# G = G 0 > G 1 > > G k = {e}

## Full text

(1)

Proposition 49.

1. A groupGis nilpotent if and only ifGappears as an element of its upper central series.

2. IfGis nilpotent, then the upper central series and the lower central series have the same length. (That is to say, the leastcsuch thatγc+1(G) ={e}, is equal to the leastcsuch thatZc(G) =G.)

Proof.

1. Suppose thatGhas nilpotency classc. Then the lower central series is a central series forGof lengthc. Now by Proposition 48 we see thatγc−i+1(G)≤Zi(G)for alli(taking the different numbering convention for the lower central series into account). In particular,γ1(G)≤Zc(G). Butγ1(G) =G, and soZc(G) =G. Suppose conversely thatZk(G) =Gfor someG. Then the series{Gi}, given byGi=Zk−i(G)for 0≤i≤k, is a central series forG. SoGis nilpotent by Corollary 47.

2. In the proof of part 1 we have seen that ifγc+1(G) ={e}, thenZc(G) =G. Suppose conversely that Zk(G) =G. Let {Gi} be the central series defined in the proof of part 1. By Proposition 46 we have γk+1(G)≤Gk ={e}. It follows that the lower central series and the upper central series have the same length.

Warning. A groupGof nilpotency classcmay have a central series of length greater thanc. For instance, ifGis abelian, then it has class 1; but any chain of subgroups

G=G0>G1>· · ·>Gk={e}

constitutes a central series, and such a series can be arbitrarily long. Proposition 46 does tell us, however, that no central series can be of lengthshorterthanc.

Theorem 50

Everyp-group is nilpotent.

Proof. LetGbe a p-group. Fori∈N, suppose thatZi(G)<G. SinceGis finite, there existsksuch that Zk(G) =Zk+1(G). Suppose thatZk(G)6=G. Then the quotientG/Zk(G)is a non-trivial p-group, and so by Proposition 21 it has a non-trivial centre. But it follows thatZk+1(G)>Zk(G), and this is a contradiction. So we must haveZk(G) =G, and soGis nilpotent by Proposition 49.1.

Example. We have seen that D2a+1 is nilpotent, even though no other dihedral group is. Thus the only nilpotent dihedral groups are 2-groups.

Proposition 51. LetGbe nilpotent of classc.

1. Any subgroupHofGis nilpotent of class at mostc. 2. IfNEG, thenG/Nis nilpotent of class at mostc.

(2)

Proof.

1. LetH≤G. It is easy to show inductively thatγi(H)≤γi(G)for alli. Soγc+1(H) ={e}.

2. Letθ:G−→G/Nbe the canonical map. It is easy to show inductively thatθ(γi(G)) =γi(G/N)for all i∈N. Soγc+1(G/N) ={eG/N}.

3. We note that

[(g1,h1),(g2,h2)] = ([g1,g2],[h1,h2])

for allg1,g2∈Gandh1,h2∈H. Soγi(G×H) =γi(G)×γi(H)for alli, and the result follows.

Example. LetG=hai⋊ϕhbi, wherea has order 12, bhas order 2, andϕb(a) =a7. We show thatGis nilpotent.

We note thathai=ha3i × ha4i(as an internal direct product), and thatϕb(a3) =a−3, whileϕb(a4) =a4. It follows thata4is central inG, and hence that

G= (ha3i⋊hbi)× ha4i ∼=D8×C3.

Now bothD8andC3are nilpotent, and soGis nilpotent.

Proposition 52. IfGis a finite nilpotent group, andHis a proper subgroup ofG, thenNG(H)6=H.

Proof. Letcbe the nilpotency class of G. Sinceγ1(G) =G, andγc+1(G) ={e}, there exists jsuch that γj(G)6≤H, butγj+1(G)≤H. Sinceγj(G) is normal inGwe see that γj(G)H is a subgroup ofG. Now H/γj+1(G)is certainly normal in the abelian groupγj(G)/γj+1(G), and soH is normal inγj(G)H. So we haveH<γj(G)H≤NG(H), hence the result.

Example. LetG=D16. Every subgroupH of rotations is normal inG, and soNG(H) =G. Let a be a rotation of order 8, and letbbe a reflection. Then we see that

hbi⊳ha4,bi⊳ha2,bi⊳G,

since each of the subgroups in the chain has index 2 in the next. Every subgroup ofGis either a rotation subgroup, or else one of the subgroups in the chain above (for some reflectionb), and so we see that every proper subgroup ofGis normalized by a strictly larger subgroup.

Lemma 53 (Frattini Argument). Let G be a finite group, and K a normal subgroup. Let P be a Sylow p-subgroup ofK. ThenG=KNG(P).

Proof. Let g∈G. Then ()gPK since K is normal, and so gPSyl

p(K). Since any two Sylow p-subgroups ofKare conjugate inK, we havegP=kPfor somekK. But now it is clear thatk−1gP=P, and g=k(k−1g)∈KN (P).

(3)

Corollary 54. LetGbe a finite group, and letPbe a Sylow p-subgroup ofG. ThenNG(NG(P)) =NG(P). Proof. SinceNG(P)is a normal subgroup ofNG(NG(P)), and sinceP∈Sylp(NG(P)), the Frattini Argument tells us thatNG(NG(P)) =NG(P)NNG(NG(P))(P). But clearlyNNG(NG(P))(P)≤NG(P), and the result follows.

Theorem 55

LetGbe a finite group, and letp1, . . . ,pk be the distinct prime divisors of|G|. LetP1, . . . ,Pk be subgroups ofGwithPi∈Sylpi(G)for alli. The following statements are equivalent.

1. Gis nilpotent. 2. PiEGfor alli. 3. G∼=P1× · · · ×Pk. Proof.

1=⇒2LetGbe nilpotent. Corollary 54 tells us thatNG(Pi) =NG(NG(Pi)). But Proposition 52 tells us that no proper subgroup ofGis equal to its own normalizer. HenceNG(Pi) =G.

2=⇒3Suppose that PiEG for all i. We argue by induction; let P(j) be the statement that P1· · ·Pj ∼= P1× · · · ×Pj. CertainlyP(1)is true. Suppose thatP(j)true for a particular j<k. LetN=P1. . .Pj. Since NandNPj+1are both normal inG, and since their orders are coprime, we have[N,Pj+1]≤N∩Pj+1={e}. Song=gnfor alln∈Nandg∈Pj+1, and soNPj+1∼=N×Pj+1. The statementP(j+1)follows inductively. This is sufficient to prove the implication, since|G|=|P1· · ·Pk|.

3=⇒1Suppose thatG=P1× · · · ×Pk. Each subgroupPi is nilpotent by Theorem 50. SinceGis the direct product of nilpotent groups, it follows from Proposition 51.3 thatGis itself nilpotent.

Corollary 56. LetGbe a finite group, and letg,h∈Gbe elements with coprime orders. Thengh=hg. Proof. LetP1, . . . ,Pk be the Sylow subgroups ofG. By Theorem 55 there exists an isomorphismθG−→ P1× · · · ×Pk. Letθ(g) = (g1, . . . ,gk)andθ(h) = (h1, . . . ,hk). Sincegandhhave coprime orders, one ofgi orhimust be the identity, for alli. Hencegandhcommute.

Remark. The converse to Corollary 56 is also true. Suppose thatGhas the property that any two elements with coprime order commute. LetP1, . . . ,Pkbe Sylow subgroups for the distinct prime divisorsp1, . . . ,pkof G. Then clearlyPi∈NG(Pj)for alli,j, since ifi6= jthen the elements ofPiandPjcommute. It follows that NG(Pi) =Gfor alli, and so the Sylow subgroups ofGare all normal; henceGis nilpotent.

Definition. LetGbe a group. A subgroup M<Gismaximal ifM≤H≤G=⇒H=MorH=G. We writeM<maxGto indicate thatMis a maximal subgroup ofG.

(4)

Remark. We note that ifGis a finite group, then it is clear that every proper subgroup ofGis contained in some maximal subgroup ofG. This is not necessarily the case in an infinite group; indeed there exist infinite groups with no maximal subgroups, for instance(Q,+).

Proposition 57. IfGis a finite nilpotent group, and ifM<maxG, thenM⊳G, and|G/M|is prime.

Proof. By Proposition 52 we see thatNG(M)6=M. But certainlyM≤NG(M)≤G, and soNG(N) =G. So M⊳G. Since there are no proper subgroups of Gstrictly containingM, it follows from Proposition 2 that G/Mhas no proper, non-trivial subgroup. HenceG/Mis cyclic of prime order.

Proposition 58. LetGbe a finite group. If every maximal subgroup ofGis normal, thenGis nilpotent. Proof. Suppose that every maximal subgroup ofG is normal. Let P be a Sylow p-subgroup of G, and suppose that NG(P)6=G. ThenNG(P)≤M for some maximal subgroup M. Now P∈Sylp(M), and M is normal inGby hypothesis. So the conditions of the Frattini argument are satisfied, and we have G=

MNG(P). But this is a contradiction, sinceNG(P)≤M. So we must haveNG(P) =G. Hence every Sylow subgroup ofGis normal, and soGis nilpotent by Theorem 55.

We summarize our main criteria for nilpotence in the following theorem. Theorem 59

LetGbe a finite group. The following are all equivalent to the statement thatGis nilpotent. 1. The lower central series forGterminates at{e}.

2. The upper central series forGterminates atG. 3. Ghas a central series.

4. Every Sylow subgroup ofGis normal. 5. Gis a direct product of p-groups.

6. Any two elements ofGwith coprime order commute.

7. Each proper subgroup ofGis properly contained in its normalizer. 8. Every maximal subgroup ofGis normal.

The first three of these conditions apply also to infinite groups.

§ 8 More on group actions

Definition. LetGandHbe isomorphic groups. LetGact on a setX, andHon a setY.

1. We say that the actions ofGandHareequivalentif there is an isomorphismθ:G−→Hand a bijection α:X−→Y such thatα(gx) =θ(g)α(x)for allg∈Gandx∈X.

(5)

2. In the special case thatG=Handθ is the identity map, we haveα :X−→Y such thatα(gx) =gα(x)

for allg∈Gandx∈X. Then we say that the actions onX andY areequivalent actions of G, and thatα is anequivalence of actions.

We revisit the Orbit-Stabilizer Theorem in the light of this definition. Theorem 60. z Orbit-Stabilizer Theorem Revisited

LetGact transitively on a setX. Letx∈X, and letH be the stabilizer ofxinG. LetY be the set of left cosets ofH inG. Then the action ofGonY by left translation, and the action ofGon X, are equivalent actions ofG.

Proof. Recall that fork1,k2∈Gwe havek1H=k2H⇐⇒k1x=k2x. So there is a well-defined mapf:Y−→

Xgiven by f(kH) =kxfork∈G. The map f is bijective (this is the substance of the proof of Theorem 10.) Now for allg∈Gwe haveg f(kH) =gkx= f(gkH), and so f is an equivalence of actions.

The significance of Theorem 60 is that the study of transitive actions of Gis reduced to the study of actions ofGon the left cosets of its subgroups.

Suppose thatGis a group acting on a setΩ. For everyk∈N, there is an action ofGonΩk given by g(x1, . . . ,xk) = (gx1, . . . ,gxk).

Definition. LetGact onΩ, and letk≤ |Ω|. We say that the action ofGonΩisk-transitive if for any two k-tuplesx= (x1, . . . ,xk) andy= (y1, . . . ,yk)inΩk, with the property thatxi 6=xj andyi 6=yj when i6= j, there existsg∈Gsuch thatgx=y(sogxi=yifor 1≤i≤k).

Another way of phrasing the definition is thatGisk-transitive onΩif it acts transitively on the subset ofΩk consisting ofk-tuples ofdistinctelements ofΩ.

Remark. IfGactsk-transitively onΩ, then it is clear that it actsℓ-transitively onΩfor everyℓ≤k. Examples.

1. Snacts n-transitively onΩ={1, . . . ,n}. For anyn-tuplex= (x1, . . . ,xn) of distinct elements ofΩwe have {x1, . . . ,xn}={1, . . . ,n}, and so the map fx:i7→xi is inSn. Now ify is anothern-tuple of distinct elements then fyfx−1:x7→y.

2. Anacts only(n−2)-transitively onΩ={1, . . . ,n}. It is not(n−1)-transitive, since the tuples(1,3, . . . ,n) and(2,3, . . . ,n) lie in distinct orbits. Letxandy be(k−2)-tuple of distinct elements ofΩ. Then there is an elementgofSnsuch thatgx=y. Now there exist two pointsi,jwhich are not in the tupley, and since

(i j)y=ywe have(i j)gx=y. Since one ofgor(i j)glies inAn, we see thatxandylie in the same orbit of An, as we claimed.

(6)

3. Ifn>3 thenD2nis only 1-transitive on the vertices of a regularn-gon. There exist verticesu,v,wsuch thatuis adjacent tov but not tow, and it is clear that there is nog∈D2n such thatg(u,v) = (u,w)since symmetries of ann-gon preserve adjacency of vertices.

Remark. IfGacts onΩandH=StabG(x)forx∈Ω, thenHacts onΩ\ {x}.

Proposition 61. LetGbe transitive on Ω, letx∈Ω, and let H=StabG(x). Letk∈N. Then G acts k-transitively onΩif and only ifHacts(k−1)-transitively onΩ\ {x}.

Proof. SupposeGisk-transitive. Lety= (y1, . . . ,yk−1)andz= (z1, . . . ,zk−1)be (k−1)-tuples of distinct elements ofΩ\ {x}. Theny′= (y1, . . . ,yk−1,x)andz′= (z1, . . . ,zk−1,x)arek-tuples of distinct elements of

Ω, and so there existsh∈Gsuch thathy′=z′. Now we see thathx=x, and soh∈H. It is also clear that hy=z. So we have shown thatHis(k−1)-transitive on\ {x}.

Conversely, suppose thatHis(k−1)-transitive onΩ\ {x}. Lety= (y1, . . . ,yk)andz= (z1, . . . ,zk) be distinctk-tuples of elements ofΩ. SinceGis transitive on Ω, there exist f,g∈Gsuch that f yk =xand gzk=x. Now we see thatu= (f y1, . . . ,f yk−1)andv= (gz1, . . . ,gzk−1)are(k−1)-tuples of distinct elements ofΩ\ {x}. So there existsh∈H such thathu=v. Now it is straightforward to check thath f yi =gzi for 1≤i≤k, and sog−1h f y=z. We have therefore shown thatGisk-transitive onΩ.

Definition. Recall that an equivalence relation on a setmay be regarded as a partition of the setinto disjoint subsets (the parts, or equivalence classes) whose union isΩ. (Ifx,y∈Ω, thenx∼yif and only ifx andylie in the same part.

1. We say that an equivalence relation istrivialif it has only one part, or if all of its parts have size 1. (So eitherx∼yfor allx,y, or elsex∼yonly ifx=y.)

2. Suppose thatGacts onΩ. We say thatG preservesΩif

x∼y⇐⇒gx∼gy, forx,y∈Ω. It is clear that any groupGacting onΩpreserves the trivial partitions. Definition. Let|Ω|>1, and letGact transitively onΩ.

1. If∼is a non-trivial equivalence relation onΩ, such thatGpreserves∼, then we say the action ofGis imprimitive, and that∼is asystem of imprimitivityforG. The parts of∼are calledblocks.

2. If there is no non-trivial equivalence relation onΩwhich is preserved byG, then we say that the action ofGisprimitive.

Examples.

1. Letgbe then-cycle(1 2. . .n)∈Sn, and lethgiact onΩ={1, . . . ,n}in the natural way. The action is primitive if and only ifnis prime. Otherwisenhas some proper divisord, and the equivalence relation on

(7)

2. Snis primitive on={1, . . . ,n}forn>1. AndAnis primitive onforn>2.

3. D2n acts primitively on the vertices of ann-gon if and only ifnis prime. Ifn=abis a proper factor-ization ofn, then we can inscribebdistincta-gons inside then-gon, and the vertices of thea-gons form the blocks of a system of imprimitivity.

Proposition 62. LetBbe a system of imprimitivity for the action ofGon. For a blockBofB, define

gB={gx|x∈B}. ThengBis a block ofB, and the mapB7→gBdefines a transitive action ofGon the set

of blocks ofB.

Proof. Ifx∈gBandy∈Ω, theng−1x∈B, and sog−1x∼g−1y⇐⇒g−1y∈B. So we havex∼y⇐⇒y∈gB, and sogBis a block ofB. It is easy to see thateB=Band thatg(hB) =ghBforg,hG, and so the map

B7→gBgives an action ofG. And sinceGis transitive onΩ, it is clear that it is transitive on the blocks. Corollary 63. IfBis a system of imprimitivity for the action ofGon, then all of the blocks ofBhave

the same size.

Proof. This is clear from the transitivity ofGon the blocks; ifBandCare blocks thenC=gBfor some g∈G.

Corollary 64. If|Ω|=p, wherepis prime, and ifGacts transitively onΩ, then the action ofGis primitive. Proof. Let∼be an equivalence relation preserved byG. It is clear from Corollary 63 that the size of each part divides|Ω|. But since pis prime, its only divisors are 1 and p. So∼must be trivial, and henceGis primitive.

Remark. Corollary 64 establishes the claim, in the examples above, that the actions ofCp andD2p on p points are primitive.

Proposition 65. IfGis2-transitive onΩ, then it is primitive.

Proof. Let∼be any non-trivial equivalence relation onΩ. it has distinct partsB1 andB2, and the partB1 comtains distinct pointsxandy. Letz∈B2. Then sinceGacts 2-transitively, there existsg∈Gsuch that g(x,y) = (x,z). So we havex∼ybutgx6∼gy. HenceGdoes not preserve∼, and soGis primitive.

Remark. Primitivity is a strictly stronger condition than transitivity, since e.g. h(1 2 3 4)i is transitive but not primitive on{1,2,3,4}. And 2-transitivity is strictly stronger than primitivity, since e.g.A3is primitive but not 2-transitive on{1,2,3}.

Proposition 66. LetGact transitively onΩ. LetH be the stabilizer of a pointx∈Ω. Then the action ofG is primitive if and only ifH<maxG.

(8)

Proof. Suppose thatBis a system of imprimitivity forGon. LetBbe the block ofBwhich contains

the pointx. Letybe point inBdistinct fromx. SinceGis transitive, there existsg∈Gwithgx=y. Since y∈gB, we havegB=B. LetL=StabG(B), the stabilizer in the action ofGon the blocks ofB. It is clear thatH≤L, and sinceg∈L\H, we haveH<L. But since there is more than one block ofB, andGis

transitive on blocks, we see thatL<G. We have thatH<L<G, and soHis not maximal inG.

Conversely, supposeHis not maximal. LetLbe a subgroup such thatH<L<G. Recall from Theorem 60 that the action ofGonΩis equivalent to its action on left cosets ofH. So we may suppose thatΩis this set of cosets. Now we define an equivalence relation∼onΩbyaH∼bH⇐⇒aL=bL. This relation is well defined, since ifaH =a′HthenaL=a′L, and it is clear that it is an equivalence relation; it is non-trivial, sinceH<L<G. Now forg∈Gwe have

aH∼bH⇐⇒aL=bL⇐⇒gaL=gbL⇐⇒gaH∼gbH, and so∼is a system of imprimitivity forG. SoGis not primitive.

Corollary 67. IfGis nilpotent andGacts primitively onΩ, then|Ω|is prime.

Proof. LetH be the stabilizer ofx∈Ω. ThenH<maxG, and so|G:H|is prime, by Proposition 57. But

|Ω|=|G:H|by Theorem 60.

Remark. LetGandΩbe as in Corollary 67. We have thatH⊳Gby Proposition 57, and it follows thatH is the kernel of the action. So we have a homomorphismG/H−→Sym(Ω)∼=Spfor some prime p. Now inSpthere is no non-p-element which commutes with ap-cycle. But sinceGis nilpotent, any two elements of coprime order commute, and it follows that the image ofG/HinSpis cyclic of order p.

Proposition 68. LetG act faithfully and primitively on a finite setΩ, and let N be a non-trivial normal subgroup ofG. ThenNis transitive onΩ.

Proof. LetM<maxGbe the stabilizer ofx∈Ω, and letNEG. ThenM≤MN≤G, and soMN is equal either toMor toG. IfMN=NthenN≤M, and sinceN is normal we haveN≤gMfor everygG. But the conjugates ofMare the point-stabilizers inG, and it follows thatN is contained in the kernel ofG. But Gis faithful, and so we must haveN={e}in this case.

We may therefore suppose thatMN=G. Now the stabilizer ofxinNisM∩N, and we have

|OrbN(x)|=

|N| |M∩N|=

|MN| |M| =

|G|

|M|=|OrbG(x)|.

(Here we have used the Orbit-Stabilizer Theorem and the Third Isomorphism Theorem.) SinceGis transi-tive, we have OrbN(x) =Ω, and soNis transitive.

(9)

Proof. First note that since 7 divides 168, there is a 7-cycle inG, and soGis transitive on{1, . . . ,7}. By Corollary 64 we see thatGacts primitively. Suppose thatNis a proper, non-trivial normal subgroup ofG. ThenNis transitive by Proposition 68, and so|N|is divisible by 7. SoNcontains a Sylow 7-subgroup ofG, and since it is normal, it must contain all Sylow 7-subgroups ofG.

Let P be a Sylow 7-subgroup of G. We observe that S7 contains 120 conjugates of P, and so the normalizer ofPinS7has order 42. SoGdoes not normalizeP. The numbern7(G)of Sylow 7-subgroups ofGis therefore greater than 1; we haven7(G)≡1 mod 7, andn7(G) divides 168, and so we must have n7(G) =8. So the subgroupNhas 8 Sylow 7-subgroups, and hence 8 divides|N|. Since 7 also divides|N|, we see that 56 divides|N|. It follows that|N|=56, since 168=56×3.

NowNcontains 48 elements of order 7, since each conjugate ofPcontains 6 such elements. It follows thatN has only 8 elements whose order is not 7, and it is clear that these must form a normal Sylow 2-subgroupK. NowNacts transitively on 7 points, and soN is primitive by Corollary 64. SoKis transitive by Proposition 68. So we have a subgroup of order 8 acting transitively on 7 points, and this is absurd since 7 does not divide 8. So we have a contradiction; no such subgroupNcan exist, and soGis simple.

We conclude the course by constructing a subgroup ofS7of order 168. More precisely, we shall construct a groupGwith a faithful action on 7 points, which can therefore be identified with a subgroup ofS7.

LetG=GL3(2), the set of invertible 3×3 matrices with entries fromZ2, under matrix multiplication. SinceZ2is a field, the standard results of linear algebra apply. A 3×3 matrix is invertible if and only if it has non-zero determinant, which is the case if and only if its columns are linearly independent overZ2.

There is an action ofGon the spaceZ23of column vectors, with basis

e1=     1 0 0    

, e2=

    0 1 0    

, e3=

    0 0 1     .

This space has size 23=8, andGacts transitively on the 7 non-zero vectors.

Note that ifAei=ci fori=1,2,3, then the columns ofAarec1,c2,c3. It follows that the action ofGis faithful, since the only element which fixes all three basis vectors is the identity.

It remains to calculate|G|. We consider how we can construct an elementAofG, column by column. There are clearly 7 possibilities for the first columnc1ofA, since any non-zero vector will do. The columns must be linearly independent, and so the second columnc2cannot lie in the span of the first. This rules out the choicesc2∈ {0,c1}, and there are 6 remaining possibilities. Now we require the last columnc3 to lie outside the span of the first two; this span contains 4 vectors, and so there are 4 possible choices forc3. So there are 7×6=168 possible choices for the matrixA, and so|G|=168.

(10)

Proof. We have shown that GL3(2)acts faithfully and transitively on 7 points. So it is isomorphic with a subgroup ofS7, of order 168, and so it is simple by Proposition 69.

Remark. The group GL3(2) is one member of an infinite family of simple groups. Let GLd(p) be the general linear group, the group of invertibled×d matrices over the fieldZp. Define SLd(p) to be the special linear group, the subgroup consisting of matrices with determinant 1. The centre ofSLd(p)is given by

Z=Z(SLd(p)) =

n

λI|λ∈Zp,λd=1o.

The quotient group PSLd(p) =SLd(p)/Z, theprojective special linear group, is simple except whend=2 andp=2,3. It has a natural 2-transitive action on thelinesofZpd.

Updating...