Order of quotient group divides order of group

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).
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Statement

Statement in terms of quotient groups

Let $G$ be a finite group and $N$ be a normal subgroup. The order of the quotient group $G/N$ divides the order of the group $G$.

Statement in terms of surjective homomorphisms

Let $G$ be a finite group and $\varphi :G\to H$ be a homomorphism of groups. Then, the order of the subgroup $\varphi (G)$ of $H$ divides the order of $G$.

Related facts

Closely 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

• Homomorphism between groups of coprime order is trivial

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

1. Lagrange's theorem
2. First isomorphism theorem

Proof

Proof of the statement in terms of quotient groups

Given: A group $G$, a normal subgroup $N$.

To prove: The order of $G/N$ divides the order of $G$.

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

$|G|=|N|[G/N|$.

This yields that the order of the quotient group $G/N$ divides the order of $G$.

Proof of the statement in terms of homomorphisms

Given: A homomorphism of groups $\varphi :G\to H$, with $G$ a finite group.

To prove: The order of $\varphi (G)$ divides the order of $G$.

Proof: Let $N$ be the kernel of $\varphi$. By fact (2), $N$ is normal in $G$ and $\varphi (G)\cong G/N$. Since the order of $G/N$ divides the order of $G$, we obtain that the order of $\varphi (G)$ also divides the order of $G$.