Normalisers of Subgroups

Normality depends on the ambient group, so it is natural to ask for the largest part of

G

in which a given subgroup behaves normally. This leads to the normaliser. The normaliser collects exactly those elements of

G

whose conjugation leaves the subgroup fixed. It turns a failed normality condition into a successful one by shrinking the ambient group from

G

N(H)

. This concept is used frequently in group actions, Sylow theory, conjugacy arguments, and the study of internal symmetry.

Let

(G,\circ)

be a group and let

H

be a non-empty subset of

G

. The \textbf{normaliser} of

H

G

is the subset

\begin{equation*} N_G(H)=\{x\in G:xHx^{-1}=H\}. \end{equation*}

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

N(H)

Normality depends on the ambient group, so it is natural to ask for the largest part of

G

in which a given subgroup behaves normally. This leads to the normaliser. The normaliser collects exactly those elements of

G

whose conjugation leaves the subgroup fixed. It turns a failed normality condition into a successful one by shrinking the ambient group from

G

N(H)

. This concept is used frequently in group actions, Sylow theory, conjugacy arguments, and the study of internal symmetry.

Let

(G,\circ)

be a group and let

H

be a non-empty subset of

G

. The \textbf{normaliser} of

H

G

is the subset

\begin{equation*} N_G(H)=\{x\in G:xHx^{-1}=H\}. \end{equation*}

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

N(H)

The normaliser is not defined by asking whether each element of

H

commutes with

x

. It asks whether conjugation by

x

sends the whole subset

H

back to itself. If

H

is a subgroup, then

N_G(H)

is the largest subgroup of

G

in which

H

becomes normal. Thus normalisers measure how much normality survives when the whole group is too large.

Let

(G,\circ)

be a group and let

H

be a non-empty subset of

G

. Then

N_G(H)

is a subgroup of

G

Given that

(G,\circ)

is a group and

H

is a non-empty subset of

G

. To prove that

N_G(H)

is a subgroup of

G

. Since

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

we get

e\in N_G(H)

. Therefore

N_G(H)

is non-empty. Let

x,y\in N_G(H)

. Then

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

and

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

Since

yHy^{-1}=H

, we also get

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

Now

\begin{align*} (x\circ y^{-1})H(x\circ y^{-1})^{-1} &=x\circ y^{-1}Hy\circ x^{-1}\\ &=xHx^{-1}\\ &=H. \end{align*}

Therefore

\begin{equation*} x\circ y^{-1}\in N_G(H). \end{equation*}

By the one-step subgroup criterion,

N_G(H)

is a subgroup of

G

. Hence,

N_G(H)\leq G

\square

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

H

is normal in

G

, then

\begin{equation*} N_G(H)=G. \end{equation*}

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and

H

is normal in

G

. To prove that

N_G(H)=G

. Since

H

is normal in

G

\begin{equation*} xHx^{-1}=H \quad \forall x\in G. \end{equation*}

Therefore every

x\in G

belongs to

N_G(H)

. Thus

G\subseteq N_G(H)

. Since

N_G(H)\subseteq G

by definition, we get

\begin{equation*} N_G(H)=G. \end{equation*}

Hence,

N_G(H)=G

\square

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Then

H

is normal in

N_G(H)

Given that

(G,\circ)

is a group and

H

is a subgroup of

G

. To prove that

H

is normal in

N_G(H)

. Let

x\in N_G(H)

. By the definition of the normaliser,

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

Therefore

\begin{equation*} xHx^{-1}\subseteq H \quad \forall x\in N_G(H). \end{equation*}

Hence,

H

is normal in

N_G(H)

\square

Let

(G,\circ)

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,\circ)

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 the previous theorem,

\begin{equation*} H\trianglelefteq N_G(H). \end{equation*}

Let

L

be a subgroup of

G

such that

H\trianglelefteq L

. To prove that

L\subseteq N_G(H)

. Let

x\in L

. Since

H

is normal in

L

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

Therefore

x\in N_G(H)

. Thus

\begin{equation*} L\subseteq N_G(H). \end{equation*}

Hence,

N_G(H)

is the largest subgroup of

G

in which

H

is normal.

\square

Let

G=S_3

and let

\begin{equation*} H=\langle(12)\rangle=\{e,(12)\}. \end{equation*}

The elements that normalise

H

must conjugate

(12)

to an element of

H

. In

S_3

, conjugation sends transpositions to transpositions, and only elements preserving the transposition

(12)

keep

H

fixed. Thus

\begin{equation*} N_{S_3}(H)=H. \end{equation*}

H

is normal in its normaliser, although

H

is not normal in all of

S_3

\square

Let

G=S_3

and let

H=A_3

. Since

A_3\trianglelefteq S_3

, the normaliser theorem gives

\begin{equation*} N_{S_3}(A_3)=S_3. \end{equation*}

This example should be compared with the previous one. The normaliser of a normal subgroup is the whole group, while the normaliser of a non-normal subgroup can be strictly smaller.

\square

Let

H\leq G

. Prove that

H\trianglelefteq G

if and only if

N_G(H)=G

Let

H\leq G

. If

H\trianglelefteq G

, then

xHx^{-1}=H

for every

x\in G

. Therefore

N_G(H)=G

. Conversely, if

N_G(H)=G

, then every

x\in G

satisfies

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

Therefore

H\trianglelefteq G

by the conjugation equality criterion.

We observe the chain

H\leq N_G(H)\leq G

. Choose one of the standard

S_3

examples and compare how large the normaliser becomes. For

A_3

, the normaliser is all of

S_3

because

A_3

is normal. For the subgroup generated by

(12)

, the normaliser is only the subgroup itself, so normality is recovered only after shrinking the ambient group.

We use orders to check the subgroup chain

H\leq N_G(H)\leq G

. Enter the orders of

G

N_G(H)

, and

H

. The calculator verifies the divisibility conditions and computes the indices. When

|N_G(H)|=|G|

, the normaliser criterion says that

H

is normal in

G

1. Define the normaliser of a non-empty subset

H

G

. 2. Prove that

N_G(H)

is a subgroup of

G

. 3. Prove that if

H\trianglelefteq G

, then

N_G(H)=G

. 4. Prove that

H\trianglelefteq N_G(H)

when

H

is a subgroup of

G

. 5. Find

N_{S_3}(A_3)

N_G(H)=\{x\in G:xHx^{-1}=H\}

. 2. The identity belongs to

N_G(H)

, and if

x,y\in N_G(H)

, then

(xy^{-1})H(xy^{-1})^{-1}=H

. 3. If

H

is normal in

G

, then every element of

G

fixes

H

under conjugation, so every element of

G

lies in

N_G(H)

. 4. If

x\in N_G(H)

, then

xHx^{-1}=H

, which is the normality condition inside

N_G(H)

. 5. Since

A_3\trianglelefteq S_3

N_{S_3}(A_3)=S_3

It measures the largest part of

G

that keeps

H

fixed under conjugation.

Yes. The normaliser of any non-empty subset of a group is a subgroup.

For a subgroup

H

, this happens exactly when

H

is normal in

G

Every element of

N_G(H)

is defined to conjugate

H

onto itself.

BMLabs Mathematics

BMLabs Mathematics

Normalisers of Subgroups

Normalisers of Subgroups

OVERVIEW

NORMALISER

EXPLORE THE NORMALISER LADDER

NORMALISER LADDER EXPLORER

COMPUTE NORMALISER INDICES

NORMALISER INDEX CALCULATOR

FREQUENTLY ASKED QUESTIONS

CONTINUE TO QUOTIENT GROUPS

Coordinates and Locus

Invariants under Orthogonal Transformation