Suppose that G is a simple group of order 720 = 8 ·9 ·10. Writing the order this way makes it plausible that G should act 3-transitively on a set with 10 elements, asPGL2(9) has this order and
acts 3-transitively on its set of 10 Sylow 3-subgroups. Since PGL2(9) is not simple, it may make
sense to look in that direction for a contradiction. So we begin by studying the3-local structure of our hypothetical simple groupGof order720. However, since there does exist a simple group of order360, the2-local structure must eventually be decisive. It is here that Burnside’s sketch (which begins with the5-local structure)10 seems to have a serious gap. Our treatment of the3-local part is based on that
of Derek Holt11 which I have recast to better highlight his essential point, with different arguments in
places. Then we will have enough information for a correct2-local argument in the spirit of Burnside. From the Sylow and Burnside Transfer theorems, the possible3-factorizations of|G|are
|G|= 32·2·40, or |G|= 32·8·10. (26) In the former case, wheren3(G) = 40, the normalizer of a Sylow3-subgroupP acts by an involution
onP with trivial fixed points, and normalizes every subgroup ofP. Holt’s crucial observation about the3-local structure is as follows.
Lemma 5.25 Every subgroup of order three inGis contained in just one Sylow3-subgroup ofG.
Proof: LetQ < Gbe a subgroup of order three, with normalizerN =NG(Q), and letP be a Sylow
3-subgroup ofGcontainingQ. The3-factorization ofN is
|N|= 32 ·ν·n3(N),
whereν = [N ∩NG(P) :P]andn3(N)is the number of Sylow3-subgroups inN. Since the Sylow
3-subgroups are abelian, any such subgroup containingQmust lie inN. So the lemma is equivalent to the assertion thatn3(N) = 1.
10Notes on the theory of groups of finite order, Bull. London Math. Soc. 1894 11http://sci.tech-archive.net/Archive/sci.math/2006-12/msg07456.html
If n3(G) = 40 then NG(P) contains an element inverting P, hence normalizing Q, so ν > 1. If
n3(G) = 10 then again ν > 1, lest we have at least [NG(P) : N ∩NG(P)] = [NG(P) : P] = 8
conjugates of Q in P, whereas the group P of order 32 can have at most four subgroups of order
three. Thus, in either case we haveν > 1. We also have n3(N) < 10, lest G ≤ S4. It follows that
n3(N)∈ {1,4}.
Assume thatn3(N) = 4. Thenν = 2, lest[G:N]≤5, soN has3-factorization
|N|= 32·2·4.
LetP = P1, P2, P3, P4 be the Sylow 3-subgroups inN, and let X = {gQ : g ∈ G}be the set of
G-conjugates of Q. We have |X| = [G : N] = 10. The groupGacts on X by conjugation and we consider the fixed points ofQ:
XQ={Q0 ∈X :Q < NG(Q0)}.
Since a group of order three admits no automorphism of order three, we have
Q < NG(Q0) ⇔ Q < CG(Q0) ⇔ Q0 < CG(Q) ⇔ Q0 < N,
in which caseQ0 < Pifor somei. Therefore
XQ =
4
[
i=1
X(Pi), (27)
where, for any Sylow p-subgroup P0 of G we define X(P0) := {Q0 ∈ X : Q0 < P0} to be the
set of conjugates of Q which are contained in P0. Note that gX(P0) = X(gP0) for any g ∈ G. In particular, the setsX(Pi)all have the same cardinality. SinceP is abelian,NG(P)acts transitively on
X(P)(by the same argument used for Lemma5.11) and the stabilizer of QinNG(P)has cardinality
|N ∩NG(P)|= 32·2, som = [NG(P) :NG(P)∩N]∈ {1,4}, according to the two possibilities for
NG(P)in (26). Finally, sincePiis generated by any two elements ofX(Pi), we haveX(Pi)∩X(Pj) =
{Q}fori6=j. It now follows from (27) that|XQ|= 1 + 4(m−1). Since|XQ| ≤ |X| = 10we must
havem = 1, soXQ ={Q}and theQ-orbits inXhave sizes1,3,3,3.
AsAut(Q) = C2, the centralizerCG(Q)has order36or72, with Sylow2-subgroupRof order 4or
8. AsRfixesQandGis simple,Racts faithfully onX− {Q}, permuting the threeQ-orbits therein, giving a homomorphismR → S3. Ifr ∈Rpreserves a Q-orbit{Q1, Q2, Q3}inX thenrnormalizes
eachQi, sincer commutes withQ. Hence the image ofRinS3 is nontrivial, so somer ∈ Rmaps to
a2-cycle inS3. This meansrhas cycle type[1323]onX− {Q}and cycle type[1423]onX. Thusris
an odd permutation onX, contradicting the simplicity ofG. It follows thatn3(N) = 1and the lemma
is proved.
LetY be the set of Sylow 3-subgroups of G and let P ∈ Y. The lemma implies that P acts simply transitively onY − {P}, so|Y| ≡1 mod 9. This rules outn3(G) = 40, so we must haven3(G) = 10
and|NG(P)|= 32·8. RegardingGas a subgroup ofA10via its action onY by conjugation shows that
P cannot be cyclic because the normalizer of a9-cycle in S10 has order 32 ·6, hence cannot contain
ChooseP0inY distinct fromP and let
H =NG(P)∩NG(P0)
be the normalizer ofP0 inNG(P). SinceNG(P)acts transitively onY − {P}, it follows that|H|= 8,
soH is a Sylow2-subgroup of NG(P)andNG(P) = P ·H. I claim thatH acts freely onP − {1}
by conjugation. For by the lemma again, any nonidentity elementt ∈ P has cycle type [1333]inY, whose centralizer inA10has order33·6/2 = 81and intersectsHtrivially.12
Lets ∈ H be an involution (an element of order two). Since s acts freely on P − {1}, it must act by inversion, hences is the unique involution in H. The only groups of order8containing a unique involution areQ8 andC8. In the latter caseH would be generated by an[8]-cycle onY which is odd,
contradicting the simplicity ofG. HenceH ' Q8. 13 SinceQ8 does not embed inSn forn < 8, the
faithful action ofH onY − {P, P0}
must also be free and transitive. Hence the elements of order four inHhave cycle type[1244]onY, andshas cycle type[1224].
Ifx ∈ P − {1}andH ∩Hx 6= 1, thens ∈ H∩Hx, says = hx for some hinH. But thenh is an involution, soh = s, meaning that xand scommute. This contradicts H acting freely onP − {1}, soH∩Hx = 1. It follows that there are exactly nine N
G(P)-conjugates ofsinH, and these are all
of the involutions inNG(P). From the cycle type of sonY, we see that NG(P)andNG(P0)are the
only conjugates ofNG(P)containings. As there are ten conjugates ofNG(P), there are9·10/2 = 45
conjugates ofs inG. Hence the centralizer S = CG(s) is a Sylow 2-subgroup of G. By the same
argument there are45conjugates ofHinG, soS =NG(H)is also the normalizer ofHinG.
I claim that distinct involutions have distinct centralizers. For suppose t 6= s is an involution with
CG(t) = CG(s). Then t ∈ NG(H)−H so preserves the fixed point setYH = {P, P0}, but tcannot
normalizeP orP0, sot must switchP andP0. Butt is a product of2-cycles and is even, while10/2
is odd, sotmust have at least two fixed-points on Y. This means t ∈ NG(P00)∩NG(P000) for some
pairP00, P000 ∈Y − {P, P0}. Buttcannot act trivially onY − {P, P0}lest it be a2-cycle onY. AsH,
which centralizest, must preserve the fixed points oftinY − {P, P0}, this contradicts the transitivity
ofHonY − {P, P0}
. 14
Now takeh∈H of order4and consider the action ofHon the setZof Sylow2-subgroups ofG. We haveh2 =s. Sincesis contained in just one Sylow2-subgroup, namelyCG(s), the cycle type ofson
Zis1222. Hence the cycle type ofhis[1411], which is an odd permutation, contradicting the simplicity
ofG.
The argument is difficult for several reasons. First,720 = 2·360, and there does exist a simple group of order360, namely the alternating groupA6. Secondly,A6 'PSL2(9)(see section8.5.2), soSL2(9)is
a group of order720surjecting ontoA6. Thirdly, there are three groups of order720containingA6with
index two (see below). All these groups of order720 flirt with the simple group A6, but themselves
just fail to be simple, for different reasons, which is why it is hard to rule them out.
12Since|H| = 8 = |P − {1}|, it follows thatH is also transitive onP− {1}, soGhas a unique conjugacy class of
elements of order three, and this class has80elements. However we do not need this.
13We do not actually need to know thatH =Q
8and notC8.
The three groups containingA6with index two can be seen as follows. We have seen that automorphism
group of S6 is Aut(S6) = S6 oC2. It follows that Out(A6) ' C2 ×C2. By the Correspondence
Theorem, there are three subgroups ofAut(A6)of order720, containing A6. One of these isS6. Via
the isomorphismA6 'PSL2(9), another one isPGL2(9). The third group is theMathieu GroupM10,
which is part of a family of highly transitive simple (or almost simple, in the case ofn= 10) subgroups
Mn≤Sn.
To seeM10more explicitly, we start withPGL2(9), which we think of as the group of permutations of
P1( F9)given by P GL2(9) = z 7→ az+b cz+d : ad−bc6= 0. .
The fieldF9 of9elements is built fromF3 =Z/3Zjust asCis built fromR, namely,
F9 ={x+iy: x, y ∈F3}, i2 =−1,
and just likeC,F9 has an automorphismx+iy=x−iy. We define the group
PΓL2(9) = z 7→ az+b cz+d : ad−bc 6= 0 ∪ z 7→ az¯+b cz¯+d : ad−bc6= 0 .
This group contains PGL2(9) with index two; a nontrivial coset representative is simply z 7→ z¯. It
turns out that
PΓL2(9)'Aut(A6). In this viewpoint,M10is the subgroup ofPΓL2(9)given by
M10= z 7→ az+b cz+d : ad−bcis a square inF × 9 ∪ z 7→ az¯+b cz¯+d : ad−bcis a nonsquare inF × 9 .
Thus,M10containsPSL2(9)with index two; a nontrivial coset representative isz 7→iz¯.
The subgroupS6 is generated byPSL2(9) andz 7→ z¯. In this guise an outer automorphism ofS6 is
conjugation byz 7→iz.
6
Solvable and nilpotent groups
A groupGissolvableif it has a chain of subgroups
1 =G0 < G1/ G2/ G3/· · ·/ Gn=G (28)
where eachGi is normal inGi+1with abelian quotientGi+1/Gi.
IfH / GthenGis solvable if and only if both H andG/H are solvable (exercise). It follows thatG
isnotsolvable precisely when there exist subgroupsH / K ≤GwithK/Hnonabelian simple. In this sense solvable groups are diametrically opposed to nonabelian simple groups.
i) Any finite groupGwith|G|< 60is solvable, because there are no simple groups of order< 60. In particularSnis solvable forn≤4. However,Snis not solvable whenn≥5, because it contains the
simple groupA5.
ii) The dihedral groupDnis solvable for everyn≥2because it has a chain of subgroups1< Cn<
GandG/Cn=C2.
iii) For any fieldF, the subgroupBof upper triangular matrices inGLn(F)is solvable. For example
ifn = 3one can take the series
1< 1 0 ∗ 0 1 0 0 0 1 / 1 ∗ ∗ 0 1 ∗ 0 0 1 / × ∗ ∗ 0 × ∗ 0 0 × =G,
with the obvious generalization to arbitraryn.
iv) Any group G of order pq, where p and q are primes, is solvable. For if p ≤ q then G has a normal subgroupCqwith quotientCp. More generally, any group whose order is divisible by at most
two primes is solvable. This is Burnside’s paqb theorem, whose original proof uses character theory and still appears to be the most accessible proof to non-experts.
v) A groupGis solvable if and only if it has thefactorization property: If|G|=m·nfor relatively prime integers m, n, then G = M N for subgroups M, N of G of orders m, n. This is P. Hall’s generalization of Burnside’s theorem.
vi) Any finite group of odd order is solvable. This is the famous Feit-Thompson theorem from 1963, whose proof is much more difficult than the previous two theorems.
vii) Letf(x)be a polynomial with rational coefficients having n distinct roots. TheGalois group
Gf is the subgroup ofSnconsisting of those permutations of the roots which preserve all polynomial
relations among them. The group Gf is solvable precisely when the roots of f can be expressed in
terms of rational and radical expressions in the coefficients off, as in the quadratic, cubic or quartic formulas. This was discovered by E. Galois, and is the origin of the term “solvable”. It explains why there is no general formula for a quintic polynomial:S5is not solvable.
viii) Same situation, where now the coefficients off lie in the fieldQp ofp-adic numbers for some
primep. ThenGf is always solvable. In fact,Ghas a canonical chain of subgroupsG1/ G0/ G,where
G1 is ap-group andG0/G1,G/G0are both cyclic.
Solvable groups occur naturally in many areas of mathematics. Being opposite to nonabelian simple groups, they are a very natural class of groups to study. However, some of the most interesting theorems about solvable groups are very difficult to prove. For example, the proof of the Feit-Thompson theorem takes 255 pages.
Among the solvable groups are thenilpotent groups, for which the first few interesting theorems are easy, including a version of Hall’s theorem above.
The definition is as follows. For any group G, the ascending central series 15 of Gis the chain of
subgroups ofG:
Z1 EZ2 EZ3 E· · · (29)
defined inductively as follows: Z1 = Z(G) is the center of G, and given Zi, define Zi+1 to be the
unique subgroup containing Zi such that Zi+1/Zi = Z(G/Zi) is the center of G/Zi. We say G is
nilpotentifZc =Gfor somec≥1. The minimal suchcis thenilpotence classofG.
A nontrivial nilpotent group must have nontrivial center, lest allZi = 1. If Gis nilpotent of class c
thenG/Z(G)has classc−1.
IfGis nilpotent andH / GthenH andG/H are nilpotent,but not conversely: The symmetric group
S3 has trivial center, so its ascending central series has Zi = 1 for all i. Hence S3 is not nilpotent.
However, its subgroupA3 and quotientS3/A3 =C2are both nilpotent.
Direct products of nilpotent groups are nilpotent. So any direct product ofp-groups is nilpotent. We will see that all nilpotent groups are of this form.
Examples of Nilpotent Groups:
i) The abelian groups are the nilpotent groups of class1. The nilpotent groupsGof class2are those for whichG/Z(G)is abelian.
ii) Any finitep-group is nilpotent, because anyp-group has nontrivial center, henceZi 6= Zi+1, so
eventuallyZihas the same order asG. If|G|=pkthen the nilpotence class ofGis at mostk−1, since
every group of orderp2 is abelian.
iii) The dihedral groupDnis nilpotent if and only ifn is a power of2(use 5. in Thm. 6.1below).
Ifn = 2` andr =stis the product of two generating reflectionssandt, thenr has order2`andZ i is
cyclic, generated byr2`−i
. AsD2`/hr2i ' C2×C2 is abelian, it follows thatD2` has nilpotence class
equal to `, the maximum possible for a group of order 2`+1. There are exactly two other families of groups of order2`+1 having maximal nilpotence class: the generalized quaternion groupQ
2`+1 and the quasidihedral groupQD2`+1.
iv) For any fieldF, the subgroupUnof strictly upper triangular matrices inGLn(F)is nilpotent of
classn−1. For example ifn= 4the ascending central series is 1 0 0 ∗ 0 1 0 0 0 0 1 0 0 0 0 1 < 1 0 ∗ ∗ 0 1 0 ∗ 0 0 1 0 0 0 0 1 < 1 ∗ ∗ ∗ 0 1 ∗ ∗ 0 0 1 ∗ 0 0 0 1 =G,
with the obvious generalization to arbitraryn. The matrices inUnare of the formI+A, whereAn= 0.
This may be the origin of the term “nilpotent”.
The main theorem on finite nilpotent groups is the following collection of characterizations.
Theorem 6.1 For a finite groupGthe following are equivalent. 1. Gis nilpotent.
2. Any proper subgroupHofGis properly contained in its normalizerNG(H). 3. Every maximal subgroup ofGis normal inG.
4. Every Sylowp-subgroup ofGis normal inG. 5. Gis the direct product of its Sylowp-subgroups.
6. For every divisormof|G|there is a normal subgroupM / Gwith|M|=m.
Proof: (1 ⇒2): By induction we assume the result for all groups of smaller order thanG. LetH be a proper subgroup ofG. SinceGis nilpotent its center Z is nontrivial and normalizesH. IfZ is not contained inH thenH 6=HZ, done. AssumeZ < H. LetG=G/Z, H =H/Z and letπ :G →G
be the projection. ThenHis a proper subgroup ofG, so by inductionH 6=NG(H). One checks that
H =π−1(H) and NG(H) = π−1(NG(H)),
henceH 6=NG(H), by the Correspondence Theorem.
(2⇒3): Apply 2 to a maximal proper subgroup ofG.
(3⇒ 4): Suppose the Sylowp-subgroupP ofGis not normal. ThenNG(P)is a proper subgroup of
G, contained in a maximal proper subgroup M ofG which is normal in G, by 3, andP is a Sylow
p-subgroup ofM. For anyg ∈G, the conjugatePg is another Sylowp-subgroup ofM, soPg =Pm
for somem ∈ M. Theng ∈ NG(P)m ⊂ M. 16 Sinceg was arbitrary, we have shown thatG = M,
contradictingM being proper.
(4 ⇒ 5): Induct on the number of primes dividing |G|. Let |G| = pr1 1 · · ·pr
n
n . By 4, the Sylow pi-
subgroupPi is unique and any product of Sylow subgroups is a normal subgroup ofG. In particular