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
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 $|G|$.

Proof 2

 * Let $G$ be of finite order.

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

All three numbers are finite, and the result follows.


 * Now Let $G$ be of infinite order.

If $\left[{G : H}\right]$ is finite, then $\left|{H}\right|$ is infinite; if $\left|{H}\right|$ is finite, then $\left[{G : H}\right]$ is infinite.

Proof 3 (using Orbit-Stabilizer Theorem)
Follows directly from the Orbit-Stabilizer Theorem applied to Group Action on Coset Space.

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 $n!$.