TITLE: Subgroups - Klein bottle
QUESTION [9 upvotes]: Let $G$ be the fundamental group of the Klein bottle,  
$G = \langle x,y \ ; \ yxy^{-1}=x^{-1} \rangle = {\mathbb Z} \rtimes {\mathbb Z} \ .$
What are the nilpotent subgroups of $G$?
I was only able to find a normal series of abelian subgroups with cyclic quotients in $G$, namely  
$1\leq \langle y^2 \ ; \ \ \rangle\leq \langle x,y^2 \ ; \ xy^2=y^2x \rangle\leq G \ .$  
Since I'm not an algebraist, I'm sorry if this a silly question. Thanks!

REPLY [7 votes]: Here's a topological version of Arturo's argument.  (I'd leave this as a comment, but I don't have enough rep.)
Every covering space of the Klein bottle is a surface, and it's fairly easy to convince yourself that the infinite-degree ones have cyclic, hence nilpotent, fundamental group.
It remains to consider the finite-degree covering spaces.  Well, these are all closed surfaces of Euler characteristic zero (because Euler characteristic is nilpotent multiplicative), so by the classification of surfaces they are either tori or Klein bottles.  As already noted, the Klein-bottle group itself is not nilpotent, so only the abelian subgroups are left.