Lagrange's Theorem (Group Theory)

Theorem

Let $G$ be a finite group.

Let $H$ be a subgroup of $G$.

Then:

$\order H$ divides $\order G$

where $\order G$ and $\order H$ are the order of $G$ and $H$ respectively.

In fact:

$\index G H = \dfrac {\order G} {\order H}$

where $\index G H$ is the index of $H$ in $G$.

When $G$ is an infinite group, we can still interpret this theorem sensibly:

A subgroup of finite index in an infinite group is itself an infinite group.

A finite subgroup of an infinite group has infinite index.

Proof 1

Let $G$ be finite.

Consider the mapping $\phi: G \to G / H^l$, defined as:

$\phi: G \to G / H^l: \map \phi x = x H^l$

where $G / H^l$ is the left coset space of $G$ modulo $H$.

For every $y H \in G / H^l$, there exists a corresponding $y \in G$, so $\phi$ is a surjection.

From Cardinality of Surjection it follows that $G / H^l$ is finite.

From Cosets are Equivalent, $G / H^l$ has the same number of elements as $H$.

We have that the $G / H^l$ is a partition of $G$.

It follows from Number of Elements in Partition that $\index G H = \dfrac {\order G} {\order H}$

$\blacksquare$

Proof 2

Let $G$ be a group.

Let $H$ be a subgroup of $G$.

From Cosets are Equivalent, a left coset $y H$ has the same number of elements as $H$, namely $\order H$.

Since left cosets are identical or disjoint, each element of $G$ belongs to exactly one left coset.

From the definition of index of subgroup, there are $\index G H$ left cosets, and therefore:

$\order G = \index G H \order H$

Let $G$ be of finite order.

All three numbers are finite, and the result follows.

Now let $G$ be of infinite order.

If $\index G H = n$ is finite, then $\order G = n \order H$ and so $H$ is of infinite order.

If $H$ is of finite order such that $\order H = n$, then $\order G = \index G H \times n$ and so $\index G H$ is infinite.

$\blacksquare$

Proof 3

Follows directly from the Orbit-Stabilizer Theorem applied to Group Action on Coset Space.

$\blacksquare$

Examples

Intersection of Subgroups of Order $25$ and $36$

Let $G$ be a group.

Let $H$ and $K$ be subgroups of $G$ such that:

$\order H = 25$

$\order K = 36$

where $\order {\, \cdot \,}$ denotes the order of the subgroup.

Then:

$\order {H \cap K} = 1$

Order of Group with Subgroups of Order $25$ and $36$

Let $G$ be a group.

Let $H$ and $K$ be subgroups of $G$ such that:

$\order H = 25$

$\order K = 36$

where $\order {\, \cdot \,}$ denotes the order of the subgroup.

Then:

$900 \divides \order G$

where $\divides$ denotes divisibility.

Order of Union of Subgroups of Order $16$

Let $G$ be a group whose identity is $e$.

Let $H$ and $K$ be subgroups of $G$ such that:

$\order H = \order K = 16$

$H \ne K$

where $\order {\, \cdot \,}$ denotes the order of the subgroup.

Then:

$24 \le \order {H \cup K} \le 31$

Also see

• Group does not Necessarily have Subgroup of Order of Divisor of its Order, where it is shown that the converse is not true in general.

Source of Name

This entry was named for Joseph Louis Lagrange.

Historical Note

Lagrange's theorem was actually proved by Camille Jordan.

Lagrange's proof merely showed that a subgroup of the symmetric group $S_n$ has an order which is a divisor of $n!$.

Sources

• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Problem $\text {GG}$
• 1972: A.G. Howson: A Handbook of Terms used in Algebra and Analysis ... (previous) ... (next): $\S 5$: Groups $\text{I}$: Subgroups

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• 2014: Mahim Ranjan Adhikari and Avishek Adhikari: Basic Modern Algebra with Applications: $\S 2$: Some Basic Facts, Proof and Converse of Lagrange's Theorem.