5
$\begingroup$

Let $p$ be a prime such that the free 2-generator group $B(2,p)$ of exponent $p$ is infinite. Consider the short exact sequence $$ 1\to K \to B(2,p) \to B_0(2,p) \to 1, $$ where $B_0(2,p)$ is the biggest finite $2$-generator group of exponent $p$, which exists by RBP.

Question. What is known about the normal structure of the kernel $K$?

More specifically, besides the obvious facts that $K$ is perfect, finitely generated, and of exponent $p$,

  1. Are there any known proper normal subgroups of $K$?

  2. Is $Z(K)=1$?

  3. Could it be that $K$ is in fact simple?

$\endgroup$
6
$\begingroup$

The answer to question 2. is yes, since the centralizer of elements in a free Burnside group (of exponent $p > 665$) are finite cyclic of order $p$.

For question 3., the answer is no (for sufficiently large $p$ at least). A theorem of Olshanskii implies that any non-elementary torsion-free hyperbolic group has periodic quotients of period $p$ for sufficiently large odd numbers $p$. If we apply this to a 2-generator hyperbolic group $G$, then there is a period $p$ quotient. Moreover, for $p$ large enough, the quotient $G/G^p$ will not be isomorphic to $B(2,p)$, since it is known that any element in $F_2$ will be non-trivial in $B(2,p)$ for large $p$ (so apply this to any non-trivial relator in a presentation of $G$). Then one has a homomorphism $B(2,p) \to G/G^p$, with infinite index kernel. Intersect this with $K$ to get an infinite index normal subgroup.

$\endgroup$

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.