# 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 \(D\)

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

We have \(D \subset A\), thus \(A D = A\) and hence

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:

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

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,

*blynn@cs.stanford.edu*💡