Product formula

Set-theoretic version
Suppose $$H_1, H_2 \le G$$ are subgroups. Then, there is a natural bijection between the left coset spaces:

$$|H_1/(H_1 \cap H_2)| \leftrightarrow |H_1H_2/H_2|$$.

Note that while $$H_1 \cap H_2$$, being an intersection of subgroups, is a subgroup, $$H_1H_2$$ is not necessarily a subgroup.

Numerical version
Let $$H_1$$ and $$H_2$$ be two subgroups of a finite group $$G$$. Then:

$$|H_1H_2||H_1 \cap H_2| = |H_1||H_2|$$

Here $$H_1H_2$$ is the product of subgroups.

Related facts

 * Second isomorphism theorem: This is a stronger formulation of the set-theoretic version, which holds when both the groups in the denominator are normal in the respective numerators. In this case, the natural bijection turns out to be an isomorphism.

Corollaries

 * Index of intersection of permuting subgroups divides product of indices: If $$HK = KH$$, then the index of $$H \cap K$$ divides the product of the index of $$H$$ and the index of $$K$$.
 * Index satisfies transfer inequality: This states that if $$H, K \le G$$, then $$[K:H \cap K] \le [G:H]$$.
 * Index satisfies intersection inequality: This states that if $$H, K \le G$$ are subgroups, then $$[G:H \cap K] \le [G:H][G:K]$$.

Facts used

 * 1) uses::Subgroup containment implies coset containment: If $$H \le K \le G$$ are subgroups, then every left coset of $$H$$ is contained in a left coset of $$K$$.
 * 2) uses::Lagrange's theorem

Proof of the set-theoretic version
Given: A group $$G$$, and subgroups $$H_1, H_2 \le G$$.

To prove: There is a natural bijection between the coset spaces $$H_1/(H_1 \cap H_2)$$ and $$H_1H_2/H_2$$.

Proof: We first define the map:

$$\varphi: H_1/(H_1 \cap H_2) \to H_1H_2/H_2$$

as follows:

$$\varphi(g(H_1 \cap H_2)) = gH_2$$.

In other words, it sends each coset of $$H_1 \cap H_2$$ to the coset of $$H_2$$ containing it.


 * The map sends cosets to cosets: Note first that if two elements are in the same coset of $$H_1 \cap H_2$$, they are in the same coset of $$H_2$$. Thus, the map sends cosets of $$H_1 \cap H_2$$ to cosets of $$H_2$$. (This is fact (1)).
 * The map is well-defined with the specified domain and co-domain: Further, if $$g \in H_1$$, then $$gH_2 \subseteq H_1H_2$$. In other words, if the original coset is in $$H_1$$, the new coset is in $$H_1H_2$$. Thus, the map $$\varphi$$ is well-defined from $$H_1/(H_1 \cap H_2)$$ to $$H_1H_2/H_2$$.
 * The map is injective: Finally, $$\varphi(a(H_1 \cap H_2)) = \varphi(b(H_1 \cap H_2))$$. That means that $$aH_2 = bH_2$$, forcing $$a^{-1}b \in H_2$$. But we anyway have $$a,b \in H_1$$, so $$a^{-1}b \in H_1 \cap H_2$$, forcing that $$a$$ and $$b$$ are in the same coset of $$H_1 \cap H_2$$. Thus, $$a(H_1 \cap H_2) = b(H_1 \cap H_2)$$.
 * The map is surjective: Any left coset of $$H_2$$ in $$H_1H_2$$ can be written as $$gH_2$$ where $$g \in H_1H_2$$. Thus, we can write $$g = ab$$ where $$a \in H_1, b \in H_2$$. Then, $$gH_2 = abH_2 = a(bH_2) = aH_2$$, with $$a \in H_1$$. Thus, $$gH_2 = \varphi(a(H_1 \cap H_2))$$.

Proof of the numerical version using the set-theoretic version
The numerical version follows by combining the set-theoretic version with Lagrange's theorem:

$$|H_1/(H_1 \cap H_2)| = |H_1H_2/H_2|$$.

Notice that the left side measures the number of cosets of $$H_1 \cap H_2$$ in $$H_1$$. Since all these cosets are disjoint and have size equal to $$H_1 \cap H_2$$, the left side is $$|H_1|/|H_1 \cap H_2|$$. Similarly, the right side is $$|H_1H_2|/|H_2|$$. This yields:

$$\frac{|H_1|}{|H_1 \cap H_2|} = \frac{|H_1H_2|}{|H_2|}$$

which, upon rearrangement, gives the product formula.

(Note: Although for the left side, we can quote Lagrange's theorem to say that $$|H_1/(H_1 \cap H_2)| = |H_1|/|H_1 \cap H_2|$$, we cannot directly quote Lagrange's theorem for the right side, because $$H_1H_2$$ is not necessarily a group. However, the reason is precisely the same in both cases: $$H_1H_2$$ is a union of left cosets of $$H_2$$, each having the same size as $$H_2$$, so the number of such cosets is $$|H_1H_2|/|H_2|$$.)