Theorems you should know UW-Madison

Guillermo Mantilla

(A) (Isaacs,5.11) Let G be a nite group and write n

p

(G) = jSyl

p

(G)j.Then n

p

divides jGj and n

p

(G) 1 mod p.In fact,

n

p

(G) 1 mod p

e

if p

e

jS:S\Tj for all S;T 2 Syl

p

(G) with S 6= T.

(A.1) Suppose that G is a nite group of order p

k

m where gcd(m;p) = 1.Suppose that n

p

(G) 6 1 (mod p

2

).Then,there are

S;T 2 Syl

p

(G) with jS\Tj = p

k1

.Moreover N = N

G

(S\T) is a subgroup,such that:

i) p

k

j jNj:

ii) p

k+1

jNj,hence jNj = p

k

l where l is a divisor of m greater than p.

iii) 2 jSyl

p

(N)j:

Proof.Theorem (A) implies the existence of T and S.Since jT:T\Sj = p we have by theorem (L) that T N.

Analogously we have that S N,hence jNj jTSj =

jTjjSj

jT\Sj

= p

k+1

.

(B) Let G be a nite group with n

p

= 1 for some p prime divisor of jGj.Then G is simple if and only if jGj = p.

(C) Let jGj = p

a

m where p is a prime with (p;m) = 1 and p > m.Then G is simple if and only if jGj = p.

(D) (Isaacs,5.27) Let jGj = p

a

q,where p and q are primes and a > 0.Then G is not simple.

(E) Let jGj = p

2

q

2

where q > p are primes.If jGj 6= 36,then G has normal Sylow qsubgroup.If G = 36 and G has not a

normal 3-Sylow subgroup,Z(G) has order 3.In particular G is not simple.

Proof.Note that n

q

2 f1;p

2

g since q > p.Suppose n

q

6= 1.Then q divides p +1,if q 6= 3 we get a contradiction.Assume

that jGj = 36.By hypothesis n

3

(G) = 4,thus by theorem (A.1) there is a subgroup H < G of size 3 with N

G

(H) of

size at least 27.Since N

G

(H) is a subgroup of G Lagrange's theorem implies that G = N

G

(H).By (NC) we have that

[G:C

G

(H)] 2.By (A) every group of order 18 has a unique group of order 9,in particular such group is characteristic.

If jC

G

(H)j = 18 every 3-Sylow subgroup of C

G

(H) would be normal in G contradicting that n

3

(G) > 1 thus G = C

G

(H) or

equivalently H Z(G).Note that we actually proved that G has no subgroups of index 2,and since every group of order

6 has a subgroup of index 2 we have that Z(G) 6= 6.Since H Z(G) we have that jZ(G)j 2 f3;9;12;18;36g.Note that

jZ(G)j can't be either of 9;18;36 otherwise n

3

(G) = 1.If jZ(G)j = 12 the group G=Z(G) is cyclic which is a contradiction

since G is not abelian,hence jZ(G)j = 3.

(E.1) If jGj = 36 and n

3

(G) > 1 then n

2

(G) = 1.

Proof.By (E) we know that Z(G) is a 3-group.Now,since all the 3-Sylow subgroups are conjugated Z(G) is inside of

every 3-Sylow subgroups,hence n

3

(G=Z(G)) = n

3

(G) = 4.On the other hand a group of order 12 having four 3-Sylow

subgroups has eight elements of order 3.Therefore the remaining four elements constitute a unique 2-Sylow subgroup,

thus n

2

(G=Z(G)) = 1.Let Q;P 2 Syl

2

(G),since PZ(G)=Z(G) = QZ(G)=Z(G) we have that PZ(G) = QZ(G).Trivially

QCQZ(G),thus Q is the unique 3-Sylow subgroup of QZ(G) = PZ(G) hence P = Q.

(F) If jGj = pqr where p;q and r are primes,then G is not simple.

Proof.We may assume that p < q < r.From this we see that n

r

2 f1;pqg,n

q

2 f1;r;prg and n

p

2 f1;q;r;qrg.If G

were simple we would have pqr = jGj pq(r 1) +r(q 1) +q(p 1) + 1 = pqr + (r 1)(q 1).Then we would have

0 (r 1)(q 1) which is clearly a contradiction.

(G) (Isaacs,6.12) Let jGj = 2m,where m is odd.Then G has a normal subgroup of order m.In particular,if m> 1,G is not

simple.

(H) Suppose that G is a nite simple group and H G.Suppose that jG:Hj = n and that jGj 6= 2.Then G is isomorphic to a

subgroup of A

n

.

Proof.This result can be seen as a corollary of the following more general result.

Theorem.Let G be a group,let H be a subgroup of G and let G act by left multiplication on the set A of left cosets of H in

G.Let

k

:G!S

A

,where S

A

denotes the groups of bijections of the set A,be the associated group homomorphism.Then:

i) G acts transitively on A.

ii) The stabilizer in G of the point 1 H 2 A is the subgroup H.

iii) The kernel of

k

is core

G

(H):=

T

x2G

xHx

1

.Furthermore core

G

(H) is the largest normal subgroup of G contained in

H.

1

The only non trivial part is iii) By denition of

k

ker

k

= fg 2 G j gxH = xH;8x 2 Gg

= fg 2 G j x

1

gxH = H;8x 2 Gg

= fg 2 G j x

1

gx 2 H;8x 2 Gg

= fg 2 G j g 2 xHx

1

;8x 2 Gg

=

\

x2G

xHx

1

The last part of the theorem follows from the fact that core

G

(H) is a normal subgroup of G contained in H.On the other

hand If N is any normal subgroup of G inside H,then N = xNx

1

xHx

1

for all x 2 G so that N core

G

(H).

(I) Let jGj = 2

e

p

n

where p is an odd prime,n is a non negative integer and 0 e 5.Then G is simple if and only if jGj is

prime.

Proof.Thanks to theorems (B) and (D)we may assume that 2 e 5 and 2 n.Suppose n

p

6= 1 then n

p

2 f4;8;16;32g.

It can't be 4,otherwise theorem (H) would imply that G injects in to A

4

which is a contradiction since G has order greater

than 36.Therefore we have the following cases:

n

p

= 8 then p = 7.

n

p

= 16 then p 2 f3;5g and e 2 4;5.

n

p

= 32 then p = 31.

Note that n

p

6 1 mod p

2

,thus by theorem (A.1) there is a subgroup N G with jNj = 2

k

p

n

and 2

k

> p.If p 6= 3 we must

have that 2

k

is at least 8 and consequently [G:N] 4.Therefore by (H) we may assume that p = 3 and n

3

= 16.Note that

[G:N] 2

e2

,since 4 3

n

jNj.Using (H) we may assume that e = 5 and in particular that G injects in to A

8

.Since

27 6j 8!we may assume that n = 2,in other words jGj = 288.We will deal with this order in one of the examples below.

(J) (Isaacs,6.17)Let n be a positive integer greater than 2.Then A

n

is a simple group for n 6= 4.

(K) Let G be a simple group of order 2

e

3 5.Then G

= A

5

.

Proof.We may assume that 2 e.Since G is simple n

2

2 f3;5;15g.By theorem (H) we see that n

2

6= 3.If n

2

= 5 again by

(H) G injects in A

5

.Since 60 jGj,G

= A

5

.If n

2

= 15 by theorem (A.1) there is a subgroup N < G of size at leats 2

e

3.

Since N can't be G,N is the normalizer of a group of size 2,N is a subgroup of G of index 5,whence arguing as above we

see that G

= A

5

.

(L) Suppose that G is a nite group and H G.Suppose that jG:Hj = p is the smallest prime divisor of jGj.Then,H E G.

Proof.Let G act on the right cosets of H.This gives a homomorphism from :G!S

p

.The kernel of this action is

core

G

(H) H.Thus,we have that G=core

G

(H)

= (G) S

p

.Thus,jG:core

G

(H)j divides p!= p(p 1)!.Since p is the

smallest prime dividing jGj,(p 1)!is relatively prime to jGj and hence jG:core

G

(H)j = p.Thus,jH:core

G

(H)j = 1 and

hence H = core

G

(H) E G.

(M) Suppose that p is prime and jGj = p(kp +1).If 1 k 2,G has a normal subgroup of order p or of order kp +1.

Proof.If n

p

(G) = 1 then G has a normal subgroup of order p.If n

p

(G) > 1 then n

p

(G) = kp +1.Since each subgroup of

order p has p 1 elements of order p,G has (kp +1)(p 1) elements of order p,so G has kp +1 elements not of order p.

Suppose that x is a nonidentity element not of order p.We will determine jC

G

(x)j.First,if P 2 Syl

p

(G) and y 2 P has

order p,then if x and y commute then P = hyi is normalized by x and hence x 2 N

G

(P).However,since n

p

(G) = kp +1,

N

G

(P) = P,a contradiction.Thus,p does not divide jC

G

(x)j.

The number of conjugates of x is

jGj

jC

G

(x)j

=

p(kp +1)

jC

G

(x)j

:

Since all conjugates of x do not have order 1 or p,and there are only kp elements in G that do not have order 1 or p,we

must have that

p(kp +1)

jC

G

(x)j

kp:

Thus,

kp +1

jC

G

(x)j

k:

Further,it cannot equal k unless k = 1.Hence,jC

G

(x)j = kp +1.Further,G has exactly kp +1 elements of order not equal

to p and C

G

(x) constitutes all of these.Thus,if H = C

G

(x),then H is a subgroup,and since H is the set of all elements

whose order doesn't divide p it is clearly characteristic.Hence,H E G.

(N) Let p be a prime number and let n = p or n = p +1.Let x 2 S

n

of order p.Then jC

G

(x)j = p and jN

G

(x)j = p(p 1).

Proof.For n = p the p-cycle can be uniquely specied by starting it with 1 and placing any arrangement of 2;:::;p following

it.There are (p1)!such arrangements and so S

n

has (p1)!elements of order p.Further,they are all conjugate and hence

if x 2 G has order p then jG:C

G

(x)j = (p 1)!so

jC

G

(x)j =

jGj

jG:C

G

(x)j

=

p!

(p 1)!

= p:

Now,a Sylow subgroup of S

p

contains (p 1) elements of order p and since there are (p 1)!elements,n

p

(S

p

) = (p 2)!.

Thus,if x 2 G has order p then

jN

G

(x)j =

jGj

n

p

(G)

=

p!

(p 2)!

= p(p 1):

For n = p + 1 the p-cycle can be uniquely specied by choosing the p elements it will contain,which can be done in

p+1

p

= p +1 ways,and arranging them beginning with the smallest.Once the smallest is placed,there are (p 1)!ways to

place the rest and hence S

p+1

contains (p +1)(p 1)!p-cycles.Again these elements are all conjugate.Similar arguments to

above show that

jC

G

(x)j =

jGj

jG:C

G

(x)j

=

(p +1)!

(p +1)(p 1)!

= p:

Also,n

p

(S

p+1

) = (p +1)(p 2)!and so

jN

G

(x)j =

jGj

n

p

(G)

=

(p +1)!

(p +1)(p 2)!

= p(p 1):

(NC) (Isaacs,)Let G be a group and H G a subgroup.Then N

G

(H)=C

G

(H) is isomorphic to a subgroup of AutH.

(O) (Isaacs,8.10)Let G be a nite group and N E G.If P 2 Syl

p

(N) then G = N

G

(P)N.

(P) Let p be an odd prime and let S be a nite group such that p j jSj.Let A be a subgroup of S of index 2 and P 2 Syl

p

(A).

Then 2jN

A

(P)j = jN

S

(P)j.In particular if x 2 S

n

of order p,notation as in (N),jN

A

n

(x)j = p(p 1)=2.Recall that n is

either p or p +1.

Proof.By (O) S = N

S

(P)A.Then

2jAj =

jN

S

(P)jjAj

jN

S

(P)\Aj

=

jN

S

(P)jjAj

jN

A

(P)j

:

(Q) Let F be a eld of size at least 4.Then PSL

2

(F) is a simple group.

(R) Let G be a nite group and p be an odd prime.Let P 2 Syl

p

(G) of order p,and x 2 N

G

(P) n C

G

(P).Then x normalizes at

most k = 1 +

n

p

(G)1

p

elements of Syl

p

(G).

Proof.Note that k is the number of orbits of the conjugation action of P on Syl

p

(G).If x normalizes more than k p-Sylow

subgroups,there exist Q and Q

1

two dierent p-Sylow subgroups,conjugated by P,such that x 2 N

G

(Q)\N

G

(Q

1

).Hence

there is a non trivial y 2 P such that Q

y

= Q

1

.Let z = [y;x],the commutator of y;x.Since x 2 N

G

(P) and y 2 P,z 2 P.

On the other hand

Q

z

= Q

yxy

1

x

1

= Q

xy

1

x

1

1

= Q

y

1

x

1

1

= Q

x

1

= Q

thus z 2 N

G

(P).If z 6= 1,z would generate P hence P N

G

(Q).Now this can't happen,otherwise P = Q and Q = Q

1

contrary to our assumption.Therefore z = 1,in other words x 2 C

G

(y) = C

G

(P) a contradiction.

(Ex) There are not simple group G of the following orders:

(Ex.1) 180 = 2

2

3

2

5

Proof.If there is such G we may assume that n

5

(G) 2 f6;36g and n

3

(G) = 10.If n

5

(G) = 6,by (H) A

6

would not be

simple.Since 36 4 +10 8 > 180 there are T;S 2 Syl

3

(G) such that jT\Sj = 3.We see,as in the proof of (A.1),that

N

G

(T\S) has index at most 5 in G which by (H) is a contradiction.

(Ex.2) 252 = 2

2

3

2

7

Proof.If there is such G we may assume that n

7

(G) = 36 and n

3

(G) 2 f7;28g.I claim that there is N subgroup of G of

order 36.If n

3

(G) = 7 take N as the normalizer of any 7-Sylow group.If n

3

(G) = 28 there must exist T;S 2 Syl

3

(G)

such that jT\Sj = 3.To see this note that 28 8+36 6 > 252.Taking N = N

G

(T\S) we see,as in the proof of (A.1),

that [G:N] 2 f1;2;4;7g.The rst three possibilities are excluded from (H) and the fact that N is a normalizer.Since

n

7

(G) = 36 there are 216 elements of order 7,which is the same to say that the set of elements of order dierent to

7 has size 36.Now since we found a group with this size,namely N,this group is precisely this set so it is obviously

characteristic in particular normal.

(Ex.3) 288 = 2

5

3

2

or 576 = 2

6

3

2

Proof.If there is a simple groups G with either of those orders we may assume that n

2

(G) = 9.I claim that any two

3-Sylow subgroups intersect trivially.Assume the opposite.Then there is Q G of order 3 with normalizer of size at

least 36

1

and therefore,by (NC),with centralizer of order at least 18.In particular there is an element x of order 6.

Now since n

2

(G) = 9 each 2-Sylow subgroup is its own normalizer.Therefore in the embedding of G in to A

9

given by

(H) x has no x points.Note that this is a contradiction,so the claim is proved,since 9 can't be written as a non zero

even multiple of 2 plus a multiple of 3.By (A.1) we may assume that n

3

6= 16,furthermore jGj = 576 and n

3

= 64.Also

from the claim we deduce that there are 576 64 8 = 64 elements of order not a power of 3,which clearly contradicts

that n

2

(G) = 9.

(Ex.4) 315 = 3

2

5 7

Proof.If there is such G we may assume that n

3

(G) = 7.By (A.1) there is a group Q of order 3 with normalizer N

of order either 45 or 315,since G is simple it can't be 315.Also by (A.1) N has at least two 3-Sylow subgroups,but

thanks to (A) this is impossible in a group of order 45.

(Ex.5) Counting Sylow subgroups for two primes at the same time is also a good method to prove non simplicity.

380 = 2

2

5 19

Proof.If there is such G we may assume that n

19

(G) = 20 and n

5

(G) = 76.Since 20 18 +4 76 > 380 we obtain

a contradiction.

495 = 3

2

5 11

Proof.If there is such G we may assume that n

11

(G) = 45 and n

5

(G) = 11.Since 45 10 +11 4 +1 = 495 there

is not enough room for elements of order 3 so we get a contradiction.

520 = 2

3

5 13

Proof.If there is such G we may assume that n

5

(G) = 26 and n

13

(G) = 40.Since 40 12 +4 26 = 584 we get a

contradiction.

552 = 2

3

3 23

Proof.If there is such G we may assume that n

23

(G) = 24 and n

3

(G) 46.Since 24 22 +46 2 > 552 we get a

contradiction.

616 = 2

3

7 11

Proof.If there is such G we may assume that n

11

(G) = 56 and n

7

(G) = 22.Since 56 10 +22 8 > 616 we get a

contradiction.

(Ex.6) 420 = 2

2

3 5 7

Proof.If there is such G we may assume that n

7

(G) = 15.By (NC) we deduce that there is x 2 G of order 14

that centralizes a 7-Sylow subgroup.Note that x

2

is a generator for that 7-Sylow subgroup,in particular that x only

normalizes that 7-Sylow subgroup.If G were simple A

15

would have an element x of order 14 that xes only one letter,

which is clearly impossible from the cycle decomposition.

(Ex.7) 540 = 2

2

3

3

5

Proof.If there is such G we may assume that n

3

(G) = 10 and n

5

(G) = 36.By (NC) there is an element x of order 15.

Since the normalizer of any 3-Sylow subgroup has size 54 x does not normalizes any of them.In other words A

10

would

have an element of order 15 that does not x any letter,which is clearly impossible.

(Ex.8) 612 = 2

2

3

2

17

Proof.If there is such G we may assume that n

3

(G) = 34 and n

17

(G) = 18.By (A.1) there is a subgroup Q of order 3

with normalizer of size at least 36.Hence,by (NC),Q has centralizer of size at least 18 and in particular there is x 2 G

of order 6.Note that jN

G

(P)j = 34 for P 2 Syl

17

(G).In particular x does not normalizes any 17-Sylow subgroup,

hence using (H) x can be seen as an element of order 6 in A

18

that does not x any letter.If x had any transposition

in its cycle decomposition,x

2

would be an element of order 3 xing at least 2 letters.In other words x

2

would be in the

normalizer of some 17-Sylow subgroup,which is impossible since these normalizers have order 34.Therefore the cycle

decomposition of x consists of a non zero even number of 6-cycles and 3-cycles,from which we see that it has to be two

6-cycles and two 3-cycles.Considering now x

3

,we have an element of order 2 xing exactly six letters.In other words

there are six 17-Sylow subgroups with intersection equal to x

3

.On the other hand,from (P) we see that jC

G

(P)j = 17

1

same argument as in the proof of (A.1)

for any P 2 Syl

17

(G).Therefore x

3

2 N

G

(P) n C

G

(P) for six dierent 17-Sylow subgroups.But this contradicts (R)

which says that x

3

normalizes at most two.

(Ex.9) 840 = 2

3

3 5 7

Proof.If there is such G we may assume that n

2

(G) 7,n

3

(G) 10,n

5

(G) 21 and n

7

(G) 2 f8;15;120g.If n

7

(G) = 8

we get a contradiction with (H) and (P) since 3 5 7 does not divide 7 3.If n

7

(G) = 120 we get a contradiction

since 840 (10 2 +21 4 +6 120) = 16,but the union of 7 dierent 2-Sylow subgroups has at least 7 8 6 4 = 32

elements.Suppose that n

7

(G) = 15 or equivalently jN

G

(P)j = 56 for P 2 Syl

7

(G).By (NC) we see that jC

G

(P)j 28,

in particular there is x 2 G of order 14.Since x

2

has order 7 and normalizes P,x

2

is a generator for P.In particular x

can't normalize any other 7-Sylow subgroup.Therefore,by (H),x would be an element of order 14 in A

15

xing only

one letter,which is clearly impossible from the cycle decomposition.

(Ex.10) 756 = 2

2

3

3

7

Proof.If there is such G we may assume that n

3

(G) 2 f4;7;28g and n

7

(G) = 36.By (H) we have that that n

3

(G) = 28.

Let P;Q 2 Syl

3

(G) dierent and let H = P\Q.Note that if jHj = 9 the order of N = N

G

(H) is at least jPQj = 81

and [G:N] is at most 7.The later is a contradiction thanks to (H) and that 756 - 7!.Therefore jHj = 3,and N has

order at least 36.In fact jNj = 36,otherwise [G:N] 7 which we already know is not possible.On the other hand

n

3

(N) > 1 otherwise N

G

(S) has order at least 4 27 for S 2 Syl

3

(N),and again we would have a subgroup of index

at most 7.By (E.1) n

2

(N) = 1,in particular if T 2 Syl

2

(N) we have that N

G

(T) has order at least 36 and in fact,

arguing as above,it has order 36.Since T is also in Syl

2

(G) we conclude that n

2

(G) = 21.Let X = fN

1

;:::;N

21

g be

the set of normalizers of the elements in Syl

2

(G).Let H

i

= Z(N

i

) then N

i

C

G

(H

i

),and jH

i

j = 3 thanks to (E).As

we argued before any subgroup of G of order 36 is maximal,thus N

i

= C

G

(H

i

) = N

G

(H

i

) and in particular H

i

6= H

j

for i 6= j.We conclude that the set fH

1

;:::;H

21

g = fZ(N

1

);:::;Z(N

21

)g is the set of 3-groups that arise as intersection

of two dierent 3-Sylow subgroups of G.

[Claim:Each H

i

is contained in exactly four 3-Sylow subgroups of G.Proof of the claim:Since H

i

is a 3-group and it

is central in N

i

we have that H

i

is contained in all four 3-Sylow subgroups of N

i

.On the other hand,as we discussed

above,every subgroup of order 9 is contained in exactly one 3-Sylow subgroup of G.Therefore each 3-Sylow subgroup

of N

i

is contained in a unique 3-Sylow subgroup of G,thus H

i

is contained in the four elements of Syl

3

(G) that contain

the four elements of Syl

3

(N

i

).Now suppose there is another R

i

2 Syl

3

(G),dierent from the four 3-Sylow subgroups

just discussed,such that H

i

R

i

.Now H

i

is contained in a subgroup P

i

R

i

of order 9,hence H

i

CP

i

.Therefore P

i

is a 3-Sylow subgroup of N

i

,in particular it is inside of one of the four 3-Sylow subgroups mentioned above.Since R

i

is dierent to all these four groups and any two 3-Sylow subgroups intersect in a group of order at most 3 we have a

contradiction.]

Let fP

i

g

1i28

:= Syl

3

(G) and S

i

= P

i

n f1g:By the claim above we have that

T

i2I

S

i

=;for all I f1 i 28g such

that jIj 5:Furthermore,there are exactly twenty one subsets I of size 4,one for each H

i

,such that

T

i2I

S

i

6=;.Let

fi

1

;::;i

4

g be one of such twenty one subsets.Then S

i

1

\:::\S

i

4

= H

i

n f1g;in particular

X

If1i28g;jIj=4

j

\

i2I

S

i

j = 21 2:

On the other hand each I of size 4 contains four subsets of size 3 and six subsets of size 2.If there is J of size 3 or 2

such that the intersection of its elements is not empty,then that J is contained in exactly one I,of size 4,such that the

intersection of its elements is not empty.Otherwise there would be at least ve S

i

that intersect non trivially.Hence

we have that

X

If1i28g;jIj=3

j

\

i2I

S

i

j = 4 21 2;

and

X

If1i28g;jIj=2

j

\

i2I

S

i

j = 6 21 2:

By the inclusion-exclusion formula we have that G contains 603 elements of order a power of 3.This,since

j

28

[

i=1

P

i

j = 1 +j

28

[

i=1

S

i

j = 1 +28 26 6 21 2 +4 21 2 21 2 = 603:

On the other hand there are 6 36 = 216 elements of order 7.Since 706 < 216 +603 there can't be simple group of

order 756.

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