# Product formula

This article describes a natural isomorphism between two structures or between a family of structures

## Statement

### 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.

## Facts used

1. 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. Lagrange's theorem

## Proof

### 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 $|H_1H_2|/|H_2|$.)