# Preissmann and Byers Theorems

I'm starting to study at the elementary level the relationship between topology and geometry of a Riemannian manifold of negative curvature. The first two theorems, simple and interesting in this direction are:

$\bf{Preissmann~~ Theorem}:$ Let be $M$ a Riemannian manifold with sectional curvature $K<0$, then every non trivial Abelian subgroup of the fundamental group $\pi_1(M)$ is cyclic infinite.

$\bf{Byers~~ Theorem}:$ Let be $M$ a Riemannian manifold with sectional curvature $K<0$, then every no trivial solvable subgroup of the fundamental group $\pi_1(M)$ is cyclic infinite, and $\pi_1(M)$ have no cyclic subgroup of finite index.

$\bf{My ~~Question}:$ I'm looking for nontrivial examples (counterexamples) for Byers Theorem, i.e, non trivial examples of a Riemannian manifold that:

1. Has a no solvable fundamental group.

EDIT:

2 Has a cyclic $\bf{infinite}$ subgroup of finite index of the fundamental group.(In this case the trivial examples are welcome.)

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simultaneous post at math.stackexchange.com/questions/156571/… – Will Jagy Jun 10 '12 at 18:08
I'm just trying to learn mathematics. – Juan Valdez Jun 10 '12 at 19:20
You need to add the hypothesis that $M$ is compact in the statement of these theorems. – Ian Agol Jun 10 '12 at 21:43
of course, thanks. – Juan Valdez Jun 11 '12 at 15:13

1) To construct a manifold with no solvable fundamental group, take for example a finite unsolvable group $G$ and embed it into $SU(n)$ as a discrete subgroup. This embedding is obtained by realizing $G$ as a subgroup of the permutations $S(G)$ of the set $G$, then note $S(G)$ is isomorphic to a subgroup of $SU(n)$ via monomial matrices ($n=|G|$). The quotient $SU(n)/G$ carries a natural manifold structure with fundamental group $G$. Note these examples are compact. Taking product with $R$, one obtains non-compact examples.
2) For manifolds with cyclic subgroups of finite index, consider products of real projective spaces $RP^n$ or lens spaces $S^{2n+1}/Z_p$ with manifolds with finite fundamental group.
Then take an $S^1$ in place of $RP^n$ or the lens space. – Malte Jun 10 '12 at 21:13