Cosets and Intersections of Subgroups

We finish this section by connecting cosets with intersections of subgroups. If

H

and

K

are subgroups of the same group, then

H\cap K

is also a subgroup. Its cosets can be described as intersections of corresponding cosets of

H

and

K

. This result is useful because it lets us move between subgroup intersections and coset intersections without guessing element by element. The formulas in this lecture also train students to keep left and right multiplication separate.

Let

(G,\circ)

be a group and let

H

and

K

be subgroups of

G

. If

a\in G

, then

\begin{equation*} (H\cap K)a=Ha\cap Ka. \end{equation*}

We finish this section by connecting cosets with intersections of subgroups. If

H

and

K

are subgroups of the same group, then

H\cap K

is also a subgroup. Its cosets can be described as intersections of corresponding cosets of

H

and

K

Let

(G,\circ)

be a group and let

H

and

K

be subgroups of

G

. If

a\in G

, then

\begin{equation*} (H\cap K)a=Ha\cap Ka. \end{equation*}

Given that

(G,\circ)

is a group,

H

and

K

are subgroups of

G

, and

a\in G

. To prove that

(H\cap K)a=Ha\cap Ka

. [1] To prove that

(H\cap K)a\subseteq Ha\cap Ka

. Let

x\in (H\cap K)a

. Then there exists

u\in H\cap K

such that

\begin{equation*} x=u\circ a. \end{equation*}

Since

u\in H\cap K

, we get

u\in H

and

u\in K

. Since

u\in H

, we get

x=u\circ a\in Ha

. Since

u\in K

, we get

x=u\circ a\in Ka

. Therefore

x\in Ha\cap Ka

. Thus

\begin{equation*} (H\cap K)a\subseteq Ha\cap Ka. \end{equation*}

[2] To prove that

Ha\cap Ka\subseteq (H\cap K)a

. Let

x\in Ha\cap Ka

. Then

x\in Ha

and

x\in Ka

. Since

x\in Ha

, there exists

h\in H

such that

\begin{equation*} x=h\circ a. \end{equation*}

Since

x\in Ka

, there exists

k\in K

such that

\begin{equation*} x=k\circ a. \end{equation*}

Therefore

\begin{align*} h\circ a=k\circ a &\implies h\circ a\circ a^{-1}=k\circ a\circ a^{-1}\\ &\implies h=k. \end{align*}

Since

h=k

h\in H

, and

k\in K

, we get

h\in H\cap K

. Therefore

\begin{equation*} x=h\circ a\in (H\cap K)a. \end{equation*}

Thus

\begin{equation*} Ha\cap Ka\subseteq (H\cap K)a. \end{equation*}

Hence,

\begin{equation*} (H\cap K)a=Ha\cap Ka. \end{equation*}

\square

Let

(G,\circ)

be a group and let

H

and

K

be subgroups of

G

. If

a\in G

, then

\begin{equation*} a(H\cap K)=aH\cap aK. \end{equation*}

Given that

(G,\circ)

is a group,

H

and

K

are subgroups of

G

, and

a\in G

. To prove that

a(H\cap K)=aH\cap aK

. [1] To prove that

a(H\cap K)\subseteq aH\cap aK

. Let

x\in a(H\cap K)

. Then there exists

u\in H\cap K

such that

\begin{equation*} x=a\circ u. \end{equation*}

Since

u\in H\cap K

, we get

u\in H

and

u\in K

. Since

u\in H

, we get

x=a\circ u\in aH

. Since

u\in K

, we get

x=a\circ u\in aK

. Therefore

x\in aH\cap aK

. Thus

\begin{equation*} a(H\cap K)\subseteq aH\cap aK. \end{equation*}

[2] To prove that

aH\cap aK\subseteq a(H\cap K)

. Let

x\in aH\cap aK

. Then

x\in aH

and

x\in aK

. Since

x\in aH

, there exists

h\in H

such that

\begin{equation*} x=a\circ h. \end{equation*}

Since

x\in aK

, there exists

k\in K

such that

\begin{equation*} x=a\circ k. \end{equation*}

Therefore

\begin{align*} a\circ h=a\circ k &\implies a^{-1}\circ a\circ h=a^{-1}\circ a\circ k\\ &\implies h=k. \end{align*}

Since

h=k

h\in H

, and

k\in K

, we get

h\in H\cap K

. Therefore

\begin{equation*} x=a\circ h\in a(H\cap K). \end{equation*}

Thus

\begin{equation*} aH\cap aK\subseteq a(H\cap K). \end{equation*}

Hence,

\begin{equation*} a(H\cap K)=aH\cap aK. \end{equation*}

\square

We observe the formula

a(H\cap K)=aH\cap aK

in a modular group. Choose two subgroups and a representative, then compare the translated intersection with the intersection of the translated cosets. The two sets match because the same element must come from both subgroup translations. This helps connect subgroup intersections with coset intersections without guessing element by element.

Let

G=\mathbb{Z}_{12}

under addition modulo

12

. Let

\begin{equation*} H=\{0,4,8\} \end{equation*}

and

\begin{equation*} K=\{0,6\}. \end{equation*}

Then

\begin{equation*} H\cap K=\{0\}. \end{equation*}

For

a=2

, we have

\begin{align*} 2+(H\cap K) &=\{2\}. \end{align*}

Also,

\begin{align*} 2+H&=\{2,6,10\},\\ 2+K&=\{2,8\}. \end{align*}

Therefore

\begin{align*} (2+H)\cap (2+K) &=\{2\}\\ &=2+(H\cap K). \end{align*}

Let

(G,\circ)

be a group and let

H,K

be subgroups of

G

. Suppose

x\in Ha\cap Ka

. Prove that

x\in (H\cap K)a

Let

(G,\circ)

be a group, let

H,K

be subgroups of

G

, and let

a\in G

. Given that

x\in Ha\cap Ka

. Then

x\in Ha

and

x\in Ka

. Hence, there exist

h\in H

and

k\in K

such that

\begin{equation*} x=h\circ a=k\circ a. \end{equation*}

Therefore

\begin{align*} h\circ a=k\circ a &\implies h\circ a\circ a^{-1}=k\circ a\circ a^{-1}\\ &\implies h=k. \end{align*}

Since

h\in H

and

h=k\in K

, we get

h\in H\cap K

. Therefore

\begin{equation*} x=h\circ a\in (H\cap K)a. \end{equation*}

We compute both sides of the modular intersection formula. Enter subgroup steps for

H

and

K

and a representative

a

, then compare

a+(H\cap K)

with

(a+H)\cap(a+K)

. The output also displays

H\cap K

so the common subgroup is visible before translation. This makes the formula a direct set calculation.

1. Let

G=\mathbb{Z}_{12}

H=\{0,3,6,9\}

, and

K=\{0,6\}

. Find

2+(H\cap K)

. 2. With the same

G,H,K

, find

(2+H)\cap (2+K)

. 3. Let

(G,\circ)

be a group and let

H,K

be subgroups of

G

. Prove that

a(H\cap K)\subseteq aH\cap aK

1. Since

\begin{equation*} H\cap K=\{0,6\}, \end{equation*}

we get

\begin{equation*} 2+(H\cap K)=\{2,8\}. \end{equation*}

2. We have

\begin{align*} 2+H&=\{2,5,8,11\},\\ 2+K&=\{2,8\}. \end{align*}

Therefore

\begin{equation*} (2+H)\cap (2+K)=\{2,8\}. \end{equation*}

3. If

x\in a(H\cap K)

, then

x=a\circ u

for some

u\in H\cap K

. Hence

u\in H

and

u\in K

. Therefore

x\in aH

and

x\in aK

, so

x\in aH\cap aK

An element is in both

Ha

and

Ka

exactly when the multiplying element before

a

belongs to both

H

and

K

Yes. We have

a(H\cap K)=aH\cap aK

Yes. The intersection of two subgroups of the same group is again a subgroup.

BMLabs Mathematics

BMLabs Mathematics

Cosets and Intersections of Subgroups

Cosets and Intersections of Subgroups

OVERVIEW

EXPLORE THE COSET INTERSECTION FORMULA

COSET INTERSECTION FORMULA EXPLORER

INTERSECTIONS OF COSETS IN INTEGERS MODULO TWELVE

A RIGHT COSET INTERSECTION

SOLUTION

CALCULATE A COSET INTERSECTION

SUBGROUP COSET INTERSECTION CALCULATOR

EXERCISES

ANSWERS

Coordinates and Locus

Invariants under Orthogonal Transformation