Order of span of product of subgroups

Question

Let $G$ be a finite group and $A,B\leq G$. Is there some formula of the sort $$|\langle AB \rangle| = f(|A|,|B|,|A\cap B|,\ldots)$$ to get the order of the span of $AB$ as a function of some known quantities?

I'm using "$\ldots$" in my formula because there must be other parameters at play than the first three: consider these two options.

• $A=\{1,(12)\},B=\{1,(13)\}$
• $A=\{1,(12)\},B=\{1,(34)\}$

In both cases, $A,B\leq \Sigma_4$, $|A|=|B|=2$ and $|A\cap B|=1$. But in the first case $|\langle AB \rangle|=6$ and in the second $|\langle AB \rangle|=4$.

So I'm asking an open question in some sort here.

Answer 1

You are asking for a formula for the order of the subgroup $\langle A,B \rangle$ generated by $A$ and $B$. There won't be such a formula, since $\langle A,B \rangle$ heavily depends on the relations between the elements of $A$ and $B$. Notice that these relations can be arbitrary long words in the free product of $A$ and $B$ (such as $a_1 b_1 a_2 b_3 a_3 = 1$). This already happening when $A$, $B$ are cyclic. See Possible order of $ab$ when orders of $a$ and $b$ are known.

The problem is only feasible if $AB$ is already a subgroup, in which case we have the usual formula.

My guess is that this question actually arises in the context of a different, manageable problem. In that case, I recommend that you post that problem instead.