Product formula

From Groupprops
Jump to: navigation, search

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


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 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|.)