# Product vs. join of subgroups of a finite group

Hypotheses:

• $$G$$ is a finite group
• $$A$$ and $$B$$ are subgroups if $$G$$.

Definitions:

• $$AB = \{ ab \mathbin{|} a \in A, b \in B \}$$ is the product of $$A$$ and $$B$$
• $$A \vee B$$ is the join of (subgroup of $$G$$ generated by) $$A$$ and $$B$$
• $$A \wedge B$$ is the intersection $$A$$ and $$B$$ (a subgroup of $$G$$)

It is known that $$|AB|$$ is equal to $$|A||B| \mathbin{/} |A \wedge B|$$, and that $$|AB|$$ is less than or equal to $$|A \vee B|$$.

Question: Is $$|AB|$$ a divisor of $$|A \vee B|$$ ? If so, I would like to have a proof. If not, I would like to have a concrete counter-example.

UPDATE (Galois theoretic motivation)

Let $$L$$ be a normal extension of the rationals, and let $$G$$ be its Galois group. Let $$\alpha$$ be some element of $$L$$, generating a subfield fixed by the subgroup $$A$$ of $$G$$. The degree of the minimal polynomial $$f(x)$$ of $$\alpha$$ over $$Q$$ is then $$[G : A]$$. Consider now the subfield $$K$$ of $$L$$ fixed by some subgroup $$B$$ of $$G$$. The minimal polynomial $$g(x)$$ of $$\alpha$$ over $$K$$ is a divisor of $$f(x)$$ and has degree $$[B : A \wedge B]$$. In particular, the degree of $$g(x)$$ is a divisor of the degree of $$f(x)$$, that is, $$[B : A \wedge B]$$ is a divisor of $$[G : A]$$.

Now, for any finite group $$G$$ and subgroups $$A$$ and $$B$$ holds: $$1 = \frac{|AB||A \wedge B|}{(|A||B|)} \leq \frac{|A \vee B||A \wedge B|}{(|A||B|)} \leq \frac{|G||A \wedge B|}{(|A||B|)} = \frac{[G : A]}{[B : A \wedge B]}$$

The full question would thus be whether all these values are integers or not, considering the case that $$G$$ is the Galois group of some normal extension of the rationals as well as the case that $$G$$ is an arbitrary finite group.

What is known:

• If $$G$$ is the Galois group of a normal extension of the rationals, then $$[G : A] \mathbin{/} [B : A \wedge B]$$ is an integer.
• $$|AB|$$ is not necessarily a divisor of $$|A \vee B|$$ (answer to this question)

UPDATE (wrong claim)

I did a mistake: $$g(x)$$ is a divisor of $$f(x)$$, but the degree of $$g(x)$$ is not necessarily a divisor of the degree of $$f(x)$$. This invalidates the question, I was on a wrong path.

Consider $$G=S_3$$, $$A=\langle (1\ 2)\rangle$$, $$B=\langle (1\ 3)\rangle$$. Then $$A\vee B=\langle (1\ 2),(1\ 3)\rangle=S_3$$, and $$AB=\{(1),(1\ 2),(1\ 3),(1\ 3\ 2)\}$$. So $$|AB|=4$$, $$|A\vee B|=6$$. Therefore $$|AB|$$ is not a divisor of $$|A\vee B|$$.