I believe the following statement is true, and I've even seen it referenced here. Could someone point me to a proof?
The fundamental group of a closed hyperbolic 3manifold is not a free product.
I believe the following statement is true, and I've even seen it referenced here. Could someone point me to a proof?



If $M$ is a closed $3$manifold and $\pi_1(M) \cong A \ast B$ with $A$ and $B$ nontrivial, then Kneser's conjecture (which is a theorem  the proof can be found in Hempel's book on 3manifolds) says that we can write $M = M' \sharp M''$ where $M'$ and $M''$ are closed 3manifolds with $\pi_1(M')=A$ and $\pi_1(M'')=B$. In particular, $M$ contains an embedded $2$sphere which does not bound a ball (namely, the sphere from the connect sum decomposition). However, all embedded $2$spheres in hyperbolic 3manifolds bound balls, as can be seen by lifting to the universal cover. 


One can also see it using the theory of ends. If $\pi_1M$ were freely decomposable, then it would follow from the easy direction of Stallings' Ends Theorem that $\pi_1M$ had two or infinitely many ends. On the other hand, by the SvarcMilnor Lemma, $\pi_1M$ is quasiisometric to hyperbolic 3space, which has one end. Because ends are a quasiisometry invariant, we are done. This argument shows that this isn't really a 3manifold fact. Indeed, it proves:


