Lagrange Theorem

The index of a subgroup counts how many cosets the subgroup produces. We now combine this with the fact that all cosets of a subgroup have the same number of elements. This gives one of the central counting results in finite group theory: Lagrange's theorem. The theorem says that the order of a subgroup must divide the order of the finite group. Many later restrictions on groups, elements, and subgroup structures come from this single theorem.

Let

(G,\circ)

be a finite group. The

\textbf{order}

G

is the number of elements in

G

and is denoted by

\begin{equation*} |G|. \end{equation*}

Let

(G,\circ)

be a finite group. The

\textbf{order}

G

is the number of elements in

G

and is denoted by

\begin{equation*} |G|. \end{equation*}

Let

(G,\circ)

be a finite group and let

H

be a subgroup of

G

. Then

|H|

divides

|G|

and

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

Equivalently,

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

Lagrange's theorem says that a finite group is divided into equal-sized coset blocks. The size of each block is

|H|

, and the number of blocks is

[G:H]

. Change the subgroup size and the index to see how their product gives the whole group order. The picture makes the formula

|G|=[G:H]|H|

behave like ordinary multiplication of equal blocks.

The cosets fill the group without overlap and without gaps. That is why the total number of elements must be a multiple of the subgroup order.

Given that

(G,\circ)

is a finite group and

H

is a subgroup of

G

. To prove that

|H|

divides

|G|

and

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

Let the distinct left cosets of

H

G

\begin{equation*} a_1H,a_2H,\ldots,a_rH. \end{equation*}

Then

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

Since the left cosets of

H

form a partition of

G

, we get

\begin{equation*} G=a_1H\cup a_2H\cup\cdots\cup a_rH, \end{equation*}

where the cosets

a_1H,a_2H,\ldots,a_rH

are pairwise disjoint. Since every left coset of

H

has the same cardinality as

H

, we get

\begin{equation*} |a_iH|=|H| \quad \text{for } i=1,2,\ldots,r. \end{equation*}

Therefore

\begin{align*} |G| &=|a_1H|+|a_2H|+\cdots+|a_rH|\\ &=|H|+|H|+\cdots+|H|\\ &=r|H|\\ &=[G:H]|H|. \end{align*}

Therefore

|H|

divides

|G|

. Also,

\begin{align*} |G|=[G:H]|H| &\implies [G:H]=\frac{|G|}{|H|}. \end{align*}

Hence,

|H|

divides

|G|

and

[G:H]=\frac{|G|}{|H|}

\square

Lagrange's theorem is a divisibility theorem, not an existence theorem. It says that if

H

is a subgroup of

G

, then

|H|

must be a divisor of

|G|

. It does not say that every divisor of

|G|

is the order of a subgroup. This distinction is important and prevents a common error in first courses on group theory.

The formula in Lagrange's theorem has three quantities:

|G|

|H|

, and

[G:H]

. Enter any two of them and leave the third blank to solve for the missing value. The solver checks divisibility when the whole group order is known. Use it to test whether proposed subgroup data can be consistent with Lagrange's theorem.

Let

(G,\circ)

be a finite group and let

H

be a subgroup of

G

. Then

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

Given that

(G,\circ)

is a finite group and

H

is a subgroup of

G

. To prove that

|G|=[G:H]|H|

. By Lagrange's theorem,

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

Therefore

\begin{align*} [G:H]=\frac{|G|}{|H|} &\implies |G|=[G:H]|H|. \end{align*}

Hence,

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

\square

Let

G=\mathbb{Z}_{12}

and let

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

Then

|G|=12

and

|H|=3

. By Lagrange's theorem,

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

This agrees with the four cosets

\begin{equation*} H,\quad 1+H,\quad 2+H,\quad 3+H. \end{equation*}

Let

(G,\circ)

be a finite group with

|G|=60

. Let

H

be a subgroup of

G

such that

|H|=12

. Find

[G:H]

Let

(G,\circ)

be a finite group with

|G|=60

and let

H

be a subgroup of

G

such that

|H|=12

. By Lagrange's theorem,

\begin{align*} [G:H] &=\frac{|G|}{|H|}\\ &=\frac{60}{12}\\ &=5. \end{align*}

Therefore

\begin{equation*} [G:H]=5. \end{equation*}

Let

(G,\circ)

be a finite group with

|G|=48

. Suppose

H

is a subgroup of

G

and

[G:H]=6

. Find

|H|

Let

(G,\circ)

be a finite group with

|G|=48

. Let

H

be a subgroup of

G

and suppose

[G:H]=6

. By Lagrange's theorem,

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

Therefore

\begin{align*} 48&=6|H|\\ &\implies |H|=8. \end{align*}

Hence,

\begin{equation*} |H|=8. \end{equation*}

1. Let

|G|=72

and let

H

be a subgroup of

G

with

|H|=9

. Find

[G:H]

. 2. Let

|G|=84

and let

[G:H]=7

. Find

|H|

. 3. Let

H

be a subgroup of a finite group

G

. If

|G|=45

and

|H|=10

, explain why this is impossible.

1. By Lagrange's theorem,

\begin{equation*} [G:H]=\frac{72}{9}=8. \end{equation*}

2. Since

|G|=[G:H]|H|

, we get

\begin{equation*} 84=7|H|. \end{equation*}

Therefore

\begin{equation*} |H|=12. \end{equation*}

3. If

H

is a subgroup of

G

, then

|H|

must divide

|G|

. But

10

does not divide

45

. Therefore such a subgroup cannot exist.

It says that the order of a subgroup divides the order of a finite group.

No. Lagrange's theorem gives a necessary condition, not a sufficient condition.

Cosets partition the group into equal-sized blocks, and each block has

|H|

elements.

BMLabs Mathematics

BMLabs Mathematics

Lagrange Theorem

Lagrange Theorem

OVERVIEW

ORDER OF A FINITE GROUP

LAGRANGE'S THEOREM

LAGRANGE BLOCK COUNTER GUIDE

LAGRANGE BLOCK COUNTER

LAGRANGE FORMULA SOLVER GUIDE

LAGRANGE FORMULA SOLVER

LAGRANGE'S THEOREM IN INTEGERS MODULO TWELVE

FINDING THE INDEX FROM ORDERS

SOLUTION

FINDING THE ORDER OF A SUBGROUP

SOLUTION

EXERCISES

ANSWERS

Coordinates and Locus

Invariants under Orthogonal Transformation