Normal Subgroups in Quotient Groups

Having seen how abelian and cyclic properties pass to quotient groups, we now examine how subgroups behave inside a quotient. If

H\trianglelefteq G

and

H\subseteq K\subseteq G

, then the cosets of

H

lying inside

K

form a subgroup

K/H

G/H

. This is the first appearance of a correspondence principle: subgroups between

H

and

G

produce subgroups of

G/H

. The most important case is normality. A subgroup

K/H

is normal in

G/H

exactly when

K

is normal in

G

Let

(G,\circ)

be a group, let

H\trianglelefteq G

, and let

K

be a subgroup of

G

such that

H\subseteq K

. The \textbf{subgroup determined by

K

G/H

} is

\begin{equation*} K/H=\{kH:k\in K\}. \end{equation*}

Having seen how abelian and cyclic properties pass to quotient groups, we now examine how subgroups behave inside a quotient. If

H\trianglelefteq G

and

H\subseteq K\subseteq G

, then the cosets of

H

lying inside

K

form a subgroup

K/H

G/H

. This is the first appearance of a correspondence principle: subgroups between

H

and

G

produce subgroups of

G/H

. The most important case is normality. A subgroup

K/H

is normal in

G/H

exactly when

K

is normal in

G

Let

(G,\circ)

be a group, let

H\trianglelefteq G

, and let

K

be a subgroup of

G

such that

H\subseteq K

. The \textbf{subgroup determined by

K

G/H

} is

\begin{equation*} K/H=\{kH:k\in K\}. \end{equation*}

Let

(G,\circ)

be a group, let

H\trianglelefteq G

, and let

K

be a subgroup of

G

such that

H\subseteq K\subseteq G

. Then

K/H

is a subgroup of

G/H

Given that

(G,\circ)

is a group,

H\trianglelefteq G

, and

K

is a subgroup of

G

such that

H\subseteq K\subseteq G

. To prove that

K/H

is a subgroup of

G/H

. Since

H\trianglelefteq G

, the quotient group

G/H

exists. Since

e\in K

, we get

\begin{equation*} H=eH\in K/H. \end{equation*}

Therefore

K/H

is non-empty. Let

aH,bH\in K/H

. Then

a,b\in K

. Since

K

is a subgroup of

G

, we get

\begin{equation*} a\circ b^{-1}\in K. \end{equation*}

Now the inverse of

bH

G/H

b^{-1}H

, and

\begin{align*} (aH)(bH)^{-1} &=(aH)(b^{-1}H)\\ &=(a\circ b^{-1})H\\ &\in K/H. \end{align*}

By the one-step subgroup criterion,

K/H

is a subgroup of

G/H

. Hence,

K/H\leq G/H

\square

Let

(G,\circ)

be a group, let

H,K

be subgroups of

G

such that

\begin{equation*} H\subseteq K\subseteq G, \end{equation*}

and let

H\trianglelefteq G

. Then

K/H

is normal in

G/H

if and only if

K

is normal in

G

Given that

(G,\circ)

is a group,

H,K

are subgroups of

G

H\subseteq K\subseteq G

, and

H\trianglelefteq G

. To prove that

K/H

is normal in

G/H

if and only if

K

is normal in

G

. [1] Let

K/H

be normal in

G/H

. To prove that

K

is normal in

G

. Let

x\in G

and let

k\in K

. Since

K/H

is normal in

G/H

\begin{equation*} (xH)(kH)(xH)^{-1}\in K/H. \end{equation*}

Thus there exists

k_1\in K

such that

\begin{equation*} (x\circ k\circ x^{-1})H=k_1H. \end{equation*}

Therefore

\begin{equation*} k_1^{-1}\circ x\circ k\circ x^{-1}\in H. \end{equation*}

Since

H\subseteq K

and

k_1\in K

, we get

\begin{equation*} x\circ k\circ x^{-1}\in K. \end{equation*}

Therefore

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

Thus

K

is normal in

G

. [2] Let

K

be normal in

G

. To prove that

K/H

is normal in

G/H

. Let

xH\in G/H

and let

kH\in K/H

. Since

K

is normal in

G

\begin{equation*} x\circ k\circ x^{-1}\in K. \end{equation*}

Therefore

\begin{align*} (xH)(kH)(xH)^{-1} &=(x\circ k\circ x^{-1})H\\ &\in K/H. \end{align*}

Thus

\begin{equation*} (xH)(K/H)(xH)^{-1}\subseteq K/H \quad \forall xH\in G/H. \end{equation*}

Therefore

K/H

is normal in

G/H

. Hence,

K/H\trianglelefteq G/H

if and only if

K\trianglelefteq G

\square

Let

G=(\mathbb{Z},+)

, let

H=6\mathbb{Z}

, and let

K=2\mathbb{Z}

. Then

\begin{equation*} 6\mathbb{Z}\subseteq 2\mathbb{Z}\subseteq \mathbb{Z}. \end{equation*}

Since

\mathbb{Z}

is abelian, both

H

and

K

are normal in

G

. Therefore

\begin{equation*} K/H=2\mathbb{Z}/6\mathbb{Z} \end{equation*}

is a normal subgroup of

\begin{equation*} \mathbb{Z}/6\mathbb{Z}. \end{equation*}

Hence, normality of

K

G

produces normality of

K/H

G/H

\square

We observe the chain

H\subseteq K\subseteq G

and its image

K/H\leq G/H

. Use the checkboxes to decide which hypotheses are available. The subgroup

K/H

exists inside the quotient when

H

is normal in

G

and contained in

K

. Normality of

K/H

inside

G/H

is equivalent to normality of

K

inside

G

The quotient

K/H

keeps only the cosets coming from elements of

K

. The normality statement is not a new accident inside the quotient; it reflects the original normality of

K

G

Let

H\trianglelefteq G

and let

K

be a subgroup of

G

with

H\subseteq K

. Prove that

K/H

contains the identity element of

G/H

Since

K

is a subgroup of

G

, we have

e\in K

. Therefore

\begin{equation*} eH=H\in K/H. \end{equation*}

The identity element of

G/H

H

. Thus

K/H

contains the identity element of

G/H

Let

H\trianglelefteq G

H\subseteq K\subseteq G

, and

K\trianglelefteq G

. Prove that

K/H\trianglelefteq G/H

Let

xH\in G/H

and let

kH\in K/H

. Since

K\trianglelefteq G

\begin{equation*} x\circ k\circ x^{-1}\in K. \end{equation*}

Therefore

\begin{align*} (xH)(kH)(xH)^{-1} &=(x\circ k\circ x^{-1})H\\ &\in K/H. \end{align*}

Thus

\begin{equation*} K/H\trianglelefteq G/H. \end{equation*}

When

G

K

, and

H

are finite with

H\subseteq K

, the subgroup quotient

K/H

has order

|K|/|H|

. Enter compatible orders to compute both

|G/H|

and

|K/H|

. This numerical check reinforces the containment picture:

K/H

is a smaller subgroup sitting inside

G/H

1. Define

K/H

when

H\trianglelefteq G

and

H\subseteq K\subseteq G

. 2. Prove that

K/H

is a subgroup of

G/H

. 3. Prove that if

K\trianglelefteq G

, then

K/H\trianglelefteq G/H

. 4. Prove the converse: if

K/H\trianglelefteq G/H

, then

K\trianglelefteq G

. 5. Give an example of

H\subseteq K\subseteq G

using subgroups of

\mathbb{Z}

K/H=\{kH:k\in K\}

. 2. If

aH,bH\in K/H

, then

(aH)(bH)^{-1}=(ab^{-1})H\in K/H

because

ab^{-1}\in K

. 3. If

K\trianglelefteq G

, then conjugating

kH

xH

gives

(xkx^{-1})H\in K/H

. 4. If

K/H\trianglelefteq G/H

, then

(xH)(kH)(xH)^{-1}\in K/H

, forcing

xkx^{-1}\in K

. 5. One example is

6\mathbb{Z}\subseteq 2\mathbb{Z}\subseteq \mathbb{Z}

The expression

K/H

uses cosets of

H

inside

K

, so

H\subseteq K

is required.

No. It is normal in

G/H

exactly when

K

is normal in

G

The identity element is

H=eH

, the same identity coset used in

G/H

BMLabs Mathematics

BMLabs Mathematics

Normal Subgroups in Quotient Groups

Normal Subgroups in Quotient Groups

OVERVIEW

SUBGROUP IN A QUOTIENT GROUP

EXPLORE THE SUBGROUP CORRESPONDENCE

QUOTIENT SUBGROUP CORRESPONDENCE EXPLORER

WHAT THE CORRESPONDENCE MEANS

COMPUTE ORDERS IN A SUBGROUP QUOTIENT

SUBGROUP QUOTIENT ORDER CALCULATOR

FREQUENTLY ASKED QUESTIONS

CONTINUE TO THE ORDER OF A QUOTIENT GROUP

Coordinates and Locus

Invariants under Orthogonal Transformation