## The Product Theorem

Product Theorem: Let $A, B$ be groups. 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,

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