# The Product Theorem

Then $$|A B| = |A| |B| / |A\cap B|$$ and $$A B$$ is a group if and only if $$A, B$$ commute.

Let $$D = A\cap B$$. Then we can decompose $$B$$ into cosets relative to $$D$$

$B = D b_1 \cup D b_2 \cup ... \cup D b_n$

where $$n = |B| / |D|$$ (and all cosets are distinct). Then left multiplying by $$A$$ gives

$A B = A D b_1 \cup A D b_2 \cup ... \cup A D b_n$

We have $$D \subset A$$, thus $$A D = A$$ and hence

$A B = A b_1 \cup A b_2 \cup ... \cup A b_n$

Note if $$A b_i$$ and $$A b_j$$ have an element in common, then we must have $$a_1 b_i = a_2 b_j$$ for some $$a_1, a_2 \in A$$ from which it follows $$a_2^{-1} a_1 = b_j b_i^{-1}$$ which then is contained in $$D$$, the intersection of $$A$$ and $$B$$.

But then $$D (b_j b_i^{-1}) = D$$, that is, $$D b_j = D b_i$$, implying $$i = j$$. Thus the sets $$A b_i$$ are disjoint so $$A B$$ contains exactly $$n |A| = |B||A| / |D|$$ elements.

Now suppose $$A B$$ is a group. Then let $$a \in A, b\in B$$. Then $$(a^{-1} b^{-1})^{-1} = b a \in A B$$ thus $$B A \subset A B$$. But from above, $$B A$$ and $$A B$$ both contain exactly $$|A||B| /|A \cap B|$$ elements thus $$A B = B A$$. Alternatively, by symmetry we have $$A B \subset B A$$.

Conversely, if $$A, B$$ commute then $$(A B)^2 = (A B)(A B) =A (B A) B =A(A B)B = A^2 B^2 = A B$$, hence $$A B$$ is a group.

Theorem: [Frobenius] Let $$A,B$$ be subgroups of a group $$G$$. Then $$G$$ admits a decomposition into disjoint sets:

$G = A g_1 B + A g_2 B +...+ A g_r B$

where $$g_i \in G$$. We have $$|A g_i B| = |A||B| / |g_i^{-1} A g_i|$$.

Proof: Suppose $$A g_1 B, A g_2 B$$ have an element in common, that is, we have $$a_1 g_1 b_1 = a_2 g_2 b_2$$ for some $$a_1,a_2\in A, b_1,b_2 \in B$$. Then

$A g_1 B = A a_1 g_1 b_1 B = A a_2 g_2 b_2 B = A g_2 B$

Note $$|g_i^{-1} A g_i B| = |A g_i B|$$. Since $$g_i^{-1} A g_i \cong A$$, the result follows after applying the product theorem.

Corollary: Using the same notation,

$|G| = \sum_{i=1}^r |A||B|/|g_i^{-1} A g_i|$

Ben Lynn blynn@cs.stanford.edu 💡