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](http://en.wikipedia.org/wiki/Burnside's_problem#Restricted_Burnside_problem).


**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?



Properties of the Burnside kernel
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 $p$](https://mathoverflow.net/a/79615/1345). 


For question 3., the answer is no (for sufficiently large $p$ at least). [A theorem of Olshanskii](http://www.ams.org/mathscinet-getitem?mr=1119008) 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. 


