I was thinking about the following problem and I hope some of you can answer me.
Let $G$ be a nilpotent group. Let $P$ be the $p$-Sylow subgroup of $G$. What about the $p$-subgroup of an homomorphic image of $G$? Is it of the form $PH/H\leq G/H$? Or do we require something else (for istance that the quotient is something like $G/Z(G)$)?
In the finite case is easy but in the infinite one?
No. Take $G=\mathbb{Z}$ with Sylow 2-subgroup $P=\{0\}$ and normal subgroup $H=2\mathbb{Z}$ with quotient $G/H = \mathbb{Z}/2\mathbb{Z}$ cyclic of order 2. The quotient group has a 2-subgroup $G/H$, but $G/H$ is not a subgroup of $PH/H = H/H$.
I'm not sure of a sufficient condition. Maybe look for something along the lines of "pure subgroups" (an abelian group concept).