Prime 3-Manifolds I'm trying to understand what the prime 3-manifolds are. Milnor's Prime Decomposition Theorem states that any compact, orientable 3-manifold is a connected sum of a unique finite collection of prime 3-manifolds.
I've read that these prime manifolds will be one of the following:

*

*$S^3$ or any quotient of $S^3$


*$S^2 \times S^1$


*Some $K(\pi, 1)$ manifold
I'm wondering where this statement comes from. Why these are the only prime 3-manifolds. My thoughts are it could be something to do with $\pi_2$, because Milnor's paper states a 3 manifold is prime $\iff$ $S^2 \times S^1$ or irreducible. The latter condition $\implies$ $\pi_2 =0$.
Thus the prime 3 manifolds are $S^2 \times S^1$ and those with trivial second homotopy group. It is here where I become stuck since I have not formally studied second homotopy groups before.
My question is how do we conclude that the the only 3 options are those stated above?
 A: Here are the details, which use a lot of additional tools. You can find all these arguments (except for the application of Perelman's theorem) in older 3-manifolds textbooks such as Hempel's "3-manifolds".
Let $M$ be a closed, orientable 3-manifold.
First, the sphere theorem of Papakyriakopolous says that if $\pi_2(M)$ is nontrivial then there exists an embedding $S^2 \approx S \subset M$ which represents a nontrivial element of $\pi_2(M)$. Nontriviality implies that $S$ does not bound a 3-ball. It follows that if $S$ is separating (or indeed if any other separating sphere exists in $M$ that does not bound a 3-ball), then $M$ is not prime. Finally, if every separating sphere in $M$ bounds a 3-ball, and if $S$ is nonseparating, then a pretty easy separate argument shows $M$ is homeomorphic to $S^2 \times S^1$. Therefore, the conclusion that you desire is true in the case that $\pi_2(M)$ is nontrivial.
So now let's assume that $\pi_2(M)$ is trivial.
Consider the universal covering space $\widetilde M$.
If $\pi_1(M)$ is finite then $\widetilde M$ is a closed, simply connected 3-manifold. Before the work of Perelman, the best conclusion one could make here was that $\widetilde M$ is homotopy equivalent to $S^3$. But nowadays, by applying Perelman's Theorem, which says that the Poincare conjecture is true, it follows that $\widetilde M$ is homeomorphic to $S^3$, and therefore $M$ is a quotient of $S^3$. (A broader statement of Perelman's theorem says that Thurston's whole geometrization conjecture is true, and therefore $M$ is a spherical space form, meaning that $M$ is a quotient of $S^3$ with a deck transformation group consising of isometries of $S^3$).
If $\pi_1(M)$ is infinite, then $\widetilde M$ is noncompact. Applying the long exact homotopy sequence of a fibration to the universal covering map $\widetilde M \to M$, it follows that $\pi_2(\widetilde M)$ is homeomorphic to $\pi_2(M)$ and hence is trivial. Since $\pi_1(\widetilde M)$ and $\pi_2(\widetilde M)$ are both trivial, it follows from the Hurewicz theorem that $\pi_3(\widetilde M)$ and $H_3(\widetilde M)$ are isomorphic. But noncompactness of $\widetilde M$ implies that $H_3(\widetilde M)$ is trivial; this is one of the side-results accompanying the Poincare duality theorem. Using that $M$ has a 3-dimensional triangulation, and hence so does $\widetilde M$, it follows that all of the higher homology groups $H_n(\widetilde M)$ are also trivial. Again, by applying Hurewicz inductively, it follows that all of the higher homotopy groups $\pi_n(\widetilde M)$ are trivial. Therefore, $\widetilde M$ is contractible and $M$ is a $K(\pi,1)$ space.
