Abstract:
Let $B(m,n)$ be a free periodic group of arbitrary rank $m$ with period $n$. In this paper, we prove that for all odd numbers $n\ge1003$ the normalizer of any nontrivial subgroup $N$ of the group $B(m,n)$ coincides with $N$ if the subgroup $N$ is free in the variety of all $n$-periodic groups. From this, there follows a positive answer for all prime numbers $n>997$ to the following problem set by S. I. Adian in the Kourovka Notebook: is it true that none of the proper normal subgroups of the group $B(m,n)$ of prime period $n>665$ is a free periodic group? The obtained result also strengthens a similar result of A. Yu. Ol'shanskii by reducing the boundary of exponent $n$ from $n>10^{78}$ to $n\ge1003$. For primes $665<n\leq997$, the mentioned question is still open.