Let $SX$ be the suspension of CW complex. What are some results available to determine the homotopy groups of $SX$?



How to determine the homotopy groups of the suspension of a space?
This, in general, an incredibly difficult problem. Even we just want to compute the *rational* homotopy groups of the suspension of $X$ and $X$ is simply connected, where we can do everything using rational homotopy theory to reduce things to commutative DGAs, this starts to involve things like free Lie algebras and the like.


Locally at the prime 2, there is actually a famous long exact sequence when $X$ is a sphere called the EHP long exact sequence. It relates the suspension homomorphism, the Hopf map, and a "whitehead product" map. This gives rise to the [EHP spectral sequence](http://en.wikipedia.org/wiki/EHP_spectral_sequence) that, funnily enough, starts with the 2-local homotopy groups of *odd*-dimensional spheres and computes the 2-local homotopy groups of spheres. Miller and Ravenel have a paper titled "Mark Mahowald's work on the homotopy groups of spheres" that covers some of this material in detail.


Another approach is to say: The "stable" homotopy groups of $X$ are a first-order approximation using the Freudenthal suspension theorem that Andrea mentioned. There is then a "quadratic" correction term that you can try to use to get an approximation of the homotopy groups that is correct out to roughly three times the connectivity, and so on. These lead into the subject of Goodwillie calculus.


For a classifying space $K(G,1)$, neither of these approaches work very well, because the higher homotopy groups are going to depend pretty intricately on your group itself. For instance, $\pi\_3$ of the suspension of the classifying space of a *free* group is the set of symmetric elements in $G\_{\text{ab}}\otimes G\_{\text{ab}}$ where $G\_{\text{ab}}$ is the abelianization of $G$, and for a general group it lives in an exact sequence between something involving such symmetric elements and the second group homology of $G$. I don't know a closed form for it but maybe someone else knows better.


**EDIT**: Let me at least be precise, there's an exact sequence
$$\pi \_4 (\Sigma BG)\rightarrow H\_3 G \rightarrow (G\_{\text{ab}}\otimes G\_{\text{ab}})^{\mathbb Z/2} \rightarrow \pi\_3 (\Sigma BG)\rightarrow H\_2 G\rightarrow 0$$
Note that for $R$ a ring, an element of $(G\_{\text{ab}}\otimes G\_{\text{ab}})^{\mathbb Z/2}$ gives rise to an $R$-valued symmetric bilinear pairing on $\mathsf{Hom}(G\_{\text{ab}},R)$.


**EDIT FOR THE FINAL TIME**: sorry for the multiple revisions, switching back and forth between homology and cohomology gave me errors. the exact sequence above should be correct now.


