# Order of Subgroup Product

## 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 $H$.

### Corollary

$\size {H \vee K} \ge \dfrac {\order H \order K} {\order {H \cap K} }$

or

$\dfrac {\size {H \vee K} } {\order H} \ge \dfrac {\order K} {\order {H \cap K} }$

where $H \vee K$ denotes join and $\order H$ denotes the order of $H$.

## Proof 1

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 $1 \le i \le r$.

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

whence the result.

$\blacksquare$

## Proof 2

### Lemma

Let $h_1, h_2 \in H$.

Then:

$h_1 K = h_2 K$
$h_1$ and $h_2$ are in the same left coset of $H \cap K$.

$\Box$

We have that $H K$ is the union of all left cosets $h K$ with $h \in H$:

$\ds H K = \bigcup_{h \mathop \in H} h K$

From Left Coset Space forms Partition, unequal $h K$ are disjoint.

From Cosets are Equivalent, each $h K$ contains $\order K$ elements.

From the Lemma, the number of different such left cosets is:

$\index H {H \cap K}$

where $\index H {H \cap K}$ denotes the index of $H \cap K$ in $H$.

First, let $\order H < + \infty$.

Then, from Lagrange's Theorem:

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

Hence:

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

Finally, let $\order H = + \infty$.

Recall that $G$, $H$ and $K$ have the same identity element $e$ by Identity of Subgroup.

By Definition of Subset Product:

$H = H \set e \subseteq H K$

In particular, $\order {H K} = + \infty$,

Hence:

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

## Proof 3

The number of product elements $h k$ that can be formed where $h \in H$ and $k \in K$ is $\order H \order K$, although this may include duplication:

$h_1 k_1 = h_2 k_2$ may be possible for $h_1, h_2 \in H, k_1, k_2 \in K$.

So, consider the Cartesian product $H \times K$.

$\size {H \times K} = \order H \times \order K$

Let us define a relation $\sim$ on $H \times K$ as:

$\tuple {h_1, k_1} \sim \tuple {h_2, k_2} \iff h_1 k_1 = h_2 k_2$

As $\sim$ is based on the equality relation it is seen that $\sim$ is an equivalence relation:

Reflexivity: $h_1 k_1 = h_1 k_1$
Symmetry: $h_1 k_1 = h_2 k_2 \implies h_2 k_2 = h_1 k_1$
Transitivity: $h_1 k_1 = h_2 k_2, h_2 k_2 = h_3 k_3 \implies h_1 k_1 = h_3 k_3$

Each equivalence class of $\sim$ corresponds to a particular element of $H K$.

Hence $\size {H K}$ is the number of equivalence classes of $\sim$.

It remains to be shown that each of these equivalence classes contains exactly $\order {H \cap K}$ elements.

Let $E$ be the equivalence class of $\tuple {h k}$.

We aim to prove that:

$(1): \quad E = \set {\tuple {h x^{-1}, x k}: x \in H \cap K}$

Let $x \in H \cap K$.

Then:

$a x^{-1} \in H$

and:

$x k \in K$

so:

$\tuple {h x^{-1}, x k} \in H \times K$
$\tuple {h, k} \sim \tuple {h_1, k_1} \implies h k = h_1 k_1$

and so:

$x = h_1^{-1} h = k_1 k^{-1} \in A \cap B$

Thus:

$h_1 = h x^{-1}$

and:

$k_1 = x k$

and $(1)$ is seen to hold.

Finally:

$\tuple {h x^{-1}, x k} = \tuple {h y^{-1}, y k} \implies x = y$

Thus $E$ has exactly $\order {H \cap K}$ elements and the proof is complete.

$\blacksquare$