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