About the normal subgroups of Burnside groups

Question

I was reading "On periodic groups of odd period $n\ge 1003$" of V. S. Atabekyan. He found that the Burnside group $B_n$ with $n\ge 1003$ has uncountably many normal subgroups. However, I was wondering if the intersection of all the non-trivial normal subgroups is trivial or not.

Clearly,using the solution to the restricted Burnside problem, the Burnside group is not residually finite at least for large $n$, but I do not know, and cannot prove if it has a monolith or not. I think this is something that should really be well known, so this is why I'm asking here, because I cannot find it.

Answer 1

The main result (stated in the abstract) of this paper implies that the intersection of proper normal subgroups of $B(m,n)$, $n$ sufficiently large, must be cyclic. Suppose $N\lhd B(m,n)$ is the intersection over all non-trivial subgroups of $B(m,n)$ and is non-cyclic. Then Ivanov’s theorem implies that there exists $H< N, H \cong B(\infty, n)$, and such that for $K\lhd H$, the normal closure $\langle K \rangle^{B(m,n)}$ of $K$ in $B(m,n)$ intersects $H$ in $K$. But this contradicts that $N \subset \langle K \rangle^{B(m,n)}$.

But $B(m,n)$ cannot have a cyclic normal subgroup. If $ a\in B(m,n)-\{1\}$, $\langle a\rangle \lhd B(m,n)$, then a finite-index subgroup of $B(m,n)$ centralizes $a$. But the centralizers of non-trivial elements are cyclic (see the properties listed on p. 2 of this paper), a contradiction. Thus for $n$ sufficiently large we conclude that the intersection of non-trivial normal subgroups of $B(m,n)$ is trivial.