Lagrange's Theorem (Group Theory)/Proof 1

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