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

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 $$n!$$.