# Examples of $X$ such that $\pi_1(X)$ not abelian?

I've come up with some examples to apply the Hurewicz theorem to compute $H_1(X)$.

This is only interesting if $\pi_1(X)$ is not abelian. The only examples of $X$ such that $\pi_1(X)$ not abelian I can come up with are $\vee_i S^1$ and $\Sigma_g$ the surface of genus $g$ for $g > 1$.

Does anyone know any other examples, preferably easy ones? Many thanks for your help!

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Very easy examples would be products of the ones you know already :) – t.b. Jul 31 '11 at 16:44
your first comment: I was looking for something more interesting, yet not too difficult. : ) – Rudy the Reindeer Jul 31 '11 at 16:48
More seriously (I temporarily confused myself): Take $S^3$ identified with the unit quaternions. The quaternions $\pm 1, \pm i, \pm j, \pm k$ form the quaternion group $Q$. The quotient $S^3/Q$ has fundamental group $Q$. Interesting examples also arise from knot complements (try to compute the fundamental group of the complement of the trefoil knot, for example). – t.b. Jul 31 '11 at 16:56
@Matt: are you aware that every group is the fundamental group of some space? It's not particularly hard to find a finite $2$-complex whose fundamental group is any finitely presented group, either (start with a wedge of circles, then glue in $2$-cells corresponding to the relations), and their homology isn't difficult to compute. – Qiaochu Yuan Jul 31 '11 at 16:58
@Qiaochu's answer here explains his comment in somewhat more detail and lhf gives the precise reference to Hatcher. – t.b. Jul 31 '11 at 17:06

One more example: complement to any non-trivial knot in $S^3$ has non-abelian $\pi_1$.

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Thank you, this is my favourite answer!! – Rudy the Reindeer Aug 3 '11 at 19:13

I am late to the party here, but take any finitely presented group, say $G=\langle x_1,\ldots , x_n: r_1,\ldots r_m \rangle$. Take a bouquet of circles embedded in 4-space and a tubular neighborhood thereof. This is a space with fund. group free of rank $n$. Now carve out a tubular neighborhood of the words represented by the relations, and sew in $D^2 \times S^2$s in their place. There is plenty of room to do this in 4-space. So you can construct a $4$-manifold with $\pi_1$ being any finitely presented group that you like.

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Late to the party or not, at least you gave the first reasonable answer in this thread. In case you don't know it, there's a beautiful and (sort of) related paper by Martin Bridson, The geometry of the word problem. – t.b. Aug 1 '11 at 0:12

Consider the figure 8. Using Van Kampen's theorem, you know that its fundamental group is the free product of the additive group integers with itself. Said group is nastily non-abelian.

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Matt already gave the bouquets of circles in his question :) – t.b. Jul 31 '11 at 19:34

Try the cube with a twist

Take the quotient space of the cube $I^3$ obtained by identifying each square face with opposite square via the right handed screw motion consisting of a translation by 1 unit perpendicular to the face, combined with a one-quarter twist of its face about it's center point.

I think it is a problem in Hatcher's book somewhere

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I believe the fundamental group of this quotient ends up being precisely the $8$-element quaternion group. – Dylan Yott Mar 29 '13 at 22:51