥nsuch thatGsatisfiesmax{2k−n,1}-largelyv(x,¯ g)¯ =c, then c=1.
Proof. This is true forn≥k, as then var(w)=varx¯(w)+var0x¯(w) ≥1+n, and c=w(x,¯ g) ∈¯ γvar(w)G≤γn+1G={1}
for some ¯x∈G.
Now suppose it is true forn+1≤k, and letv(x,¯ z)¯ be a product of supercommutators wwith varx¯(w) >0 and var0x¯ ≥ n, such thatH satisfies 2k−n-largelyv(x,¯ g)¯ = c. By Lemma 6.3 there isΦ, a product of supercommutators whose factorswsatisfy varx¯(w)>0 and var0x¯(w)>n, such that
v(y¯·x,¯ z)¯ =v(x,¯ z)¯ v(y,¯ z)¯ Φ(x,¯ y,¯ z).¯
Choose ¯h∈Gwithv(h,¯ g)¯ =c. IfX ={x¯∈G:v(x,¯ g)¯ =c}, thenXis 2k−n-large, and Y =X∩h¯−1Xis 2k−n−1-large. Moreover, for ¯x∈Ywe have
Φ(x,¯ h,¯ g)¯ =v(h,¯ g)¯ −1v(x,¯ g)¯−1v(h¯·x,¯ g)¯ =c−1c−1c=c−1.
By hypothesisc−1=1 and we are done.
Theorem 6.5. IfGis nilpotent of classkand satisfies2k-largely an equationv(x,¯ g)¯ =c, then it satisfiesv(x,¯ g)¯ =c.
Proof. Bringing all the constants to the right-hand side, we may assume thatv(x,¯ z)¯ is a product of supercommutatorswwith varx¯(w)>0. By Lemma 6.3 there isΦ, a product of supercommutators whose factorswsatisfy varx¯(w)>0 and var0x¯(w)>0, such that
v(y¯·x,¯ z)¯ =v(x,¯ z)¯ v(y,¯ z)¯ Φ(x,¯ y,¯ z).¯ Fix ¯h ∈G. Then
Φ(x,¯ h,¯ g)¯ =v(h,¯ g)¯−1c−1c=v(h,¯ g)¯ −1
2k−1-largely onG. By Theorem 6.4 we havev(h,¯ g)¯ =1. Sov(x,¯ g)¯ is constant.
Corollary 6.6. IfGis nilpotent of classkandxn=cis true2k-largely, thenc=1and the exponent ofGdividesn.
Proof. Immediate from Theorem 6.5.
Corollary 6.7. IfGis nilpotent of classkandµ∗(xn =c)>1−2−k, thenc=1and the exponent ofGdividesn.
7. Autocommutativity
The notion of autocommutativity has been introduced by Sherman in 1975 [19].
Definition 7.1. LetGbe a finite group,Σa group of automorphisms ofG, andH a subgroup ofG. Thedegree of autocommutativity relative to(H;Σ)is given by
ac(H;Σ)= |{(σ,g) ∈Σ×H:σ(g)=g}|
|Σ| · |H| .
It gives the probability that a random element ofHis fixed by a random automorphism inΣ.
Note that ac(H;Σ)=µ({(σ,g) ∈Σ×H:σ(g)=g}), whereµis the counting measure onΣ×H.
Theorem 7.2. LetH≤Gbe finite groups,Σa group of automorphisms ofG, and suppose that{(σ,g) ∈Σ×H:σ(g)=g}is4-large inΣ×H. ThenH ≤Fix(Σ).
Proof. Givenσ∈Σandg ∈H, by 4-largeness there arex∈Handτ∈Σwith
τ(x)=x, (σ◦τ)(x)=x, τ(gx)=gx and (σ◦τ)(gx)=gx.
Then
gx=σ(τ(gx))=σ(gx)=σ(g)σ(x)=σ(g)σ(τ(x))=σ(g)x,
whenceg =σ(g).
Corollary 7.3. IfH ≤Gare finite groups andΣis a group of automorphisms ofGwith H6≤Fix(Σ), thenac(H;Σ) ≤ 34.
Proof. If ac(H;Σ)> 34then{(σ,g) ∈Σ×H:σ(g)=g}is 4-large inΣ×Hby Lemma 2.5.
HenceH≤Fix(Σ)by Theorem 7.2.
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Khaled Jaber
Department of Mathematics Faculty of Sciences
Lebanese University, Lebanon [email protected]
Frank O. Wagner
Université de Lyon; Université Lyon 1 CNRS UMR 5208, Institut Camille Jordan 21 avenue Claude Bernard
69622 Villeurbanne-cedex, France [email protected]
