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, and $$\left[{G : H}\right]$$ 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
First, we need the following lemma:

Lemma: Let $$G$$ be a finite group, $$H \leq G$$. Let $$\sim$$ be the relation on $$G$$ such that $$a\sim b \leftrightarrow a\in Hb$$

Then $$\sim$$ is an equivalence relation, with the cosets of $$H$$ as its equivalence classes.

Proof of Lemma
Reflexivity:As $$H$$ is a subgroup, $$e\in H$$, so $$a = ea\in Ha$$. Thus $$a\sim a$$.

Symmetry: $$ a\sim b\rightarrow b\in Ha \rightarrow b=ha,~h\in H \rightarrow h^{-1}b = a\rightarrow a\in Hb~ \mbox{as}~h^{-1}\in H\rightarrow b\sim a$$

Transitivity: $$a\sim b~\mbox{and}~b\sim c\mbox{, then}~a=h_1b,~b=h_2c;~h_1,h_2\in H$$

$$\rightarrow a = h_1b = h_1(h_2c) = (h_1h_2)c \rightarrow a\in Hc \rightarrow a\sim c$$

For the equivalence classes:

$$ [a] = \{b\in G~|~b\sim a\} = \{b\in G~|~b\in Ha\} = Ha$$.

Q.E.D.

Proof of Lagrange's Theorem
By the above lemma, the cosets of $$H$$ partition $$G$$. Note that $$ |H| = |Hg|~\forall~g\in G$$ since multiplication by group elements induces an injective map. That is, $$gh_1=gh_2\rightarrow h_1=h_2$$. Thus, presuming there are $$k$$ distinct cosets of $$H$$, we have

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

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

Q.E.D.

Proof 2 (using previously obtained results)

 * 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.