I'd like to ask can we characterize the structure of finitely generated infinite pgroup which has a unique subgroup of order p?
Can we say that these group are residually nilpotent? Any comments are welcome.
I'd like to ask can we characterize the structure of finitely generated infinite pgroup which has a unique subgroup of order p? Can we say that these group are residually nilpotent? Any comments are welcome. 


Here is a counterexample. Take $p>1001$, a prime. Take the free Burnside group $B(2,p)$ of exponent $p$ with $2$ generators $x,y$. It is infinite and has a central extension $\tilde B$ described in Aryan's book "Burnside problems and identities of groups" such that the intersection of any two cyclic subgroups of $\tilde B$ is infinite. The center $Z$ of $\tilde B$ is an infinite cyclic group. Let $H$ be the subgroup of index $p$ in $Z$. Then $Z/H$ is the unique subgroup of order $p$ in $\tilde B/H$, and every element of $\tilde B/H$ outside $Z/H$ has order $p^2$. By Zelmanov's theorem, since $\tilde B/H$ is infinite, it is not residually finite. Another, more elementary, counterexample, of unbounded exponent can be found in Erschler, Anna, Not residually finite groups of intermediate growth, commensurability and nongeometricity. J. Algebra 272 (2004), no. 1, 154–172. 

