# Order of quotient group divides order of group

From Groupprops

This fact is an application of the following pivotal fact/result/idea:Lagrange's theorem

View other applications of Lagrange's theorem OR Read a survey article on applying Lagrange's theorem

This article states a result of the form that one natural number divides another. Specifically, the (order) of a/an/the (quotient group) divides the (order) of a/an/the (group).

View other divisor relations |View congruence conditions

## Contents

## Statement

### Statement in terms of quotient groups

Let be a finite group and be a normal subgroup. The order of the quotient group divides the order of the group .

### Statement in terms of surjective homomorphisms

Let be a finite group and be a homomorphism of groups. Then, the order of the subgroup of divides the order of .

## Related facts

- Variety of groups is congruence-uniform: This is a stronger version of the fact that the order of the quotient group divides the order of the group. It says that the fibers of the quotient map are all of equal size.
- First isomorphism theorem
- Normal subgroup equals kernel of homomorphism

### Applications

### Other facts about order dividing

- Lagrange's theorem: This states that the order of any subgroup divides the order of the group.
- Size of conjugacy class divides order of group
- Size of conjugacy class divides index of center
- Degree of irreducible representation divides group order
- Degree of irreducible representation divides index of center
- Degree of irreducible representation divides index of Abelian normal subgroup

## Facts used

## Proof

### Proof of the statement in terms of quotient groups

**Given**: A group , a normal subgroup .

**To prove**: The order of divides the order of .

**Proof**: By Lagrange's theorem (fact (1)), we have:

.

This yields that the order of the quotient group divides the order of .

### Proof of the statement in terms of homomorphisms

**Given**: A homomorphism of groups , with a finite group.

**To prove**: The order of divides the order of .

**Proof**: Let be the kernel of . By fact (2), is normal in and . Since the order of divides the order of , we obtain that the order of also divides the order of .