§13 Sylow Theorems

§13.1 Counting

We have already seen that the orbit-stabilizer theorem can answer some nontrivial questions. For example: how large is the symmetry group of a tetrahedron?

Recall that the theorem says that for any group $G$ acting on a set $X$ , and any element $x\in X$ , there is a bijection

G/G_x\cong\mathcal{O}_x.

In particular, if $G$ is finite, then

|\mathcal{O}_x| = |G|/|G_x|.

Counting theorems of this kind are extremely useful in mathematics. They are like a “layup” in basketball—the easiest way to score. Once you reduce a difficult problem to a counting problem, you have made progress.

In the proof of Lagrange's theorem, we used the fact that any set is the union of its orbits.

Thus, given an action of a group $G$ on a finite set $X$ , we obtain

|X| = \sum_{\text{orbits}} |\mathcal{O}_x|.

Let us make further use of this observation.

The equation above is called the counting formula.

§13.2 $p$ -Groups

Definition 13.1 Let $p$ be a prime number. A finite group $G$ is called a $p$ -group if its order is a power of $p$ , i.e.

|G|=p^n

for some integer $n\geqslant1$ .

Definition 13.2 Let $G$ act on a set $X$ . If an element $x\in X$ satisfies

gx=x

for every $g\in G$ , then $x$ is called a fixed point of the group action.

Proposition 13.1 Fix a $p$ -group $G$ , and let $X$ be a finite set whose cardinality is not divisible by $p$ . Then every action of $G$ on $X$ has at least one fixed point.

Example 13.1 If someone claims to have a $p$ -group acting on a tetrahedron, you can consider the induced action of $G$ on the set of vertices of the tetrahedron.

If $p$ is not $2$ , then this group action must fix at least one vertex.

Proof: By the orbit-stabilizer theorem, the order of every orbit $\mathcal{O}_x$ divides the order of the group $G$ .

Therefore,

|\mathcal{O}_x| = p^k

for some $k\geqslant0$ .

We need to show that for some $x\in X$ ,

|\mathcal{O}_x| = p^0 = 1,

which gives a fixed point.

Such an $x$ must exist.

Otherwise, every orbit $\mathcal{O}_x$ would have size $p^k$ with $k\geqslant1$ , and hence every orbit size would be divisible by $p$ .

Then the right-hand side of the counting formula

|X| = \sum_{\text{orbits}}\mathcal{O}_x

would be divisible by $p$ .

But by assumption, $|X|$ is not divisible by $p$ .

Therefore, some $\mathcal{O}_x$ must have size $1$ .

~\tag*{$\square$}

Here is another application.

Proposition 13.2 Let $G$ be a $p$ -group. Then the center of $G$ is nontrivial; that is, it contains more than just the identity element.

In what follows, we write

Z

for the center of $G$ .

Proof: Consider the conjugation action of $G$ on itself.

The orbits of this action are precisely the conjugacy classes of $G$ .

Thus the counting formula becomes

|G| = \sum_{\text{conjugacy classes}} |[x]|,

where $[x]$ denotes the conjugacy class of $x$ , i.e. the set of all elements of the form

gxg^{-1}, \qquad g\in G.

We have

|[x]|=1

if and only if $x$ lies in the center of $G$ .

Indeed, if the only element of $\mathcal{O}_x$ is $x$ itself, then

gxg^{-1}=x

for all $g\in G$ , which is equivalent to

gx=xg.

Finally, we know that

1_G\in G

always lies in the center.

Therefore, the counting formula becomes

|G| = 1+ \sum_{\text{conjugacy classes}\neq[1_G]} |[x]|.

|[x]|\geqslant2

for every $x\neq1_G$ , then the right-hand side would not be divisible by $p$ , since it would have the form

1+ \sum_{\text{various }k\geqslant1} p^k.

This contradicts the fact that $|G|$ is divisible by $p$ .

Therefore, there must exist some

x\neq1_G

such that

|[x]|=1.

That is, the center must contain some element

x\neq1_G.

~\tag*{$\square$}

This has a very nice consequence.

Corollary 13.3 Every group of order

p^2

is Abelian.

This is a highly nontrivial consequence.

For example, imagine trying to prove directly that every group of order $49$ must be Abelian.

We already know that every group of order $p$ is Abelian because it must be cyclic.

This is the next power.

Proof: The center $Z$ of $G$ is a subgroup, so by Lagrange's theorem we must have

|Z|=1,\quad p,\quad\text{or}\quad p^2,

since these are the only divisors of $p^2$ .

On the other hand, the proposition tells us that

|Z|\neq1.

Therefore,

|Z|=p

|Z|=p^2.

Suppose

|Z|=p.

We derive a contradiction.

Fix

x\in G, \qquad x\notin Z.

Consider the stabilizer of $x$ under the conjugation action of $G$ .

This is the centralizer of $x$ , which we previously denoted by

Z(x).

It is the set of all $y\in G$ such that

xy=yx.

Since the stabilizer of any group action is a subgroup, Lagrange's theorem tells us that

|Z(x)|

must divide $p^2$ .

On the other hand,

Z\subset Z(x),

because every element of the center commutes with $x$ by definition.

Also,

x\in Z(x),

because $x$ commutes with itself.

This shows

|Z|<|Z(x)|.

Therefore, $|Z(x)|$ must be a divisor of $p^2$ strictly larger than $p$ .

Hence

|Z(x)|=p^2.

But this means that every element of $G$ commutes with $x$ .

Therefore, $x$ must lie in the center.

~\tag*{$\square$}

Thus this “counting” strategy has already been very successful.

Let us push it as far as possible.

One beautiful outcome of all this effort is the Sylow theorems.

§13.3 The First Sylow Theorem

Suppose $p$ divides $|G|$ .

Write

|G| = p^e\cdot m,

where $p^e$ is the highest power of $p$ dividing $|G|$ .

In particular,

\operatorname{gcd}(m,p)=1.

Definition 13.3 A Sylow $p$ -subgroup, or a $p$ -Sylow subgroup, is a subgroup

H\subset G

such that

|H|=p^e.

In other words, it is a subgroup of maximal possible $p$ -power order.

Thus, if several different primes $p$ divide $|G|$ , we can try to find a Sylow $p$ -subgroup for each such prime.

At this point, we do not yet know whether such subgroups exist or how many there can be in $G$ .

Example 13.2 Let

G=S_3.

Since

6=3\cdot2,

a Sylow $3$ -subgroup is a subgroup of order $3$ .

There is only one such subgroup:

H = \{\mathrm{id},(123),(132)\}.

There are three Sylow $2$ -subgroups:

\{\mathrm{id},(12)\}, \qquad \{\mathrm{id},(13)\}, \qquad \{\mathrm{id},(23)\}.

Theorem 13.4 (First Sylow Theorem)

Let $p$ divide $|G|$ .

Then there exists a Sylow $p$ -subgroup.

Before proving the theorem, let us introduce two lemmas that will be used in the proof.

Lemma 13.5

p\nmid|\mathcal{P}_{p^e}(G)|.

Proof: The number of subsets of size $p^e$ is not divisible by $p$ .

Recall that

\begin{pmatrix} a\\ b \end{pmatrix}

is the number of ways to choose $b$ unordered elements from a set of size $a$ .

Therefore,

\begin{pmatrix} a\\ b \end{pmatrix} = \frac{a!}{b!(a-b)!}.

Hence

\begin{pmatrix} p^em\\ p^e \end{pmatrix} = \frac{ (p^em)(p^em-1)\cdots(p^em-p^e+1) }{ p^e(p^e-1)\cdots1 }.

Notice that if $p$ divides a numerator term

p^em-k,

then it also divides

p^e-k

in the denominator, with the same multiplicity.

Why?

k=p^il, \qquad p\nmid l,

then

\begin{aligned} p^em-k &= p^i(p^{e-i}m-l),\\ p^e-k &= p^i(p^{e-i}-l). \end{aligned}

Notice that

i<e,

otherwise

p^e-k

would be negative.

Thus

\frac{p^em-k}{p^e-k}

is not divisible by $p$ .

~\tag*{$\square$}

Remark We have argued that

\begin{pmatrix} p^em\\ p^e \end{pmatrix} = \frac{ p^em(p^em-1)\cdots(p^em-p^e+1) }{ p^e(p^e-1)\cdots1 }

is not divisible by $p$ .

This reduces to the fact that

p^i\mid p^em-k \quad\Rightarrow\quad p^i\mid p^e-k.

Why is

i<e?

Because otherwise,

k=p^{e+a}l \quad\Rightarrow\quad p^e-k = p^e(1-p^al) \leqslant0 \quad \text{if }a\geqslant0.

More concretely, by the definition of the binomial coefficient, $k$ runs from $0$ to $p^e-1$ , so

k<p^e.

Proposition 13.6 Let

U\subset G

be a subset.

Then the order of the stabilizer $H$ of $U$ divides $|U|$ .

Proof: If $U$ is fixed by $H$ , then

U = \bigcup_{u\in U}Hu,

and $U$ is partitioned into cosets.

Thus

U = \bigsqcup Hu.

Therefore,

|U| = |H|+\cdots+|H|.

~\tag*{$\square$}

Let us now try to prove the First Sylow Theorem.

What is our strategy?

Counting.

A better question is:

what should we count?

Proof: Let $G$ act on

\mathcal{P}_{p^e}(G)

by left multiplication:

S\mapsto gS.

By the lemma,

p\nmid|\mathcal{P}_{p^e}(G)|.

Therefore, there must exist an orbit

\mathcal{O}_U

whose size is not divisible by $p$ , since

|\mathcal{P}_{p^e}(G)| = \sum_{\text{orbits}} |\mathcal{O}_U|.

Let

U\in\mathcal{O}_U.

By the orbit-stabilizer theorem,

|G|/|G_U| = |\mathcal{O}_U|.

By the other lemma,

U = \bigcup_{\text{cosets}}G_U.

Thus,

|G_U|

divides

|U|=p^e.

Therefore,

p^em = |G| = |G_U| \cdot |\mathcal{O}_U|,

where $|G_U|$ is a power of $p$ and $|\mathcal{O}_U|$ is not divisible by $p$ .

Hence

|G_U| = p^e.

~\tag*{$\square$}

Corollary 13.7 If $p$ divides $|G|$ , then there exists an element

x\in G

of order $p$ .

You may not have considered this consequence before.

By Lagrange's theorem, we know that the order of every element $x\in G$ must divide $|G|$ .

But if a prime $p$ divides $|G|$ , is it obvious that there must exist a specific element of order $p$ ?

Proof: Since $p$ divides $G$ , we have

|H|\geqslant2.

Therefore, we can choose

x\in H

with

x\neq1_G.

By Lagrange's theorem, the order of $x$ divides $|H|$ .

Hence

x^{p^k} = 1_G

for some

k\geqslant1.

Let

y=x^{p^{k-1}}.

Then

y^p = 1_G.

~\tag*{$\square$}

Example 13.3 Let

G = S_7\times\mathbb{Z}/14\mathbb{Z}.

Then

\begin{aligned} |G| &= 7!\times14\\ &= 7^2\times5\times3^2\times2^5. \end{aligned}

The First Sylow Theorem guarantees that this group contains:

a subgroup of order $49$ ;
a subgroup of order $5$ ;
a subgroup of order $9$ ;
a subgroup of order $32$ .

Example 13.4 Can you find a subgroup of order $16$ in $S_7$ ?

Proof:

Find the appropriate prime power.

The order of $S_7$ is

7!=5040,

and

16=2^4.

We want to find a subgroup of order $16$ in $S_7$ , which would be a Sylow $2$ -subgroup.

Look for elements of order $2$ .

In $S_7$ , elements of order $2$ are transpositions, i.e. $2$ -cycles, which exchange two elements and fix the others.

For example,

(12)\in S_7

has order $2$ .

A product of disjoint transpositions, such as

(12)(34),

also has order $2$ .

In general, a product of $k$ disjoint transpositions has order

2^k.

Construct a subgroup of order $16$ .

To obtain a subgroup of order $16$ , we need a group generated by four disjoint transpositions, since

2^4=16.

Consider the following disjoint transpositions in $S_7$ :

(12), \qquad (34), \qquad (56), \qquad (78).

These four disjoint transpositions generate a subgroup of order $16$ .

This subgroup is isomorphic to

\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2,

because each transposition generates a cyclic group of order $2$ , and the transpositions are independent, i.e. they commute with one another.

Therefore, a subgroup of order $16$ in $S_7$ can be generated by the disjoint transpositions

\{(12),(34),(56),(78)\},

and is isomorphic to

\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2.

~\tag*{$\square$}

The First Sylow Theorem tells us that Sylow $p$ -subgroups exist.

Given a group $G$ with

|G|=p^em, \qquad p\nmid m,

|H|=p^e,

then

H\subset G

is a Sylow $p$ -subgroup.

§13.4 The Second Sylow Theorem

Theorem 13.8 (Second Sylow Theorem)

Fix a finite group $G$ and a prime $p$ such that $p$ divides $|G|$ .

(1) Any two Sylow $p$ -subgroups are conjugate.

(2) For any $p$ -subgroup

H\subset G,

there exists a Sylow $p$ -subgroup containing $H$ .

Proof: Let

K\subset G

be a subgroup.

Let $K$ act on

G/H,

the set of left cosets ${gH}$ of $H$ in $G$ , by

gH \mapsto (kg)H.

If $K$ is a $p$ -group and $H$ is a Sylow $p$ -subgroup, then

|K|=p^l

and

|G/H| = |G|/|H| = p^em/p^e = m.

Since

p\nmid m,

the action of $K$ on $G/H$ has a fixed point.

Thus there exists some $g$ such that

k\cdot gH = gH, \qquad \forall k\in K.

That is, for every $k\in K$ and every $h\in H$ , there exists $h^{\prime}\in H$ such that

k\cdot g\cdot h = g\cdot h^{\prime}.

Therefore,

k = gh^{\prime}h^{-1}g^{-1}.

Hence,

k\in gHg^{-1}.

Therefore,

K \subset gHg^{-1}.

~\tag*{$\square$}

Notice that if there is only one Sylow $p$ -subgroup, then it must be normal.

Why?

For every

g\in G,

the map

\begin{aligned} C_g:G&\to G\\ x&\mapsto gxg^{-1} \end{aligned}

is a group isomorphism.

Therefore, it sends a subgroup of order $k$ to another subgroup of order $k$ .

If there is only one such subgroup $H$ , then $C_g$ must map $H$ to itself for every $g\in G$ .

Thus

gHg^{-1} = H, \qquad \forall g\in G.

Therefore, $H$ is normal.

The Second Sylow Theorem tells us that the converse is also true.

Corollary 13.9 If a Sylow $p$ -subgroup

H

is normal in $G$ , then $H$ is the unique Sylow $p$ -subgroup of $G$ .

§13.5 Normalizers

Before stating the Third Sylow Theorem, we need another important concept: the normalizer.

Definition 13.4 Let

K\subset G

be a subgroup.

Then

N(K) = \{g\in G\mid gKg^{-1}=K\}

is called the normalizer of $K$ .

Proposition 13.10

(1)

N(K)

is a subgroup of $G$ .

(2)

K\triangleleft N(K),

i.e. $K$ is normal in its normalizer.

(3)

N(K)

is the stabilizer of $K$ under the conjugation action of $G$ on

\mathcal{P}_{|K|}(G),

where $\mathcal{P}_{|K|}(G)$ denotes the set of all subgroups of $G$ of order $|K|$ .

§13.6 The Third Sylow Theorem

Before stating the Third Sylow Theorem, let us briefly review a useful fact.

We know that semidirect products can be identified with split short exact sequences:

H\cong L\rtimes R

if and only if there exists a split short exact sequence

Question: When can we identify such a group with a direct product?

Proposition 13.11 The following statements are equivalent:

(1)

H\cong L\times R.

(2)

H\cong L\rtimes_{\phi}R,

where

\phi:R\to\mathrm{Aut}(L)

is the trivial map.

(3) There exists a split short exact sequence such that

j(R)\triangleleft H.

Proof: We prove the chain

(1)\Rightarrow(3)\Rightarrow(2)\Rightarrow(1).

$(1)\Rightarrow(3)$ :

H\cong L\times R,

we can define the inclusion map

\begin{aligned} i:L&\to H\\ l&\mapsto(l,1_R) \end{aligned}

and the projection map

\begin{aligned} p:H&\to R\\ (l,r)&\mapsto r. \end{aligned}

Both maps are group homomorphisms.

Thus we obtain a split short exact sequence

where

\begin{aligned} j:R&\to H ~\text{is the other ``inclusion'' homomorphism,}\\ r&\mapsto(1_L,r). \end{aligned}

Moreover,

j(R)\triangleleft H.

Why?

\begin{aligned} (l^{\prime},r^{\prime}) (1_L,r) (l^{\prime},r^{\prime})^{-1} &= (l^{\prime},r^{\prime}) (1_L,r) (l^{\prime-1},r^{\prime-1})\\ &= (l^{\prime}1_Ll^{\prime-1}, r^{\prime}rr^{\prime-1})\\ &= (1_L,r^{\prime}rr^{\prime-1}) \in j(R). \end{aligned}

$(3)\Rightarrow(2)$ :

Let

r\in j(R), \qquad l\in i(L).

Then

rlr^{-1}l^{-1} \in i(L)\cap j(R) = \{1_H\}.

Therefore,

\phi(r) = \mathrm{id}_L.

$(2)\Rightarrow(1)$ :

\phi(r) = \mathrm{id}_L, \qquad \forall r,

then

\begin{aligned} (l_1,r_1)\cdot(l_2,r_2) &= (l_1\cdot\phi_{r_1}(l_2),r_1r_2)\\ &= (l_1l_2,r_1r_2). \end{aligned}

This is exactly the multiplication on

L\times R.

~\tag*{$\square$}

We now state the Third Sylow Theorem, which will greatly help us determine the structure of groups.

Theorem 13.12 (Third Sylow Theorem)

Let $G$ be a finite group and let $p$ be a prime dividing $|G|$ .

Write

\mathrm{Syl}_p(G)

for the set of Sylow $p$ -subgroups of $G$ .

Then:

(1)

|\mathrm{Syl}_p(G)|

divides $m$ .

(2)

|\mathrm{Syl}_p(G)| \equiv 1 \pmod p.

Remark Part (1) says that $G$ acts on

\mathrm{Syl}_p(G)

by conjugation.

By the Second Sylow Theorem, this action has only one orbit.

Thus

|\mathrm{Syl}_p(G)| = |G|/\text{stabilizer }G_H,

where $H$ is a Sylow $p$ -subgroup.

Since

H\subset G_H,

we have

|G_H| = p^e\cdot m^{\prime}.

Therefore,

|\mathrm{Syl}_p(G)| = p^em/p^e\cdot m^{\prime} = m/m^{\prime}.

For part (2), we will prove that for any

H\in\mathrm{Syl}_p(G),

$H$ is the unique fixed point of the conjugation action of $H$ on

\mathrm{Syl}_p(G).

Hence

|\mathrm{Syl}_p(G)| = 1+\sum\mathcal{O},

where each $\mathcal{O}$ has size divisible by $p$ .

Proof: Let $H$ be a Sylow $p$ -subgroup.

The group $G$ acts on

\mathrm{Syl}_p(G)

by conjugation:

K \mapsto gKg^{-1}.

By the orbit-stabilizer theorem,

|G|/|G_H| = |\mathcal{O}_H| = |\mathrm{Syl}_p(G)|.

The second equality follows from the Second Sylow Theorem.

Since

H\subset G_H,

we have

p^e \mid |G_H|.

Therefore,

p^em/p^em^{\prime} = |\mathrm{Syl}_p(G)| \quad\Rightarrow\quad m = |\mathrm{Syl}_p(G)| \cdot m^{\prime}.

Since $G$ acts on $\mathrm{Syl}_p(G)$ , $H$ is a fixed point.

If $K$ is another fixed point, then

H\subset N(K)

and

K\subset N(K).

By the Second Sylow Theorem, $H$ is conjugate to $K$ inside $N(K)$ .

But

K\triangleleft N(K),

H=K.

~\tag*{$\square$}

Here is an application.

Proposition 13.13 If

|G|=15,

then $G$ is cyclic.

Proof: Our strategy is to determine the structure of

\mathrm{Syl}_5(G)

and

\mathrm{Syl}_3(G).

For $p=5$ , we have

|G| = p^em = 5^1\cdot3.

By the Third Sylow Theorem:

(a)

|\mathrm{Syl}_5(G)|

divides $3$ .

(b)

|\mathrm{Syl}_5(G)| \equiv 1 \pmod5.

Condition (a) tells us that

|\mathrm{Syl}_5(G)| = 1 \quad\text{or}\quad 3.

Condition (b) forces

|\mathrm{Syl}_5(G)| = 1,

because $1$ is the only value less than $5$ satisfying

\equiv1\pmod5.

Therefore, $G$ has a unique subgroup

H_5

of order $5$ , and

H_5\triangleleft G.

Similarly:

(c)

|\mathrm{Syl}_3(G)|

divides $5$ .

(d)

|\mathrm{Syl}_3(G)| \equiv 1 \pmod3.

From (c), we obtain

|\mathrm{Syl}_3(G)| = 1.

Therefore, $G$ contains a unique subgroup

H_3

of order $3$ , and

H_3\triangleleft G.

Since

\gcd(|H_3|,|H_5|) = \gcd(3,5) = 1,

we have

H_3\cap H_5 = \{1_G\}.

Therefore, we can construct the short exact sequence

Since both $H_3$ and $H_5$ are normal subgroups of $G$ , we obtain

G \cong H_3\times H_5 \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \cong \mathbb{Z}/15\mathbb{Z}.

~\tag*{$\square$}

Proposition 13.14 Let

|G|=p\cdot P,

where $p\neq P$ are primes and $P$ is the larger prime.

Then

G \cong \mathbb{Z}/P\mathbb{Z} \rtimes \mathbb{Z}/p\mathbb{Z}.

Proof: By the Third Sylow Theorem:

(a)

|\mathrm{Syl}_P(G)|

divides $p$ .

(b)

|\mathrm{Syl}_P(G)| \equiv 1 \pmod P.

From (a),

|\mathrm{Syl}_P(G)| = 1 \quad\text{or}\quad p<P.

From (b),

|\mathrm{Syl}_P(G)| = 1,

since the only number less than $P$ satisfying

\equiv1\pmod P

is $1$ .

Therefore,

\exists!\, H_P\subset G

of order $P$ .

Hence,

H_P\triangleleft G.

Therefore, we can consider the short exact sequence

1 \to H_P \to G \to G/H_P \to 1.

Since

|G/H_P| = p,

we have

G/H_P \cong \mathbb{Z}/p\mathbb{Z}.

Since

|H_P|=P,

we have

H_P \cong \mathbb{Z}/P\mathbb{Z}.

Thus we obtain the short exact sequence

1 \to \mathbb{Z}/P\mathbb{Z} \to G \to \mathbb{Z}/p\mathbb{Z} \to 1.

By the First Sylow Theorem, there exists a splitting map $j$ because $p$ and $P$ are relatively prime.

~\tag*{$\square$}

Contents

§13 Sylow Theorems

§13.1 Counting

§13.2 ppp-Groups

§13.3 The First Sylow Theorem

§13.4 The Second Sylow Theorem

§13.5 Normalizers

§13.6 The Third Sylow Theorem

§13.2 $p$ -Groups