Let $G$ be a nilpotent finitely generated torsion-free group. Is it true that $\underset{n}{\cap}G^{p^n}=\{1\}$, $p$ fixed prime?
I tried induction on the class of nilpotency but what I can get is only that the intersection is contained in the center of $G$. Any hints?
The problem arise since I was reading the following paper
MR0607962 Bruno, Brunella; Phillips, Richard E., "Groups with restricted nonnormal subgroups.", Math. Z. 176 (1981), no. 2, 199–221.
and it seems to me that in the proof of the Corollary 2 it uses this fact.
I'll put here something. Maybe it will be useful to someone else.
Lemma If $G$ is a finitely generated nilpotent group and every non-trivial normal subgroup contains a given element $g\neq 1$, then $G$ is finite and of prime-power order.
Proof. The proof just use the very know fact that a polycyclic group is residually finite.
Theorem Every finitely generated, torsion-free, nilpotent group is residually $p$-finite, where $p$ is any given prime number.
Proof. Let $G$ be a finitely generated, locally infinite, nilpotent group of class $c>0$, and $p$ any given prime number. We shall prove $G$ to be residually $p$-finite by induction on $c$.
When $c=1$, $G$ is a direct product of infinite cyclic groups and it is sufficient to prove (it's obvious!) the result for a single infinite cyclic group.
Now assume the result proved when $c=n-1$ and consider the case $c=n$. If $g\neq 1$ is any given element of $G$ and $g\not\in \gamma_n(G)$, then, by the induction hypothesis, $g$ can be excluded from a normal subgroup of index a power of $p$. If, however, $g\in \gamma_n(G)$, then, by the case $c=1$, we can find a subgroup $H$, contained in $\gamma_n(G)$ (and therefore normal in $G$) such that $g\not\in H$ and $|G\,:\,H|$ is a power of $p$. because $G$ satisfies the maximal condition, there exists a maximal normal subgroup $K$, containing $H$ but not $g$, and, by lemma above, $|G\, :\,K|$ is a power of $p$. Hence $G$ is residually 'of order a power of $p$'.