What Is the Index of a Subgroup in a Finite Group?

The phrase “index of a finite group” is incomplete unless a subgroup is specified. In group theory, the index usually means the index of a subgroup $H$ in a group $G$, written

$$[G:H].$$

It measures how many distinct copies of $H$—more precisely, cosets of $H$—fit inside $G$. When $G$ is finite, the index is especially easy to calculate:

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

Here $|G|$ and $|H|$ denote the numbers of elements in the groups.

Definition of subgroup index

Let $H$ be a subgroup of $G$. A left coset of $H$ is a set of the form

$$gH=\{gh:h\in H\},$$

where $g\in G$. The left cosets partition $G$: every element of $G$ belongs to exactly one left coset. The index of $H$ in $G$ is the number of distinct left cosets:

$$[G:H]=\text{the number of left cosets of }H\text{ in }G.$$

You can equivalently count right cosets, sets of the form $Hg$. The numbers of left and right cosets are always equal, even when $G$ is not abelian.

If $G$ is finite, every coset has exactly $|H|$ elements. Since the cosets partition the $|G|$ elements of $G$,

$$|G|=[G:H]|H|,$$

and therefore

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

This is the index form of Lagrange’s theorem.

Fully worked example: a subgroup of $S_3$

The symmetric group $S_3$ consists of all permutations of three objects, so

$$|S_3|=3!=6.$$

Consider the subgroup

$$H=\{e,(12)\},$$

where $e$ is the identity permutation. Its order is $|H|=2$.

The index is

$$[S_3:H]=\frac{|S_3|}{|H|}=\frac{6}{2}=3.$$

There are indeed three distinct left cosets. They can be represented by

$$H,\qquad (13)H,\qquad (23)H.$$

Each coset contains two elements, and the three cosets together contain all six elements of $S_3$. Thus the index is $3$.

For comparison, the alternating subgroup $A_3$ has order $3$, so

$$[S_3:A_3]=\frac{6}{3}=2.$$

In this case, the two cosets are $A_3$ itself and one coset containing the odd permutations.

Special cases

Several indices occur frequently:

Subgroup $H$	Index in a finite group $G$
$H=G$	$[G:G]=1$
$H=\{e\}$	$[G:{e}]=
$	H
$H$ has order $1$	$[G:H]=

The index is always a positive integer. Consequently, the order of a subgroup must divide the order of the finite group:

$$|H|\mid |G|.$$

This is the usual statement of Lagrange’s theorem. Conversely, not every divisor of $|G|$ must occur as the order of a subgroup; the divisibility condition is necessary, not sufficient.

How to calculate an index directly

If you know the orders of the groups, use this procedure:

1. Verify that $H\leq G$.
2. Find $|G|$ and $|H|$.
3. Divide $|G|$ by $|H|$.
4. Interpret the quotient as the number of cosets.

For example, suppose $G$ is a finite group of order $24$ and $H$ is a subgroup of order $6$. Then

$$[G:H]=\frac{24}{6}=4.$$

So $H$ has four left cosets and four right cosets in $G$. Each coset has six elements, and the four cosets contain all $24$ elements of $G$ without overlap.

If instead $|G|=30$ and $|H|=8$, such an $H$ cannot exist, because

$$\frac{30}{8}=\frac{15}{4}$$

is not an integer. This detects a contradiction with Lagrange’s theorem.

Relation to normal subgroups and quotient groups

The index is defined whether or not $H$ is normal. However, if $H$ is normal in $G$, the set of cosets forms a quotient group $G/H$. In that case,

$$|G/H|=[G:H].$$

For example, $A_3$ is normal in $S_3$, and

$$|S_3/A_3|=[S_3:A_3]=2.$$

If $H$ is not normal, the cosets still exist and the index is still defined, but the collection of cosets does not generally have a group structure.

A subgroup of index $2$ is always normal. Indeed, there are only two cosets: $H$ and the one remaining coset. The left and right coset partitions must therefore coincide. This makes index-$2$ subgroups especially useful.

Common mistakes

Treating “the index of $G$” as a standalone quantity

There is no single index attached to a finite group without reference to a subgroup. The notation must specify both groups, such as $[G:H]$.

Dividing in the wrong direction

The formula is

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

not $|H|/|G|$. The index counts how many $H$-sized cosets make up $G$, so it cannot be smaller than $1$.

Confusing index with subgroup order

The order $|H|$ counts elements inside $H$. The index $[G:H]$ counts cosets of $H$ in $G$. They are different quantities, although they satisfy

$$|G|=|H|[G:H].$$

Assuming normality is required

Normality is not needed to define an index. It is needed only if you want the cosets to form a quotient group.

For an infinite group, the same definition applies, but the numerical formula involving finite cardinal division may not be useful. For finite groups, however, the answer is simply the order of the group divided by the order of the subgroup.