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Let $X$ be any topological space with the first homology group $H_1(X)=\mathbb{Z}$ . I claim $\pi_1(X)=\mathbb{Z}$.

By Hurewicz, we know that $H_1(X)$ is the abelianization of $\pi_1(X)$. $H_1(X)=\mathbb{Z}=\langle a | \rangle$ has one generator, so $\pi_1(X)$ has to have one generator too. The abelianization functor introduces the new relations on $\pi_1(X)$ in order to make it abelian. But $H_1(X)=\mathbb{Z}$ does not have any non-trivial relations so no relations could be introduced and hence $\pi_1(X)$ has to have one generator of infinite order with no non-trivial relations.

There does not exist a non-abelian group $G$ whose abelianization is a trivial group $\mathbb{Z}_1$. Hence $\pi_1(X)$ cannot be a product of another non-abelian group and $\mathbb{Z}$.

Hence $\pi_1(X) \cong H_1(X) = \mathbb{Z}.$

Are my reasoning and the claim correct?

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    $\begingroup$ This new fashion of writing the trivial group $\mathbb Z_1$ is quite silly... $\endgroup$ – Mariano Suárez-Álvarez Apr 25 '13 at 4:51
  • $\begingroup$ @MarianoSuárez-Alvarez On en.wikipedia.org/wiki/Trivial_group, they denote the trivial group by $\mathbb{Z}_1$ or by $C_1$, which one should I use? $\endgroup$ – Dávid Natingga Apr 25 '13 at 4:53
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    $\begingroup$ $0$ is a great notation. $1$ is also used in a context where groups are non-abelian. $\endgroup$ – Mariano Suárez-Álvarez Apr 25 '13 at 4:56
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Pick any non-abelian group $G$ whose abelianization is $\mathbb Z$; to keep things simple, take $G=H\times\mathbb Z$, for example with $H$ any perfect group. There exist spaces $X$ such that $\pi_1(X)\cong G$ (one such space is the so called classifying space $BG$ of the discrete group $G$) and then $H_1(X)=\mathbb Z$.

Every such space is a counterexample to your claim.

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    $\begingroup$ In particular, your claim that «There does not exist a non-abelian group G whose abelianization is a trivial group $\mathbb Z_1$» is clearly false, for there exist plenty of perfect groups; all simple groups are perfect, for example. $\endgroup$ – Mariano Suárez-Álvarez Apr 25 '13 at 4:48
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    $\begingroup$ *non-abelian simple $\endgroup$ – Qiaochu Yuan Apr 25 '13 at 6:46
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As a simple to describe topological space where this fails I proffer the complement $X$ of a trefoil knot in $\mathbb{R}^3$. IIRC $\pi_1(X)=\langle a,b\mid a^2=b^3\rangle$ (a proof based on van Kampen is in Massey's book or a fun exercise). In the abelianization we have (abusing notation) $b=b^3b^{-2}=a^2b^{-2}= (ab^{-1})^2$ and therefore also $a=(ab^{-1})b=(ab^{-1})^3$, so the abelianization is generated by (the coset of) $ab^{-1}$.

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