As is mentioned in the title, I have some group $H$ and I know that its abelianization is $\mathbb{Z}_2$. Does this imply that $H$ has torsion?

Edit: Since people want more context, here's some context. Basically I'm looking at the fundamental group of the Klein bottle and I want to show that it can't split as $\pi_1(K) \cong \mathbb{Z} \oplus H$ for any group $H$. I know if I abelianize then I end up getting $\mathbb{Z} \oplus \mathbb{Z}_2$. So if the abelianization splits over direct sums (which I'm not sure about) then if somehow I could say that $H$ must be going to $\mathbb{Z}_2$ through this abelianization map and $H$ had to have torsion I'd have a contradiction to the klein bottle being a manifold and thus having torsion-free fundamental group. I think there are a number of holes in this argument though, but I'm still interested in the particular question I asked above.

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    $\begingroup$ How did this problem arise? $\endgroup$
    – Shaun
    Oct 29, 2019 at 16:42
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    $\begingroup$ Why are people voting to close the question? Do they know they answer? I would guess that the answer is no but I have not thought of a counterexample. $\endgroup$
    – Derek Holt
    Oct 29, 2019 at 21:30
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    $\begingroup$ @DerekHolt: At a guess, "lack of context", but I think that's inappropriate here. That's meant to cut down on "please do my homework for me" questions, and this is unlikely to be that. That said, some context would be nice. $\endgroup$ Oct 29, 2019 at 21:54
  • $\begingroup$ Fundamental group of Klein bottle doesn't have torsion by the way $\endgroup$
    – user29123
    Oct 30, 2019 at 0:36
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    $\begingroup$ Okay, but manifold groups can have torsion (it doesn't really matter much for your question but it seems sort of implied by the edit) $\endgroup$
    – user29123
    Oct 30, 2019 at 0:46

2 Answers 2


No. Take $G$ to be an infinite, torsion-free perfect group. Let $x$ be a nontrivial element of $G$. Take the amalgam of $G$ with $\mathbf{Z}$ by identifying $x$ with $2$: $$H=\langle G,c\mid c^2=x\rangle.$$ As an amalgam of two torsion-free groups, it is torsion-free. From the presentation, the abelianization is clearly of order 2, generated by the image of $c$.

There are many choices for $G$, including finitely presented ones, for instance $\langle s,t\mid (st)^2=s^3=t^7\rangle$.

  • $\begingroup$ May I ask why is G torsion-free? A torsion-free group is a group whose only element of finite order is the identity, but the element $s$ has order 3? $\endgroup$
    – ghc1997
    Apr 10, 2022 at 14:12
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    $\begingroup$ @David no, $s$ has infinite order. The element $s^3=t^7=(st)^2$ actually generates the center, which is infinite cyclic. $\endgroup$
    – YCor
    Apr 10, 2022 at 16:15

I believe the answer should be “no”. What I have is not an example, but I suspect that something along these lines could be massaged into an example.

The group $E\Pi\operatorname{Aut}(F_3)$, is torsion-free (a theorem of Glover and Jensen) and given by the following presentation. Throughout, $i,j,k$ will be pairwise distinct elements of $\{1,2,3\}$.

$$E\Pi\operatorname{Aut}(F_3) = \langle \rho_{ij} \mid [\rho_{ij},\rho_{ik}], \rho_{ik}\rho_{jk}\rho_{ij} = \rho_{ij}\rho_{jk}\rho_{ik}^{-1}\rangle.$$

Notice that the latter relator abelianizes to saying the image of $\rho_{ik}$ is equal to the image of $\rho_{ik}^{-1}$, so the abelianization of $E\Pi\operatorname{Aut}(F_3)$ is $C_2^6$.


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