Lagrange's Theorem (Group Theory)

Theorem

Let $G$ be a group of finite order.

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

Then $\left|{H}\right|$ divides $\left|{G}\right|$.

In fact:

$\displaystyle \left[{G : H}\right] = \frac {\left|{G}\right|} {\left|{H}\right|}$

where:

• $\left|{G}\right|$ and $\left|{H}\right|$ are the order of $G$ and $H$ respectively;
• $\left[{G : H}\right]$ is the index of $H$ in $G$.

When $\left|{G}\right|$ is infinite, we can still interpret this theorem sensibly:

• A subgroup of finite index in a group of infinite order is itself of infinite order;
• A finite subgroup of a group of infinite order has infinite index.

Proof 1

Let $G$ be finite.

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

$\phi: G \to G / H^l: \phi \left({x}\right) = x H^l$

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 $\displaystyle \left[{G : H}\right] = \frac {\left|{G}\right|} {\left|{H}\right|}$

$\blacksquare$

Proof 2

For any group $G$:

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

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

From the definition of index of a subgroup, there are $\left[{G : H}\right]$ left cosets, and therefore $\left|{G}\right| = \left[{G : H}\right] \left|{H}\right|$.

Let $G$ be of finite order.

All three numbers are finite, and the result follows.

Now Let $G$ be of infinite order.

If $\left[{G : H}\right] = n$ is finite, then $\left|{G}\right| = n\left|{H}\right| \implies \left|{H}\right|$ is infinite.

If $\left|{H}\right| = n$ is finite, then $\left|{G}\right| = \left[{G : H}\right] n \implies \left[{G:H}\right]$ is infinite.

$\blacksquare$

Proof 3

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

$\blacksquare$

Proof 4

By Congruence Modulo a Subgroup is an Equivalence, the cosets of $H$ partition $G$.

Note that $\forall g \in G: |H| = |Hg|$ since multiplication by group elements induces an injective map: see Cancellable iff Regular Representation Injective.

That is, $g h_1 = g h_2 \implies h_1 = h_2$.

Thus, presuming there are $k$ distinct cosets of $H$, we have:

$|G| = |H|\cdot k$

Thus $|H|$ divides $|G|$.

$\blacksquare$

Source of Name

This entry was named for Joseph Louis Lagrange.

This result, however, was actually due to Camille Jordan. Lagrange's proof merely showed that a subgroup of the symmetric group $S_n$ has order dividing $n!$ Gosh!

Sources

• Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.9$: Theorem $12$
• George McCarty: Topology: An Introduction with Application to Topological Groups (1967): Chapter $\text{II}$: Problem $\text{GG}$