Normalizers and Counting Conjugate Subgroups

Conjugate subgroups lead naturally to the question of which elements leave a subgroup fixed under conjugation. The answer is the normalizer. The normalizer collects exactly those elements under which a subgroup is invariant, and it is the largest subgroup in which the given subgroup becomes normal. This is the subgroup-level analogue of the centralizer, and it plays the same counting role: centralizers count conjugates of elements, while normalizers count conjugates of subgroups.

Let

G

be a group and let

H,K

be subgroups of

G

. The \textbf{normalizer of

H

K

} is the set

\begin{equation*} N_K(H)=\{k \in K \mid kHk^{-1}=H\}. \end{equation*}

Let

G

be a group and let

H,K

be subgroups of

G

. The \textbf{normalizer of

H

K

} is the set

\begin{equation*} N_K(H)=\{k \in K \mid kHk^{-1}=H\}. \end{equation*}

Let

G

be a group and let

H,K

be subgroups of

G

. Then

\begin{equation*} N_K(H)=N_G(H)\cap K. \end{equation*}

Given that

G

is a group and

H,K

are subgroups of

G

. To prove that

N_K(H)=N_G(H)\cap K

. Let

x \in N_K(H)

. Then

x \in K

and

xHx^{-1}=H

. Therefore

x \in N_G(H)

and

x \in K

. Hence

x \in N_G(H)\cap K

. Thus

\begin{equation*} N_K(H)\subseteq N_G(H)\cap K. \end{equation*}

Conversely, let

x \in N_G(H)\cap K

. Then

x \in K

and

xHx^{-1}=H

. Therefore

x \in N_K(H)

. Thus

\begin{equation*} N_G(H)\cap K\subseteq N_K(H). \end{equation*}

Hence

N_K(H)=N_G(H)\cap K

\square

Let

G

be a group and let

H,K

be subgroups of

G

. Then

N_K(H)

is a subgroup of

K

Given that

G

is a group and

H,K

are subgroups of

G

. To prove that

N_K(H)

is a subgroup of

K

. Since

e \in K

and

eHe^{-1}=H

, we have

e \in N_K(H)

. Therefore

N_K(H)

is non-empty. Let

k_1,k_2 \in N_K(H)

. Then

\begin{equation*} k_1Hk_1^{-1}=H \end{equation*}

and

\begin{equation*} k_2Hk_2^{-1}=H. \end{equation*}

From

k_2Hk_2^{-1}=H

, multiplying on the left by

k_2^{-1}

and on the right by

k_2

gives

\begin{equation*} k_2^{-1}Hk_2=H. \end{equation*}

Therefore

\begin{align*} (k_1k_2^{-1})H(k_1k_2^{-1})^{-1} &= k_1(k_2^{-1}Hk_2)k_1^{-1} \\ &= k_1Hk_1^{-1} \\ &= H. \end{align*}

Also

k_1k_2^{-1}\in K

because

K

is a subgroup of

G

. Hence

k_1k_2^{-1}\in N_K(H)

. Hence

N_K(H)

is a subgroup of

K

\square

Let

G

be a group and let

H

be a subgroup of

G

. The \textbf{normalizer of

H

G

} is

\begin{equation*} N_G(H)=\{g \in G \mid gHg^{-1}=H\}. \end{equation*}

When the ambient group is understood, it may be denoted by

N(H)

Let

G

be a group and let

H

be a subgroup of

G

. Then

N_G(H)

is the largest subgroup of

G

in which

H

is normal.

Given that

G

is a group and

H

is a subgroup of

G

. To prove that

N_G(H)

is the largest subgroup of

G

in which

H

is normal. By definition,

\begin{equation*} N_G(H)=\{g \in G \mid gHg^{-1}=H\}. \end{equation*}

Therefore

gHg^{-1}=H

for every

g \in N_G(H)

. Hence

H

is normal in

N_G(H)

. Let

L

be a subgroup of

G

such that

H

is normal in

L

. Then

lHl^{-1}=H

for every

l \in L

. Therefore

l \in N_G(H)

for every

l \in L

. Thus

L\subseteq N_G(H)

. Hence

N_G(H)

is the largest subgroup of

G

in which

H

is normal.

\square

Let

G

be a group and let

H

be a subgroup of

G

. If

H

is normal in

G

, then

N_G(H)=G

. If

G

is abelian, then

N_G(H)=G

Given that

G

is a group and

H

is a subgroup of

G

. To prove that the two stated conclusions hold. [1] Suppose that

H

is normal in

G

. Then

gHg^{-1}=H

for every

g \in G

. Therefore every

g \in G

belongs to

N_G(H)

. Hence

N_G(H)=G

. [2] Suppose that

G

is abelian. For

g \in G

and

h \in H

\begin{align*} ghg^{-1} &= hgg^{-1} \\ &= h. \end{align*}

Therefore

gHg^{-1}=H

for every

g \in G

. Hence

N_G(H)=G

\square

Let

G

be a group and let

H,K

be finite subgroups of

G

. Then the number of distinct conjugates of

H

induced by elements of

K

\begin{equation*} [K:N_K(H)]. \end{equation*}

Given that

G

is a group and

H,K

are finite subgroups of

G

. To prove that the number of distinct conjugates of

H

induced by elements of

K

[K:N_K(H)]

. Let

\begin{equation*} \mathcal{T}=\{kHk^{-1}\mid k \in K\} \end{equation*}

be the set of distinct conjugates of

H

induced by

K

. Let

\begin{equation*} \mathcal{S}=\{kN_K(H)\mid k \in K\} \end{equation*}

be the set of left cosets of

N_K(H)

K

. Define

f:\mathcal{S}\to\mathcal{T}

\begin{equation*} f(kN_K(H))=kHk^{-1}. \end{equation*}

To prove that

f

is well-defined, let

k_1N_K(H)=k_2N_K(H)

. Then

k_2^{-1}k_1\in N_K(H)

. Therefore

\begin{equation*} (k_2^{-1}k_1)H(k_2^{-1}k_1)^{-1}=H. \end{equation*}

Multiplying on the left by

k_2

and on the right by

k_2^{-1}

gives

k_1Hk_1^{-1}=k_2Hk_2^{-1}

. Therefore

f(k_1N_K(H))=f(k_2N_K(H))

. To prove that

f

is one-one, suppose that

f(k_1N_K(H))=f(k_2N_K(H))

. Then

k_1Hk_1^{-1}=k_2Hk_2^{-1}

. Therefore

\begin{equation*} (k_2^{-1}k_1)H(k_2^{-1}k_1)^{-1}=H. \end{equation*}

Thus

k_2^{-1}k_1\in N_K(H)

, so

k_1N_K(H)=k_2N_K(H)

. By definition of

\mathcal{T}

, every element of

\mathcal{T}

has the form

kHk^{-1}

for some

k \in K

. Therefore

f

is onto. Therefore

f

is a bijection. Hence

|\mathcal{T}|=|\mathcal{S}|=[K:N_K(H)]

\square

Let

G

be a group and let

H,K

be finite subgroups of

G

. If

H

is invariant under exactly

n

elements of

K

, then the number of conjugates of

H

induced by

K

\begin{equation*} \frac{|K|}{n}. \end{equation*}

Given that

G

is a group,

H,K

are finite subgroups of

G

, and

H

is invariant under exactly

n

elements of

K

. To prove that the number of conjugates of

H

induced by

K

\frac{|K|}{n}

. The elements of

K

under which

H

is invariant are precisely the elements of

N_K(H)

. Therefore

|N_K(H)|=n

. By the theorem on counting conjugate subgroups, the number of conjugates induced by

K

\begin{equation*} [K:N_K(H)]. \end{equation*}

Since

K

is finite,

\begin{equation*} [K:N_K(H)]=\frac{|K|}{|N_K(H)|}=\frac{|K|}{n}. \end{equation*}

Hence the number of conjugates of

H

induced by

K

\frac{|K|}{n}

\square

Let

G=S_3

and let

H=\{e,(12)\}

. The conjugates of

H

are

\begin{equation*} \{e,(12)\},\quad \{e,(13)\},\quad \{e,(23)\}. \end{equation*}

Thus there are three conjugates. Since

|S_3|=6

, the counting formula gives

\begin{equation*} 3=[S_3:N_{S_3}(H)]. \end{equation*}

Therefore

|N_{S_3}(H)|=2

. Since

H\subseteq N_{S_3}(H)

and

|H|=2

, we get

N_{S_3}(H)=H

This preview shows the subgroup version of the orbit-stabilizer idea. The orbit consists of all conjugates of

H

, and the stabilizer is the normalizer

N_K(H)

, the elements of

K

that leave

H

unchanged. Choose a model and compare the size of

K

, the size of the normalizer, and the number of distinct conjugates. Start with the

S_3

transposition subgroup, then compare the normal subgroup

A_3

where every element normalizes it.

The normalizer is the stabilizer of the subgroup under conjugation. A larger normalizer means more elements fix the subgroup, so fewer distinct conjugates appear. If the normalizer is the whole group, the subgroup has only one conjugate and is normal.

This calculator checks the formula for counting conjugate subgroups induced by a finite subgroup

K

. Enter

|K|

and

|N_K(H)|

. The calculator validates divisibility and then computes

[K:N_K(H)]

. Use the preset from the exercise,

|K|=24

and

|N_K(H)|=6

, and compare it with the

S_3

example.

The index counts how many different positions the subgroup can occupy under conjugation by elements of

K

. This is the subgroup analogue of counting conjugates of an element by using the centralizer index.

Let

G

be a group, let

H

be a subgroup of

G

, and let

a \in G

. Prove that

\begin{equation*} N_G(aHa^{-1})=aN_G(H)a^{-1}. \end{equation*}

Let

G

be a group, let

H

be a subgroup of

G

, and let

a \in G

. Let

x \in N_G(aHa^{-1})

. Then

\begin{equation*} x(aHa^{-1})x^{-1}=aHa^{-1}. \end{equation*}

Multiplying on the left by

a^{-1}

and on the right by

a

gives

\begin{equation*} (a^{-1}xa)H(a^{-1}x^{-1}a)=H. \end{equation*}

Since

a^{-1}x^{-1}a=(a^{-1}xa)^{-1}

, we have

a^{-1}xa \in N_G(H)

. Therefore

x \in aN_G(H)a^{-1}

. Thus

\begin{equation*} N_G(aHa^{-1})\subseteq aN_G(H)a^{-1}. \end{equation*}

Conversely, let

x \in aN_G(H)a^{-1}

. Then

x=aya^{-1}

for some

y \in N_G(H)

. Since

yHy^{-1}=H

\begin{align*} x(aHa^{-1})x^{-1} &= (aya^{-1})(aHa^{-1})(ay^{-1}a^{-1}) \\ &= ayHy^{-1}a^{-1} \\ &= aHa^{-1}. \end{align*}

Therefore

x \in N_G(aHa^{-1})

. Thus

\begin{equation*} aN_G(H)a^{-1}\subseteq N_G(aHa^{-1}). \end{equation*}

Hence

N_G(aHa^{-1})=aN_G(H)a^{-1}

\square

[1] Let

H

be a subgroup of

G

. Prove that

H\subseteq N_G(H)

. [2] Let

H

be a subgroup of

G

. Prove that

H

is normal in

G

if and only if

N_G(H)=G

. [3] Let

H,K

be finite subgroups of

G

and suppose

|K|=24

and

|N_K(H)|=6

. How many conjugates of

H

are induced by

K

[1] If

h \in H

, then

hHh^{-1}=H

because

H

is a subgroup and is closed under conjugation by its own elements. Hence

h \in N_G(H)

, so

H\subseteq N_G(H)

. [2] If

H

is normal in

G

, then

gHg^{-1}=H

for every

g \in G

, so

N_G(H)=G

. Conversely, if

N_G(H)=G

, then every

g \in G

satisfies

gHg^{-1}=H

, so

H

is normal in

G

. [3] The number is

[K:N_K(H)]=24/6=4

A centralizer fixes an element by commuting with it, while a normalizer fixes a subgroup as a set under conjugation.

It is the largest subgroup in which the given subgroup is normal, and it counts conjugate subgroups through its index.

It counts the distinct subgroups of the form

kHk^{-1}

with

k \in K

BMLabs Mathematics

BMLabs Mathematics

Normalizers and Counting Conjugate Subgroups

Normalizers and Counting Conjugate Subgroups

NORMALIZER OF A SUBGROUP IN ANOTHER SUBGROUP

NORMALIZER OF A SUBGROUP

COUNTING CONJUGATE SUBGROUPS

EXPLORE NORMALIZER ORBIT COUNTING

NORMALIZER ORBIT COUNTING EXPLORER

INTERPRET THE ORBIT COUNT

COMPUTE A NORMALIZER INDEX

NORMALIZER INDEX CALCULATOR

CONNECT THE CALCULATION

FREQUENTLY ASKED QUESTIONS

SECTION COMPLETE

Coordinates and Locus

Invariants under Orthogonal Transformation