Order of Subgroup Product/Proof 3

Proof
The number of products $h k$ 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 $H \times K$.

From Cardinality of Cartesian Product:
 * $\size {H \times K} = \order H \times \order K$

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:
 * $\tuple {h x^{-1}, x k} \in H \times K$

Conversely:


 * $\tuple {h, k} \sim \tuple {h_1, k_1} \implies h k = h_1 k_1$

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.