Maximal normal locally nilpotent subgroup in an infinite extension of a Prüfer group Let $P$ be a Prüfer $p$-group, let $x$ be an infinite automorphism of $P$, namely a $p$-adic unit, and consider the semidirect product $G$ between $P$ and $\langle x\rangle$ via the action of $x$ on $P$. Find the maximal (normal) locally nilpotent subgroup of $G$.
Now, I know that such a subgroup is not $P$ itself, but I hardly see how to find a bigger, locally nilpotent subgroup of $G$, given the fact that $x$ (or any of its powers) could not centralize $P$.
 A: (1) If $x\equiv 1\mod p$ (which is automatic for $p=2$), then then the whole group $G=P\rtimes\langle x\rangle$ is locally nilpotent (because modulo $p^n$ $x$ belongs to the kernel of $(\mathbf{Z}/p^n\mathbf{Z})^\ast\to(\mathbf{Z}/p\mathbf{Z})^*$, which is a $p$-group.
(2) If $x\equiv\!\!\!\!/ 1\mod p$, then $G=P\rtimes\langle x\rangle$ is not locally nilpotent, because its finite subquotient $((\mathbf{Z}/p\mathbf{Z})\rtimes\langle x\rangle)/\langle x^{p-1}\rangle$ is finite and not nilpotent (as its $p$-Sylow is abelian and not central).
In general, consider the residual ring homomorphism $\mathbf{Z}_p\to\mathbf{Z}/p\mathbf{Z}$ and let $k$ be the order of the image of $x$ in $(\mathbf{Z}/p\mathbf{Z})^*$. Then (1) implies that the normal subgroup $P\rtimes\langle x^k\rangle$ is locally nilpotent. A subgroup properly containing the latter has the form $P\rtimes\langle x^\ell\rangle$ where $\ell$ properly divides $k$. By (2), it is not locally nilpotent. We conclude that:

The locally nilpotent radical of $G=P\rtimes\langle x\rangle$ is equal to $P\rtimes\langle x^k\rangle$, where $k$ is the order of the image of $x$ in $(\mathbf{Z}/p\mathbf{Z})^*$.


Edit: I initially claimed that the Fitting subgroup (generated by nilpotent normal subgroups) is the same. As pointed out by Alex, it's not true. Although it's not part of the question, let me include the description:

The Fitting subgroup of $G=P\rtimes\langle x\rangle$ is equal to $P\rtimes\langle x^\ell\rangle (=P\rtimes\times \langle x^\ell\rangle)$, where $\ell$ is the order of $x$ in $\mathbf{Z}_p^*$, which when $\ell=\infty$ is meant to denote $P$.

Indeed, this is a normal abelian subgroup, hence contained in the Fitting subgroup. Any subgroup properly containing the latter one has the form $P\rtimes\langle x^n\rangle$ with $x^n\neq 1$. This is not nilpotent: indeed, we have $[x^n,\mathbf{Q}_p]=\mathbf{Q}_p$, and this passes to the quotient $P=\mathbf{Q}_p/\mathbf{Z}_p$ to yield $[x^n,P]=P$.
