21.

Sylow Theorems and applications

In general the problem of classifying groups of every order is com-
pletely intractable. Given any group G, the first thing to do to under-
stand G is to look for subgroups H. In particular if H is normal in G,
then one can take the quotient G/H and one can think of G as being
built up from the two smaller groups H and G/H.
In turn, one can then consider H, or G/H, and try to break it up
into pieces.

Definition 21.1. Let G be a group.

We say that G is simple if it contains no proper normal subgroups.

In the sense outlined above, we can think of simple groups as be-

ing the building blocks of creating an arbitrary group. (In fact even
this approach is ridiculously optimistic; even the problem of finding all
groups with a normal subgroup isomorphic to a cyclic group of order
a power of a prime, whose quotient is cyclic of order the same prime,
seems too hard to solve.) Thus we would like to classify all finite simple
groups.
Turning this problem onto its head, we would like to find ways of
producing normal subgroups of a group G.
If one thinks about Lagrange’s Theorem, and its implications, two
things are obvious.
First of all, the key part of the proof of Lagrange’s Theorem, is to
use the decomposition of G into the left cosets of H in G and to prove
that each coset has the same size (namely the cardinality of H).
Secondly, in terms of applications, the problem of classifying sub-
groups of a group G turns into a problem of counting and considering
the divisors of the order of the group.
As the problem of finding normal subgroups is so much harder than
the problem of finding subgroups, the plan is to pick a prime p dividing
the order of G and look for normal subgroups of order a power of p.

Definition 21.2. Let G be a finite group of order n = pk m, where p

is prime and p does not divide m.
A subgroup H of order pk is called a Sylow p-subgroup of G.

Theorem 21.3. Let G be a finite group of order n = pk m, where p is

prime and p does not divide m.
(1) The number of subgroups is conqruent to 1 modulo p and divides
n.
(2) Any two Sylow p-subgroups are conjugate.
1
 With the Sylow Theorem in hand, let us prove one of the basic facts
about simple groups.
Proposition 21.4. Let G be a simple group of order less than sixty.
Then the order of G is prime.
To prove (21.4), it clearly suffices to assume that we have a simple
group G of composite order n, less than sixty and derive a contradic-
tion.
First an easy, but useful Lemma.
Lemma 21.5. Let G be a group of finite order and let p be a prime
dividing the order of G.
(1) G has at least one Sylow p-subgroup P .
(2) If P is the only Sylow p-subgroup then P is normal in G (in
fact, characteristically normal).
Proof. (1) follows from (1) of (21.3), as zero is not congruent to 1.
Suppose that P is the unique Sylow p subgroup of G. Let g ∈ G and
let Q = gP g −1 . Then Q is a subgroup of G, of the same order as P .
Thus Q is another Sylow p-subgroup of G. By uniqueness Q = P and
so P is normal in G. 
To give a flavour of the method of attack, and to illustrate the
strength of (21.3), suppose first that n = 15. Let p = 5.
We count the number of Sylow 5-subgroups of G. Suppose there are
x. What do we know about x? Well x is supposed to congruent to one
modulo 5. Thus
x = 1, 6, 11, 16 . . .
On the other hand x is supposed to divide 15. Since x does not
divide 5, x must divide 3. But then x = 1 and there is one Sylow
5-subgroup, which is automatically normal in G. Thus G has a normal
subgroup of order 5 and index 3.
Proposition 21.6. Let G be a group of order pq the product of two
primes, where p < q.
Then G has a normal subgroup of order q. In particular G is not
simple.
Proof. Let x be the number of Sylow q-subgroups. Then x is congruent
to 1 modulo q. In particular x does not divide q. As x divides pq, it
must divide p. If x > 1 then x ≥ q + 1 > q > p. Thus x = 1 and there
is exactly one subgroup P of order q. But then P is normal in G. 
We will also need the following Proposition, whose proof we omit.
2
Proposition 21.7. Let G be a group of order a power of a prime.
Then the centre of G has order greater than one. In particular if G
is simple then its order is prime.
Now consider the numbers from 1 to 60. Eliminating those that are
prime, a power of a prime or the product of two primes, leaves the
following cases. n = 12, 18, 20, 24, 28, 30, 34, 36, 40, 42, 45, 48, 50,
52, 54, 56, 58.
We do some illustrative cases; the rest are left as an exercise for the
reader.
Pick n = 30 = 2 · 3 · 5. Let p = 5. How many Sylow 5-groups are
there? Suppose that there are x. Then x is congruent to 1 modulo 5.
In this case,
x = 1, 6, . . . .
On the other hand, x must divide 30, so that x = 1 or x = 6. If
G is simple, then x 6= 1 and so x = 6. Let H and K be two Sylow
5-subgroups. Then |H| = |K| = 5. On the other hand H ∩ K is a
subgroup of H and so by Lagrange, |H ∩ K| = 1. Since there are 6
Sylow 5-subgroups and each such group contains 4 elements of order 5
that are not contained in any other subgroup, it follows that there are
24 elements of order 5.
Let y be the number of Sylow 3-subgroups. Then y is congruent to
1 modulo 3, so that
y = 1, 4, 7, 10 . . .
As y divides 30 and y 6= 1, it follows that y = 10. As before there must
therefore be 20 elements of order 3. But 24 + 20 > 30, impossible.
Let us deal with one of the most tricky cases. Suppose that n =
48 = 24 · 3. We count the number of Sylow 2-subgroups. Anyone of
these must have order 16. Suppose that there are x such groups. Then
x is congruent to one modulo two. The possibilities are then
x = 1, 3, 5, . . . .
On the other hand x is supposed to divide 48, so that the only
possibilities are 1 and 3. If G is simple, then there must be 3 subgroups
of order sixteen. Let S be the set of Sylow 2-subgroups. Define a
homomorphism
φ : G −→ A(S) ' S3
by sending g ∈ G to the permutation σ = φ(g),
σ : S −→ S,
−1
where σ(H) = gHg . It is not hard, as in the proof of Cayley’s
Theorem, to prove that φ is a homomorphism. As G has order 48
3
and A(S) has order six, φ cannot be injective. Thus the kernel is a
non-trivial normal subgroup.
In fact all finite simple groups have been classified. Finite simple
groups come into two classes. There are those that belong to an infinite
series of well-understood examples. There are 15 of these series, two of
which are the cyclic groups of prime order and the alternating groups
An , n ≥ 5. Then there are the sporadic groups. There are 26 sporadic
groups.
The monster group is the largest sporadic group; it has order
808, 017, 424, 794, 512, 875, 886, 459, 904, 961, 710, 757, 005, 754, 368, 000, 000, 000.
The prime factorisation of this number is
246 · 320 · 59 · 76 · 112 · 133 · 17 · 19 · 23 · 29 · 31 · 41 · 47 · 59 · 71.
It is a subgoup of the group of linear symmetries of a real vector
space of dimension 196883 = 47 · 59 · 71,
R196883 ,
in other words it is a subgroup of
GL(196883, R),
the group of 196883 × 196883 invertible matrices.
All finite groups appear as subgroups of a finite permutation group.
The smallest n such that the monster is a subgroup of Sn is
24 · 37 · 53 · 74 · 11 · 132 · 29 · 41 · 59 · 71.