SYLOW’S THEOREMS
In 1872 the Norwegian Ludwig Sylow proved very fundamental theorems
on p-subgroups of a finite group. His results are usually divided into three
theorems (or sometimes two), although this does not exactly correspond to
his three original theorems. In any case, what is nowadays usually presented
as Sylow I, the existence of Sylow p-subgroups, was part of his original
first theorem. The proofs we give here are different from Sylow’s original
proofs. These proofs use two actions of G, one by multiplication and one by
conjugation, and we divide them accordingly in two different sections.
The first Sylow theorem
Theorem 1 (Sylow I). Let |G| = pm r, where p is a prime and p - r. Then
G has a subgroup of order pm .
Proof. Let S denote the set of all subsets U of G with |U | = pm . The
m
number of such subsets is ppmr , which is not divisible by p (easy). For
U ∈ S and g ∈ G, the subset U g is also in S, and so G acts on S by
right-multiplication. Because p - |S|, this orbit has at least one orbit T with
p - |T|. Choose V ∈ T and let H be its stabilizer in the action. We claim
that |H| = pm .
In fact, the orbit-stabilizer theorem tells us that |T| · |H| = |G| = pm r,
and because p - |T| we deduce that pm dividesS|H|. However, because H is
the stabilizer of V we have V H = V , that is, v∈V vH = V . Thus, V is a
union of left cosets of H in G, and hence |V | = pm is a multiple of H. We
conclude that |H| = pm as desired.
By continuing the argument we obtain the following strengthening of the
above theorem (because np ≡ 1 (mod p) implies that np > 0). We will also
prove it again in the next section in a different way, and hence Theorem 2
and Lemma 3 may be omitted, in principle. However, it is interesting how
Theorem 2 can be proved using two completely different actions of G.
Theorem 2. Let |G| = pm r, where p is a prime and p - r. Let np be the
number of subgroups of G of order pm . Then np ≡ 1 (mod p).
Proof. In the above proof we have shown that if V ∈ S belongs to a G-orbit
of length not divisible by p then V is a left coset of a subgroup of G of order
pm . Conversely, if P is a subgroup of G of order pm then all left cosets of
P in G are clearly stabilized by P in the action (actually, the stabilizer of
any such coset equals P ), and so belong to G-orbits of length prime to p,
because of the orbit-stabilizer theorem and because p - |G : P | = r (actually
each such orbit will have length exactly r). Hence the set S0 of elements of
S whose orbit has length not a multiple of p coincides with the set of left
cosets of subgroups of G of order pm . Since the latter has cardinality r np
we have r np = |S0 | ≡ |S| (mod p). If are willing to use the information
m that |S| = ppmr ≡ r (mod p) (see Lemma 3 below) we obtain the desired
1
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SYLOW’S THEOREMS
conclusion that np ≡ 1 (mod p). But we can also avoid that and proceed as
follows.
We have shown that r np ≡ |S| (mod p). However, |S| does not depend
on the particular group of order pm r under consideration (in fact, it equals
|G|
, which depends only on the order of G). Hence the value of r np modulo
m
p
p does not depend on the particular group G but only on its order pm r, and
we may as well compute that for our favorite group of that order. Taking
as G the cyclic group of order pm r we find that r np ≡ r (mod p), because
np = 1 for that group. Therefore, it must be r np ≡ r (mod p) for any group
G of order pm r, and we conclude that np ≡ 1 (mod p).
In the above proof we have used the following lemma (but then also shown
how to avoid using it, if we prefer).
m Lemma 3. If p is a prime we have ppmr ≡ r (mod p) (whether p divides r
or not).
Proof. Recall that (a+b)p = ap +bp in any
commutative ring of characteristic
p, because the binomial coefficients pi are divisible by p for 0 < i < p.
m
m
m
More generally, (a + b)p = ap + bp follows inductively. In particular,
this holds in Fp [x], the polynomial ring over the field Fp of p elements (thus,
Fp = Z/pZ). Consequently, we have
m
p r m r X
X
p r i
r pm j
m
m
x
x = (1 + x)p r = (1 + xp )r =
i
j
i=0
j=0
in Fp [x] (or in Z[x] provided we replace the middle equality with a congruence
m m modulo p). We conclude that p i r ≡ 0 (mod p) if pm - i, and that ppm rj ≡
r
j (mod p), of which the desired conclusion is a special case.
Corollary 4. (Cauchy’s Lemma) If a prime p divides the order of a finite
group G, then G contains an element of order p.
Proof. If |G| = pm r with p - r then, according to Theorem 1, G has a
subgroup P of order pm . If g is any element of P different from 1, then
its order is a power of m by Lagrange, and hence a suitable power of g has
order p.
Remark. Note that the case p = 2 of Cauchy’s Lemma admits a much easier
proof: in a group of even order, pair together each element with its inverse
and conclude that there is at least one element g, besides the identity element
e, such that g −1 = g, that is, g 2 = e.
A consequence of Cauchy’s Lemma is that if a p-group (which means a
group where each element has order some power of the prime p) is finite,
then its order is a power of p. (In fact, if it were not then its order would
be divisible by some other prime q as well, and hence according to Cauchy’s
Lemma the group would have some element of order q, a contradiction because q is not a power of p.) Combined with Lagrange’s Theorem this shows
that a finite p-group is the same thing as a group of order a power of p.
SYLOW’S THEOREMS
3
The other Sylow theorems
If |G| = pm r with p - r, a subgroup of G of order pm is called a Sylow
p-subgroup of G. This is the most frequent definition in finite group theory,
as it correctly extends to the definition of a Hall π-subgroup, where π is
a set of primes. However, a different definition is in use, which is more
flexible because differently from the one just given it also applies to infinite
groups: a Sylow p-subgroup of any group is a p-subgroup which is not
contained in a larger p-subgroup. Clearly a subgroup of G of order pm
has the property of not being contained in a larger p-subgroup, because
of Lagrange’s theorem, hence any subgroup satisfying the former definition
also satisfies the latter, but the converse is not clear yet. However, for
finite groups the two definitions will be seen below to be equivalent. Note,
however, that using the former definition Sylow I could be simply stated as
for p a prime, every finite group has at least one Sylow p-subgroup, but this
would be incorrect using the latter definition, as the existence of p-subgroups
of a finite group not contained in a larger psubgroup is automatic.
We start with a lemma concerning the normalizer of a subgroup of order
pm . Recall that the normalizer NG (H) of a subgroup H of G is defined as
NG (H) := {g ∈ G : H g = H}. This is the stabilizer of the subgroup H in
the action of G by conjugation on the set of its subsets (that is, on P(G)).
Note that this is different from the centralizer of HTin G, which is defined
as CG (H) := {g ∈ G : hg = h for all h ∈ H} = h∈H CG (h). Certainly
CG (H) ≤ NG (H), but NG (H) always contains H while CG (H) does exactly
when H is abelian. Also note that H E NG (H), in fact NG (H) is the largest
subgroup of G of which H is a normal subgroup.
Lemma 5. Let |G| = pm r, where p is a prime and p - r. Let P be a subgroup
of G of order pm and let H be any p-subgroup of G contained in NG (P ).
Then H ≤ P .
Proof. (Recall from the previous section that H, as any finite p-group, has
order a power of p.) Since H is a subgroup of NG (P ) and P is a normal
subgroup of NG (P ) it follows that HP is also a subgroup, and HP/P ∼
=
H/(H ∩ P ) by the second isomorphism theorem. Hence HP is isomorphic
to a quotient group of H, and so has order a power of p. However, |HP | =
|HP : P | · |P | and |P | = pm is the largest power of p which divides |G|.
Hence |HP : P | = 1 and so H ≤ P .
Theorem 6 (The rest of the Sylow theorems). Let |G| = pm r, where p is a
prime and p - r.
(1) Any two subgroups of G of order pm are conjugate.
(2) The number of subgroups of G of order pm is a divisor of the index
in G of any of them (that is, of r) and is congruent to 1 modulo p.
(3) Any p-subgroup of G is contained in one of order pm .
Proof. Let Π be the set of subgroups of G of order pm (that is, according to
a useful notation, Π = Sylp (G)). According to the first theorem of Sylow Π
is not empty. Let G act on Π by conjugation and let Σ be one of the orbits
of this action. (We want to show that Σ = Π.) If P ∈ Π we can restrict the
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SYLOW’S THEOREMS
action of G on Σ to P and decompose Σ into P -orbits. Each one of these
has cardinality length (cardinality) a power of p because it divides |P | = pm .
Suppose that Σ contains a P -orbit {P 0 } of length one. Then P ≤ NG (P 0 )
and so P ≤ P 0 by Lemma 5. But then P 0 = P because they have the same
order. Hence if P ∈ Σ then {P }, which is a P -orbit of length one, is the only
one with this property. Since all other P -orbits in Σ have length multiple of
p we have |Σ| ≡ 1 (mod p). Similarly, if P 6∈ Σ then all P -orbits in Σ have
length multiple of p, and so |Σ| ≡ 0 (mod p). Since P ∈ Π was arbitrary
|Σ| should satisfy both conclusion, which is a contradiction unless there is
only one orbit and Σ = Π. This proves (1), and also that |Π| ≡ 1 (mod p),
which is part of (2). The other part of (2), that |Π| divides |G : P |, also
follows because |Π| = |G : NG (P )| and P ≤ NG (P ).
Now let H be a p-subgroup of G, hence of order a power of p, and restrict
the action of G on Π to H. Since each H-orbits has length a power of p and
|Π| ≡ 1 (mod p), there exists at least one orbit {P } of length one. Then
H ≤ NG (P ), and so H ≤ P , again by Lemma 5.
Remark. A generalization of statement (2) of Theorem 6 due to Georg Frobenius, 1895, says that the number of subgroups of G order pk is congruent to
1 modulo p, for any 0 ≤ k ≤ m. Note that, unlike Sylow’s theorems, this
statement is nontrivial even for G a p-group.