# Order of Subgroup Product/Proof 1

## Theorem

Let $G$ be a group.

Let $H$ and $K$ be subgroups of $G$.

Then:

$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$

where:

$H K$ denotes subset product
$\order H$ denotes the order of $H$.

## Proof

From Intersection of Subgroups is Subgroup, we have that $H \cap K \le H$.

Let the number of left cosets of $H \cap K$ in $H$ be $r$.

Then the left coset space of $H \cap K$ in $H$ is:

$\set {x_1 \paren {H \cap K}, x_2 \paren {H \cap K}, \ldots, x_r \paren {H \cap K} }$

So each element of $H$ is in $x_i \paren {H \cap K}$ for some $1 \le i \le r$.

Also, if $i \ne j$, we have:

$x_j^{-1} x_i \notin H \cap K$

Let $h k \in H K$.

We can write $h = x_i g$ for some $1 \le i \le r$ and some $g \in K$.

Thus:

$h k = x_i \paren {g k}$

Since $g, k \in K$, this shows $h k \in x_i K$.

Aiming for a contradiction, suppose the left cosets $x_i K$ are not all disjoint.

Then by Left Coset Space forms Partition:

$x_i K = x_j K$ for some $i, j$.
$x_j^{-1} x_i \in K$

Since $x_i, x_j \in H$, we have:

$x_j^{-1} x_i \in H \cap K$

Therefore the left cosets $x_i K$ are disjoint for $1 \le i \le r$.
$\dfrac {\order H} {\order {H \cap K} } = \dfrac {\order {H K} } {\order K} = r$
$\blacksquare$