Order of Subgroup Product/Proof 3
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 $H$.
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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:
- $\tuple {h x^{-1}, x k} \in H \times K$
- $\tuple {h, k} \sim \tuple {h_1, k_1} \implies h k = h_1 k_1$
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
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Theorem $21$