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Are there any groups (besides $\mathbb{Z}$ itself) which have $\mathbb{Z}$ for their abelianization?

That is, is there any non-abelian group $G$ such that $G/[G,G] \cong \mathbb{Z}$?

I would appreciate some examples (if yes) or some hint as to why not (if no).

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up vote 5 down vote accepted

Take your favorite group $G$ whose abelianization is trivial. For example, $G=A_5$ works. Then $\mathbb{Z}\times G$ has abelianization $\mathbb{Z}$.

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And none of these looks like those in Mariano's examples. – Jason DeVito Feb 5 '13 at 0:13
(Except $\mathbb{Z}$ itself, that is!) (Sorry - couldn't resist!) – Jason DeVito Feb 5 '13 at 0:13
:-) ${}{}{}{}{}$ – Mariano Suárez-Alvarez Feb 5 '13 at 0:13

All knot groups have $\mathbb Z$ as abelianization.

That gives a nice supply of examples.

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And none of these looks like those in Jason's examples. – Mariano Suárez-Alvarez Feb 5 '13 at 0:09
(Except $\mathbb Z$ itself, that is!) – Mariano Suárez-Alvarez Feb 5 '13 at 0:11
This actually prompted the question—a friend and I were discussing why knot complements are supposed to be interesting. We convinced ourselves that $H_1$ must always be $\mathbb{Z}$ (with some Mayer-Vietoris.) Then we wondered if $\pi_1$ is interesting, but couldn't come up with any non-abelian groups with abelianization $\mathbb{Z}$. Knowing that they're called knot groups will probably help! – Kundor Feb 5 '13 at 0:41
There is a very simple way to get a presentation from a knot diagram (there are a few, in fact): you can simply draw a not and compute the so called Wirtinger presentation of the fundamental group. That way you can get lots of examples. – Mariano Suárez-Alvarez Feb 5 '13 at 1:30

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