Lagrange's Theorem (Group Theory)/Proof 2

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Let $G$ be a finite group.

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


$\order H$ divides $\order G$

where $\order G$ and $\order H$ are the order of $G$ and $H$ respectively.

In fact:

$\index G H = \dfrac {\order G} {\order H}$

where $\index G H$ is the index of $H$ in $G$.

When $G$ is an infinite group, we can still interpret this theorem sensibly:

A subgroup of finite index in an infinite group is itself an infinite group.
A finite subgroup of an infinite group has infinite index.


Let $G$ be a group.

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

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

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

From the definition of index of subgroup, there are $\index G H$ left cosets, and therefore:

$\order G = \index G H \order H$

Let $G$ be of finite order.

All three numbers are finite, and the result follows.

Now let $G$ be of infinite order.

If $\index G H = n$ is finite, then $\order G = n \order H$ and so $H$ is of infinite order.

If $H$ is of finite order such that $\order H = n$, then $\order G = \index G H \times n$ and so $\index G H$ is infinite.


Source of Name

This entry was named for Joseph Louis Lagrange.