Theorems you should know UWMadison
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 3Sylow 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 3Sylow 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 3group.Now,since all the 3Sylow subgroups are conjugated Z(G) is inside of
every 3Sylow subgroups,hence n
3
(G=Z(G)) = n
3
(G) = 4.On the other hand a group of order 12 having four 3Sylow
subgroups has eight elements of order 3.Therefore the remaining four elements constitute a unique 2Sylow 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 3Sylow 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 pcycle 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 pcycle 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)!pcycles.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 pSylow
subgroups,there exist Q and Q
1
two dierent pSylow 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 7Sylow 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
3Sylow 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 2Sylow 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 3Sylow 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 7Sylow subgroup.Note that x
2
is a generator for that 7Sylow subgroup,in particular that x only
normalizes that 7Sylow 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 3Sylow 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 17Sylow 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 17Sylow subgroup,which is impossible since these normalizers have order 34.Therefore the cycle
decomposition of x consists of a non zero even number of 6cycles and 3cycles,from which we see that it has to be two
6cycles and two 3cycles.Considering now x
3
,we have an element of order 2 xing exactly six letters.In other words
there are six 17Sylow 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 17Sylow 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 2Sylow 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 7Sylow 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 3groups that arise as intersection
of two dierent 3Sylow subgroups of G.
[Claim:Each H
i
is contained in exactly four 3Sylow subgroups of G.Proof of the claim:Since H
i
is a 3group and it
is central in N
i
we have that H
i
is contained in all four 3Sylow subgroups of N
i
.On the other hand,as we discussed
above,every subgroup of order 9 is contained in exactly one 3Sylow subgroup of G.Therefore each 3Sylow subgroup
of N
i
is contained in a unique 3Sylow 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 3Sylow 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 3Sylow subgroup of N
i
,in particular it is inside of one of the four 3Sylow subgroups mentioned above.Since R
i
is dierent to all these four groups and any two 3Sylow 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 inclusionexclusion 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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