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.

  • $\begingroup$ @pisco $G_n$ is the subgroup generated by the powers (hence the $\langle\rangle$) $\endgroup$ – Maxime Ramzi Jun 7 '18 at 17:10
  • $\begingroup$ Have you checked any of the 'weird groups'? In particular, the Baumslag-Solitar groups seem like a plausible place to find a counterexample. $\endgroup$ – Steven Stadnicki Jun 7 '18 at 17:16

Let $F$ be the free group on $X_{n,i}$ for $n\in \mathbb{N}, i\in \{0,1,2\}$; and let $G$ be the quotient of $F$ by the normal subgroup generated by the relations $X_{n,1}^p=X_{n,0}=X_{n,2}^q$ for each $n\in \mathbb{N}$.

Hence in $G$, each $X_{n,0} \in G_p\cap G_q$.

Claim: $X_{n,0}G_{pq} \neq X_{m,0}G_{pq}$ for $m\neq n$.

Indeed, assume $X_{n,0} = X_{m,0}w_1^{pq}...w_k^{pq}$ for some $w_i\in G$.

Then this implies, back in $F$ that $X_{n,0}=X_{m,0}w_1^{pq}...w_k^{pq} h_1r_1h_1^{-1}...h_lr_lh_l^{-1}$ for some $h_i \in F$ and $r_i$ among the relations.

Modding out by all the $X_{a,i}$'s for $a\neq m,n$ we may assume that the only letters appearing are $X_{n,i}, X_{m,j}$ and their inverses, and this equation holds in the free group on these guys modulo the obvious relations.

But now it suffices to find a group with two distinct elements $x,y \in G_p\cap G_q\setminus G_{pq}$ such that $y^{-1}x\notin G_{pq}$ to show that this isn't true. But for this, taking for instance $y=e$ it suffices to find $x\in G_p\cap G_q \setminus G_{pq}$: your example in $F[a,b]$ for instance works): so this equation cannot hold for $m\neq n$.

This proves the claim. But this implies that $[G_p\cap G_q: G_{pq}]$ is infinite.


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.


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

  2. 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 :-)


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