The Product Theorem

Product Theorem: Let $A,B$ be subgroups of the same group. Then $|AB|=|A||B|/|A\cap B|$ and $AB$ 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 $AD=A$ 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 $AB$ contains exactly $n|A|=|B||A|/|D|$ elements.

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

Conversely, if $A,B$ commute then $(AB{)}^2=(AB)(AB)=A(BA)B=A(AB)B=A^2{B}^2=AB$, hence $AB$ 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,