Order of Subgroup Product/Proof 3

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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 $H$.


The validity of the material on this page is questionable.
In particular: The right hand side does not make sense if $\order {H \cap K} = +\infty$, because it is then $\dfrac {+\infty}{+\infty}$. Can someone check the source?
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Proof

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 $H \times K$.

From Cardinality of Cartesian Product of Finite Sets:

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


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.

$\blacksquare$


Sources

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