TITLE: Is $[G_p \cap G_q:G_{pq}]$ always finite?
QUESTION [13 upvotes]: Suppose $G$ is a group, $G_n = \langle\{g^n| g \in G\}\rangle$. Suppose $p$ and $q$ are coprime integers. It is not hard to notice, that $G_{pq} \leq G_p \cap G_q$ ("$\leq$" sign means here "Is a subgroup of") and there are cases where inequality holds ($G_{pq}$ is a proper subgroup). For example in a free group $G = F[a, b]$ the element $a^3(ab^2)^3$ is an element of both $G_2$ and $G_3$ but not of $G_6$. 
However, I failed to find any example where $G_{pq}$ is of infinite index in $G_q \cap G_p$. The proof that there isn't one, didn't come to my mind either.  And thus I am asking a question:
Is the index of $G_{pq}$ in $G_p \cap G_q$ always finite?
Any help will be appreciated.

REPLY [3 votes]: In your example, $G_6$ has finite index in $G_3$ and $G_2$. This answer explains why, but also links to some serious research which is related to your question.
Suppose $G$ is free of rank $m$. Then $G/G_n$ has a name: it is the free Burnside group $B(m, n)$. See, for example, here. Serious people think about these groups, and Zelmanov was awarded his fields medal for answering a question related to these groups (the "restricted Burnside problem").
A special case of your question is then: does there exist primes $p$ and $q$ and an integer $m>1$ such that $B(m, p)$ and $B(m, q)$ are finite but $B(m, pq)$ is infinite?
So far as I know, this is unknown. But probably false (and hard!). However, Marshal Hall Jr. proved that $B(2, 6)$ is finite (see above link). It follows that $G_6$ has finite index in both $G_3$ and $G_2$, and so clearly has finite index in their intersection.

Questions:

Is $B(2, 5)$ finite? (This is a well-known open problem.)

Is one of $B(2, 10)$ or $B(2, 15)$ infinite?



If the answer to both these questions is "yes" then you have answered your question in a really nice way :-)