Index and Intersections of Subgroups

The index multiplication formula studies subgroup chains. We now study what happens when two subgroups meet. If

H

and

K

are subgroups of a group

G

, then

H\cap K

is a subgroup of both

H

and

K

. Lagrange's theorem can therefore be applied inside

H

and inside

K

. A useful comparison is that the index of

H\cap K

H

cannot exceed the index of

K

G

. This result is a precise way to measure how much of

H

is missed by

K

Let

(G,\circ)

be a group and let

H

and

K

be subgroups of

G

. The

\textbf{intersection subgroup}

H

and

K

\begin{equation*} H\cap K=\{x\in G:x\in H \text{ and } x\in K\}. \end{equation*}

The index multiplication formula studies subgroup chains. We now study what happens when two subgroups meet. If

H

and

K

are subgroups of a group

G

, then

H\cap K

is a subgroup of both

H

and

K

. Lagrange's theorem can therefore be applied inside

H

and inside

K

. A useful comparison is that the index of

H\cap K

H

cannot exceed the index of

K

G

. This result is a precise way to measure how much of

H

is missed by

K

Let

(G,\circ)

be a group and let

H

and

K

be subgroups of

G

. The

\textbf{intersection subgroup}

H

and

K

\begin{equation*} H\cap K=\{x\in G:x\in H \text{ and } x\in K\}. \end{equation*}

Let

(G,\circ)

be a group and let

H

and

K

be subgroups of

G

. Then

H\cap K

is a subgroup of

G

Given that

(G,\circ)

is a group and

H,K

are subgroups of

G

. To prove that

H\cap K

is a subgroup of

G

. Since

H

and

K

are subgroups of

G

, we have

e\in H

and

e\in K

. Therefore

\begin{equation*} e\in H\cap K. \end{equation*}

Let

a,b\in H\cap K

. Then

a,b\in H

and

a,b\in K

. Since

H

is a subgroup of

G

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

Since

K

is a subgroup of

G

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

Therefore

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

By the subgroup test,

H\cap K

is a subgroup of

G

. Hence,

H\cap K

is a subgroup of

G

\square

Let

(G,\circ)

be a group and let

H

and

K

be subgroups of

G

. If

[G:K]

is finite, then

\begin{equation*} [H:H\cap K]\leq [G:K]. \end{equation*}

Given that

(G,\circ)

is a group,

H,K

are subgroups of

G

, and

[G:K]

is finite. To prove that

\begin{equation*} [H:H\cap K]\leq [G:K]. \end{equation*}

Let

\mathcal{A}

be the set of all left cosets of

H\cap K

H

, and let

\mathcal{B}

be the set of all left cosets of

K

G

. Define

\Phi:\mathcal{A}\to\mathcal{B}

\begin{equation*} \Phi(h(H\cap K))=hK. \end{equation*}

First, we prove that

\Phi

is well-defined. Let

\begin{equation*} h_1(H\cap K)=h_2(H\cap K). \end{equation*}

Then

\begin{equation*} h_2^{-1}\circ h_1\in H\cap K. \end{equation*}

Therefore

\begin{equation*} h_2^{-1}\circ h_1\in K. \end{equation*}

By the left coset equality criterion,

\begin{equation*} h_1K=h_2K. \end{equation*}

Thus

\Phi

is well-defined. To prove that

\Phi

is one-to-one, let

\begin{equation*} \Phi(h_1(H\cap K))=\Phi(h_2(H\cap K)). \end{equation*}

Then

\begin{equation*} h_1K=h_2K. \end{equation*}

By the left coset equality criterion,

\begin{equation*} h_2^{-1}\circ h_1\in K. \end{equation*}

Since

h_1,h_2\in H

, we also have

\begin{equation*} h_2^{-1}\circ h_1\in H. \end{equation*}

Therefore

\begin{equation*} h_2^{-1}\circ h_1\in H\cap K. \end{equation*}

By the left coset equality criterion,

\begin{equation*} h_1(H\cap K)=h_2(H\cap K). \end{equation*}

Thus

\Phi

is one-to-one. Therefore the number of left cosets of

H\cap K

H

is less than or equal to the number of left cosets of

K

G

. Hence,

\begin{equation*} [H:H\cap K]\leq [G:K]. \end{equation*}

\square

We observe the comparison between the cosets of

H\cap K

inside

H

and the cosets of

K

inside

G

. Choose the two index values and watch how each coset on the left is sent into a distinct coset on the right. The map can fit only when the left side has no more cosets than the right side. Try making

[H:H\cap K]

larger than

[G:K]

and observe why the theorem rejects that data.

The theorem is not only a numerical inequality; it comes from a concrete map between coset spaces. The one-to-one property prevents two different cosets of

H\cap K

H

from landing in the same coset of

K

G

Let

(G,\circ)

be a finite group and let

H,K

be subgroups of

G

. Then

\begin{equation*} \frac{|H|}{|H\cap K|}\leq \frac{|G|}{|K|}. \end{equation*}

Given that

(G,\circ)

is a finite group and

H,K

are subgroups of

G

. To prove that

\begin{equation*} \frac{|H|}{|H\cap K|}\leq \frac{|G|}{|K|}. \end{equation*}

By the previous theorem,

\begin{equation*} [H:H\cap K]\leq [G:K]. \end{equation*}

By Lagrange's theorem,

\begin{equation*} [H:H\cap K]=\frac{|H|}{|H\cap K|} \end{equation*}

and

\begin{equation*} [G:K]=\frac{|G|}{|K|}. \end{equation*}

Therefore

\begin{equation*} \frac{|H|}{|H\cap K|}\leq \frac{|G|}{|K|}. \end{equation*}

Hence,

\begin{equation*} \frac{|H|}{|H\cap K|}\leq \frac{|G|}{|K|}. \end{equation*}

\square

Let

G=\mathbb{Z}_{12}

under addition modulo

12

. Let

\begin{equation*} H=\{0,3,6,9\} \end{equation*}

and

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

Then

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

Thus

\begin{align*} [H:H\cap K] &=\frac{|H|}{|H\cap K|}\\ &=\frac{4}{1}\\ &=4. \end{align*}

Also,

\begin{align*} [G:K] &=\frac{|G|}{|K|}\\ &=\frac{12}{3}\\ &=4. \end{align*}

Therefore

\begin{equation*} [H:H\cap K]\leq [G:K]. \end{equation*}

Let

(G,\circ)

be a group and let

H,K

be subgroups of

G

. Suppose

[G:K]=6

. What can be said about

[H:H\cap K]

Let

(G,\circ)

be a group and let

H,K

be subgroups of

G

. Given that

\begin{equation*} [G:K]=6. \end{equation*}

By the intersection index theorem,

\begin{equation*} [H:H\cap K]\leq [G:K]. \end{equation*}

Therefore

\begin{equation*} [H:H\cap K]\leq 6. \end{equation*}

We compute the two sides of the finite version of the intersection index theorem. Enter the orders of

G

H

K

, and

H\cap K

to calculate

[H:H\cap K]

and

[G:K]

. The divisibility checks matter because these ratios represent subgroup indices. When all inputs are compatible, the comparison shows exactly how the intersection controls the size of

H

relative to

K

1. Let

G=\mathbb{Z}_{18}

H=\{0,3,6,9,12,15\}

, and

K=\{0,6,12\}

. Find

H\cap K

. 2. With the same

G,H,K

, compute

[H:H\cap K]

. 3. Let

H,K

be subgroups of a finite group

G

. Prove that

|H\cap K|

divides

|H|

and

|K|

1. Since every element of

K

belongs to

H

, we get

\begin{equation*} H\cap K=K=\{0,6,12\}. \end{equation*}

2. We have

|H|=6

and

|H\cap K|=3

. Therefore

\begin{equation*} [H:H\cap K]=\frac{6}{3}=2. \end{equation*}

3. Since

H\cap K

is a subgroup of

H

, Lagrange's theorem gives

\begin{equation*} |H\cap K|\mid |H|. \end{equation*}

Since

H\cap K

is a subgroup of

K

, Lagrange's theorem gives

\begin{equation*} |H\cap K|\mid |K|. \end{equation*}

It contains the identity and is closed under the subgroup test because both

H

and

K

are subgroups.

It measures how many cosets of the common part

H\cap K

are needed to cover

H

No, when

[G:K]

is finite, there is an injection from the cosets of

H\cap K

H

into the cosets of

K

G

BMLabs Mathematics

BMLabs Mathematics

Index and Intersections of Subgroups

Index and Intersections of Subgroups

OVERVIEW

INTERSECTION SUBGROUP

EXPLORE THE COSET INJECTION

INTERSECTION INDEX INJECTION EXPLORER

WHY THE INJECTION MATTERS

INTERSECTION INDEX IN INTEGERS MODULO TWELVE

BOUNDING AN INTERSECTION INDEX

SOLUTION

CHECK THE INTERSECTION INEQUALITY

INTERSECTION INDEX CALCULATOR

EXERCISES

ANSWERS

Coordinates and Locus

Invariants under Orthogonal Transformation