Product formula

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

Statement

Set-theoretic version

Suppose $H_{1},H_{2}\leq G$ are subgroups. Then, there is a natural bijection between the left coset spaces:

$|H_{1}/(H_{1}\cap H_{2})|\leftrightarrow |H_{1}H_{2}/H_{2}|$.

Numerical version

Let $H_{1}$ and $H_{2}$ be two subgroups of a finite group $G$. Then:

$|H_{1}H_{2}||H_{1}\cap H_{2}|=|H_{1}||H_{2}|$

Here $H_{1}H_{2}$ 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.
• Index satisfies transfer inequality: This states that if $H,K\leq G$, then $[K:H\cap K]\leq [G:H]$.
• Index satisfies intersection inequality: This states that if $H,K\leq G$ are subgroups, then $[G:H\cap K]\leq [G:H][G:K]$.

Facts used

1. Subgroup containment implies coset containment: If $H\leq K\leq G$ 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 $H_{1},H_{2}\leq G$.

To prove: There is a natural bijection between the coset spaces $H_{1}/(H_{1}\cap H_{2})$ and $H_{1}H_{2}/H_{2}$.

Proof: We first define the map:

$\varphi :H_{1}/(H_{1}\cap H_{2})\to H_{1}H_{2}/H_{2}$

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 $gH_{2}\subseteq H_{1}H_{2}$. In other words, if the original coset is in $H_{1}$, the new coset is in $H_{1}H_{2}$. Thus, the map $\varphi$ is well-defined from $H_{1}/(H_{1}\cap H_{2})$ to $H_{1}H_{2}/H_{2}$.
• 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 $H_{1}H_{2}$ can be written as $gH_{2}$ where $g\in H_{1}H_{2}$. 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:

$|H_{1}/(H_{1}\cap H_{2})|=|H_{1}H_{2}/H_{2}|$.

By Lagrange's theorem, the left side is $|H_{1}|/|H_{1}\cap H_{2}|$ and the right side is $|H_{1}H_{2}|/|H_{2}|$. This yields:

${\frac {|H_{1}|}{|H_{1}\cap H_{2}|}}={\frac {|H_{1}H_{2}|}{|H_{2}|}}$

which, upon rearrangement, gives the product formula.