Central Sylow Subgroups and Normality Applications

The next application strengthens normality into centrality. A Sylow subgroup may be normal, but sometimes conjugation forces every element of the group to centralize it. In this lesson, we focus in the central Sylow subgroups and normality applications. The central idea is to study the conjugation homomorphism from

G

into the automorphism group of a normal Sylow subgroup. If the image has order forced to be

1

, then the Sylow subgroup lies in the centre.

Let

G

be a finite group and let

P

be a Sylow

p

-subgroup of

G

. Then

P

is called a \textbf{central Sylow subgroup} if

P\leq Z(G)

G

into the automorphism group of a normal Sylow subgroup. If the image has order forced to be

1

, then the Sylow subgroup lies in the centre.

Let

G

be a finite group and let

P

be a Sylow

p

-subgroup of

G

. Then

P

is called a \textbf{central Sylow subgroup} if

P\leq Z(G)

Let

P

be a normal cyclic subgroup of prime order

p

in a finite group

G

. If

\gcd(|G|,p-1)=1

, then

P\leq Z(G)

Given that

P\trianglelefteq G

|P|=p

P

is cyclic, and

\gcd(|G|,p-1)=1

. To prove that

P\leq Z(G)

. Conjugation defines a homomorphism

\begin{equation*} \varphi:G\to \operatorname{Aut}(P) \end{equation*}

\begin{equation*} \varphi(g)(x)=gxg^{-1}. \end{equation*}

Since

P

is cyclic of prime order

p

, we have

\begin{equation*} |\operatorname{Aut}(P)|=p-1. \end{equation*}

The image

\operatorname{Im}\varphi

has order dividing both

|G|

and

p-1

. Since

\gcd(|G|,p-1)=1

, we get

\begin{equation*} |\operatorname{Im}\varphi|=1. \end{equation*}

Thus conjugation by every element of

G

fixes every element of

P

. Therefore every element of

P

commutes with every element of

G

. Hence

P\leq Z(G)

\square

This preview displays the conjugation action that decides centrality. First, Sylow counting checks whether the Sylow

p

-subgroup is unique and normal. Then the automorphism-group test compares

|G|

with

p-1

. If the Sylow subgroup has prime order and

\gcd(|G|,p-1)=1

, the conjugation image is trivial and the subgroup lies in the center.

Centrality is stronger than normality. Normality says conjugation preserves the Sylow subgroup as a set, while centrality says conjugation fixes every element of that subgroup.

This calculator checks the two conditions used in the theorem. It first determines whether Sylow counting forces a unique Sylow

p

-subgroup. It then checks whether that Sylow subgroup has prime order and whether

\gcd(|G|,p-1)=1

. When all conditions hold, the calculator reports that the Sylow subgroup lies in the center.

The calculator mirrors the proof: unique Sylow gives normality, and the gcd condition kills the conjugation action. A normal Sylow subgroup may fail to be central when the action into its automorphism group is nontrivial.

Let

G

be a group of order

385

. Show that a Sylow

7

-subgroup of

G

lies in

Z(G)

Let

|G|=385=5\cdot 7\cdot 11

, and let

P

be a Sylow

7

-subgroup of

G

. Then

\begin{align*} n_7 &\equiv 1 \pmod 7, \\ n_7 &\mid 55. \end{align*}

The divisors of

55

are

1,5,11,55

. Among these, only

1

is congruent to

1

modulo

7

. Hence

\begin{equation*} n_7=1. \end{equation*}

Therefore

P\trianglelefteq G

. Since

|P|=7

P

is cyclic and

\begin{equation*} |\operatorname{Aut}(P)|=6. \end{equation*}

Conjugation gives a homomorphism

\begin{equation*} G\to \operatorname{Aut}(P). \end{equation*}

The image has order dividing both

385

and

6

. Since

\begin{equation*} \gcd(385,6)=1, \end{equation*}

the image is trivial. Thus every element of

G

centralizes

P

. Hence

P\leq Z(G)

\square

Let

G

be a group of order

5\cdot 11\cdot 19

. Show that a Sylow

19

-subgroup lies in the centre of

G

and that a Sylow

11

-subgroup is normal.

Let

P

be a Sylow

19

-subgroup of

G

. Then

\begin{align*} n_{19} &\equiv 1 \pmod {19}, \\ n_{19} &\mid 55. \end{align*}

The divisors of

55

are

1,5,11,55

, and only

1

is congruent to

1

modulo

19

. Hence

n_{19}=1

, so

P\trianglelefteq G

. Since

|P|=19

, we have

\begin{equation*} |\operatorname{Aut}(P)|=18. \end{equation*}

The conjugation image has order dividing both

5\cdot 11\cdot 19

and

18

. Their greatest common divisor is

1

. Therefore the conjugation image is trivial, and

\begin{equation*} P\leq Z(G). \end{equation*}

Now let

Q

be a Sylow

11

-subgroup. Then

\begin{align*} n_{11} &\equiv 1 \pmod {11}, \\ n_{11} &\mid 95. \end{align*}

The divisors of

95

are

1,5,19,95

, and only

1

is congruent to

1

modulo

11

. Thus

n_{11}=1

. Hence

Q\trianglelefteq G

\square

Let

G

be a group of order

100

. Suppose that

G

has a unique Sylow

2

-subgroup. Show that

G

is abelian.

Let

|G|=100=2^2\cdot 5^2

. Let

P

be the unique Sylow

2

-subgroup. Then

P\trianglelefteq G

and

|P|=4

. Every group of order

4

is abelian. Let

Q

be a Sylow

5

-subgroup. Then

\begin{align*} n_5 &\equiv 1 \pmod 5, \\ n_5 &\mid 4. \end{align*}

Therefore

n_5=1

, so

Q\trianglelefteq G

. Also

|Q|=25

, and every group of order

25

is abelian. Since

P

and

Q

are normal subgroups of relatively prime orders,

P\cap Q=\{e\}

, and elements of

P

commute with elements of

Q

. Moreover,

\begin{equation*} |PQ|=|P||Q|=4\cdot 25=100. \end{equation*}

Thus

G=PQ

. Since

P

and

Q

are abelian and commute elementwise with each other,

G

is abelian. Hence

G

is abelian.

\square

Let

G

be a group of order

p^m n

, where

p

is prime,

m\geq 1

, and

p>n

. If

|G|

is not prime, then

G

is not simple.

Given that

|G|=p^m n

, where

p

is prime,

m\geq 1

p>n

, and

|G|

is not prime. To prove that

G

is not simple. Let

P

be a Sylow

p

-subgroup of

G

. Then

\begin{equation*} [G:P]=n. \end{equation*}

Let

G

act on the left cosets of

P

by left multiplication. This gives a homomorphism

\begin{equation*} \varphi:G\to S_n. \end{equation*}

Since

p>n

, the prime

p

does not divide

n!

. Hence the image of

\varphi

cannot contain the full

p

-part of

|G|

. By the first isomorphism theorem,

\begin{equation*} |G|=|\ker\varphi|\,|\operatorname{Im}\varphi|. \end{equation*}

Therefore

\ker\varphi

is nontrivial. If

G

were simple, then

\ker\varphi=G

. But then the action would be trivial, forcing every coset to be fixed, which is impossible when the action on more than one coset is the left multiplication action on cosets. Thus

G

has a nontrivial proper normal subgroup. Hence

G

is not simple.

\square

1. Let

G

be a group of order

3\cdot 11\cdot 19

. Show that a Sylow

11

-subgroup lies in

Z(G)

. 2. Let

G

be a group of order

5\cdot 11\cdot 19

. Show that a Sylow

11

-subgroup is normal. 3. Explain why the automorphism group of a cyclic group of prime order

p

has order

p-1

1. The Sylow

11

-subgroup is unique because

n_{11}\equiv 1 \pmod {11}

and

n_{11}\mid 57

. The conjugation image has order dividing both

3\cdot 11\cdot 19

and

10

, so it is trivial. 2. Since

n_{11}\equiv 1 \pmod {11}

and

n_{11}\mid 95

, we get

n_{11}=1

. 3. A generator of

C_p

can be sent to any of the

p-1

nonidentity elements, and each such choice gives an automorphism.

If conjugation by every element fixes every element of

P

, then every element of

P

commutes with every element of

G

An automorphism is determined by the image of a generator, and there are

p-1

possible generator images.

No. Centrality requires additional restrictions on the conjugation action.

BMLabs Mathematics

BMLabs Mathematics

Central Sylow Subgroups and Normality Applications

Central Sylow Subgroups and Normality Applications

CENTRAL SYLOW SUBGROUP

EXPLORE CENTRAL SYLOW ACTIONS

CENTRAL SYLOW ACTION EXPLORER

INTERPRET THE ACTION

TEST CENTRAL SYLOW CRITERIA

CENTRAL SYLOW TEST CALCULATOR

CONNECT THE CENTRALITY TEST

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