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!

  • 2
    $\begingroup$ Does nilpotence of a subgroup correspond to something interesting in the covering space? $\endgroup$ Oct 11, 2010 at 0:22
  • $\begingroup$ @da Fonseca: Every cyclic subgroup is abelian, hence nilpotent. $\endgroup$ Oct 11, 2010 at 1:05
  • $\begingroup$ @Aaron: I don't know. $\endgroup$
    – da Fonseca
    Oct 11, 2010 at 4:32
  • $\begingroup$ Is there a shorter (yet elementary) argument showing that $G$ is not nilpotent? $\endgroup$ Oct 11, 2010 at 4:32
  • $\begingroup$ Shorter than what? The commutator $[x^a,y] = x^{-a}y^{-1}x^ay = x^{-2a}$, so $[x,y,{}_n x]\neq 1$ for all positive integers $n$ (where $[a,{}_1b] = [a,b]$ and $[a,{}_{n+1}b] = [a,{}_n b,b]$). So the group is not nilpotent. $\endgroup$ Oct 11, 2010 at 6:05

2 Answers 2


Every cyclic subgroup is abelian, hence nilpotent. Every such subgroup is generated by an element of the form $x^ay^b$ with $a$ and $b$ integers.

In addition, $y^2$ commutes with $x$. Since $y^2$ is central, to determine the commutator of two elements we only need to consider the parity of the exponents of $y$ in their normal forms. Elements of the form $x^ay^{2b}$ and $x^ry^{2s}$ commute; elements of the form $x^ay^{2k+1}$ and $x^r$ commute if and only if $r=0$; and elements of the form $x^ay^{2k+1}$ and $x^ry^{2s+1}$ commute if and only if $a=r$. That gives you lots of abelian subgroups that are isomorphic to $\mathbb{Z}\times\mathbb{Z}$.

What about higher nilpotency? Suppose you have two noncommuting elements in your subgroup. It is not hard to check that $[x^ay^{2b+1},x^r] = x^{2r}$ and $[x^ay^{2b+1},x^r y^{2s+1}] = x^{2a-2r}$. But no nontrivial power of $x$ commutes with an element of the form $x^a y^{2b+1}$; and further commutators will just yield further nontrivial powers of $x$ that still do not commute with an element of the form $x^ay^{2b+1}$. So if your subgroup $H$ has an element of the form $h_1=x^a y^{2b+1}$, and $h$ is any nontrivial element of $H$ that is not a power of $h_1$, then $h_1$ and $h$ do not commute, and $h_1$ does not commute with any of $[h_1,h]$, $[h_1,h,h_1]$, $[h_1,h,h_1,h_1],\ldots,[h_1,h,h_1,\ldots,h_1]$, etc. But if $H$ is nilpotent of class $c$, then any commutator of weight $c$ would be central in $H$. Thus, $H$ cannot be nilpotent.

So the only nilpotent subgroups of $H$ are abelian, and they are given as above.


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.

  • $\begingroup$ Very nice answer! Forgive me, but what do you mean by "Euler characteristic is nilpotent". Thanks. $\endgroup$
    – da Fonseca
    Oct 11, 2010 at 4:38
  • $\begingroup$ Sorry, that was a typo. I meant 'multiplicative'. $\endgroup$
    – HJRW
    Oct 11, 2010 at 11:43

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