Index Two and Coset Criteria for Normality

The conjugation tests are powerful, but many normal subgroups are recognised faster through cosets. The most important quick test is the index two test: if a subgroup has exactly two left cosets, then it is automatically normal. This happens because the subgroup itself accounts for one coset, and everything outside it accounts for the other. In this lesson we also study two useful coset criteria and a square condition that forces normality. These results are especially helpful in examinations, where a full conjugation calculation may be longer than necessary.

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Then

H

is called an \textbf{index two subgroup} of

G

\begin{equation*} [G:H]=2. \end{equation*}

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Then

H

is called an \textbf{index two subgroup} of

G

\begin{equation*} [G:H]=2. \end{equation*}

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

[G:H]=2

, then

H

is normal in

G

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and

[G:H]=2

. To prove that

H

is normal in

G

. Let

x\in G

. If

x\in H

, then

\begin{equation*} xH=H=Hx. \end{equation*}

x\notin H

, then

xH

is the unique left coset different from

H

, and

Hx

is the unique right coset different from

H

. Since the two left cosets are

H

and

G-H

, and the two right cosets are

H

and

G-H

, we get

\begin{equation*} xH=G-H=Hx. \end{equation*}

Therefore

\begin{equation*} xH=Hx \quad \forall x\in G. \end{equation*}

Hence,

H

is normal in

G

\square

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

[G:H]=2

, then

\begin{equation*} a^2\in H \quad \forall a\in G. \end{equation*}

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and

[G:H]=2

. To prove that

a^2\in H

for every

a\in G

. Let

a\in G

. If

a\in H

, then

a^2\in H

. If

a\notin H

, then

\begin{equation*} aH=G-H. \end{equation*}

Since

a^{-1}\notin H

, we get

\begin{equation*} a^{-1}\in G-H. \end{equation*}

Therefore

a^{-1}\in aH

. Thus there exists

h\in H

such that

\begin{equation*} a^{-1}=a\circ h. \end{equation*}

Hence

\begin{align*} a^{-1}=a\circ h &\implies e=a^2\circ h\\ &\implies a^2=h^{-1}\in H. \end{align*}

Hence,

a^2\in H

for every

a\in G

\square

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

\begin{equation*} x^2\in H \quad \forall x\in G, \end{equation*}

then

H

is normal in

G

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and

x^2\in H

for every

x\in G

. To prove that

H

is normal in

G

. Let

x\in G

and let

h\in H

. Since

x\in G

\begin{equation*} x^2\in H. \end{equation*}

Since

x\circ h\in G

\begin{equation*} (x\circ h)^2\in H. \end{equation*}

Since

x^{-1}\in G

\begin{equation*} x^{-2}\in H. \end{equation*}

Now

\begin{align*} x\circ h\circ x^{-1} &=x\circ h\circ x\circ x^{-2}\\ &=(x\circ h)^2\circ h^{-1}\circ x^{-2}. \end{align*}

Since

(x\circ h)^2\in H

h^{-1}\in H

, and

x^{-2}\in H

, we get

\begin{equation*} x\circ h\circ x^{-1}\in H. \end{equation*}

Therefore

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

Hence,

H

is normal in

G

\square

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If every left coset of

H

is a right coset of

H

, then

H

is normal in

G

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and every left coset of

H

is a right coset of

H

. To prove that

H

is normal in

G

. Let

a\in G

. Since

aH

is a left coset of

H

, there exists

b\in G

such that

\begin{equation*} aH=Hb. \end{equation*}

Since

a\in aH

, we get

a\in Hb

. Also

a\in Ha

. Therefore the right cosets

Hb

and

Ha

have non-empty intersection. Therefore

\begin{equation*} Hb=Ha. \end{equation*}

Thus

\begin{equation*} aH=Ha. \end{equation*}

Therefore

\begin{equation*} aH=Ha \quad \forall a\in G. \end{equation*}

Hence,

H

is normal in

G

\square

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If the product of two left cosets of

H

is always a left coset of

H

, then

H

is normal in

G

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and the product of two left cosets of

H

is always a left coset of

H

. To prove that

H

is normal in

G

. Let

a\in G

. By the given condition,

\begin{equation*} (a^{-1}H)(aH) \end{equation*}

is a left coset of

H

. Since

\begin{equation*} e=a^{-1}\circ e\circ a\circ e\in (a^{-1}H)(aH), \end{equation*}

this left coset contains

e

. Therefore

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

Let

h\in H

. Then

\begin{equation*} a^{-1}\circ h\circ a\in (a^{-1}H)(aH). \end{equation*}

Thus

\begin{equation*} a^{-1}\circ h\circ a\in H. \end{equation*}

Therefore

\begin{equation*} a^{-1}Ha\subseteq H \quad \forall a\in G. \end{equation*}

Replacing

a

a^{-1}

gives

\begin{equation*} aHa^{-1}\subseteq H \quad \forall a\in G. \end{equation*}

Hence,

H

is normal in

G

\square

Prove that

A_3

is normal in

S_3

Let

G=S_3

and

H=A_3

. Since

|S_3|=6

and

|A_3|=3

\begin{equation*} [S_3:A_3]=\frac{6}{3}=2. \end{equation*}

By the index two criterion,

\begin{equation*} A_3\trianglelefteq S_3. \end{equation*}

We observe why index two forces normality. Enter compatible orders for

G

and

H

, and watch how the partition collapses into only two pieces. When the index is two, every element outside

H

must lie in the same outside coset on both the left and the right. This leaves no room for a left coset and a right coset to differ.

The proof uses uniqueness. Once

H

has index two, every element outside

H

must generate the same outside part whether we multiply on the left or on the right.

We use the numerical index and the square condition as fast tests for normality. Enter the order of

G

and

H

to compute the index. Then mark whether the square condition

x^2\in H

is known for every

x\in G

. The calculator separates two sufficient conditions: index two implies normality, and the square condition also implies normality.

1. Prove that every subgroup of index two is normal. 2. If

[G:H]=2

, prove that

a^2\in H

for every

a\in G

. 3. Suppose

x^2\in H

for every

x\in G

. Prove that

H

is normal in

G

. 4. Give an example of an index two subgroup.

1. There are only two left cosets and two right cosets. For

x\notin H

, both

xH

and

Hx

equal

G-H

, so

xH=Hx

. 2. If

a\in H

, then

a^2\in H

. If

a\notin H

, then

a^{-1}\in aH

, so

a^{-1}=ah

for some

h\in H

, and hence

a^2=h^{-1}\in H

. 3. For

x\in G

and

h\in H

xhx^{-1}=(xh)^2h^{-1}x^{-2}\in H

, so

xHx^{-1}\subseteq H

. 4. The subgroup

A_3

has index two in

S_3

Outside

H

, there is only one possible left coset and only one possible right coset, so they must both be

G-H

Not necessarily. The square condition is sufficient for normality, not a complete characterisation of index two subgroups.

Quotient groups require multiplication of cosets, and that multiplication behaves well only when the subgroup is normal.

BMLabs Mathematics

BMLabs Mathematics

Index Two and Coset Criteria for Normality

Index Two and Coset Criteria for Normality

OVERVIEW

INDEX TWO SUBGROUP

EXPLORE THE TWO COSET ARGUMENT

INDEX TWO COSET EXPLORER

WHY THE OUTSIDE COSET MATTERS

CHECK INDEX TWO AND SQUARE CRITERIA

INDEX TWO NORMALITY CALCULATOR

FREQUENTLY ASKED QUESTIONS

CONTINUE TO CENTRAL AND CYCLIC SOURCES

Coordinates and Locus

Invariants under Orthogonal Transformation