13
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @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
5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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:

  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 :-)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.