TITLE: Homotopy groups of Moore spaces
QUESTION [11 upvotes]: Is there anything known about the homotopy groups of the Moore spaces $M(\mathbb Z_m,n)$ if $m\neq 2$ and $n \geq 2$?

REPLY [5 votes]: Let's consider $M(\mathbb{Z}/p^r, n)$ for large $n$, so that $\pi_{i+n}(M(\mathbb{Z}/p^r, n) \cong \pi_{i+n+1}(M(\mathbb{Z}/p^r, n+1)$ i.e. we are in the stable range. I think, the homotopy groups of $M(\mathbb{Z}/p^r, n)$ stabilizes for $n \geq 4$. Then the problem of computing the homotopy group of $M(\mathbb{Z}/p^r, n)$ is essentially computing the stable homotopy groups of a spectrum, called the $r$-th Moore spectrum and is denoted by $M_p(r)$. By a Theorem of Hopkins and Smith (see, nilpotence and stable homotopy theory II) there exists a periodic family of self-maps 
$$ (v)^k: \Sigma^{2kt(p-1)}M_p(r) \to M_p(r) $$
where $t$ is a function of $r$ and $k$ is any positive integer. This map has the property that the composite
$$\overline{v}(k):S^{2kt(p-1)} \to \Sigma^{2kt(p-1)}M_p(r) \to M_p(r) $$
is non-trivial element in stable homotopy groups of $M_p(r)$ for all $k$. Moreover, $\overline{v}(k)$ is $p^r$-torsion for all $k$. It is also known that $t$ gets arbitrarily large as $r$ gets large. 
Therefore, we can conclude that 
$$ \pi_{n+2kt(p-1)}(M(\mathbb{Z}/p^r, n)) \neq 0 $$  for all $n \geq 4$ and $k \geq 1$ as there is an infinite family of generators
 $$ v(k) \in \pi_{n+2kt(p-1)}(M(\mathbb{Z}/p^r, n)) $$
which are $p^r$-torsion. 
I do not know if $t$ is known for all values of $r$, but is known for small values of $r$. For example at prime $2$, $t=4$ when $r=1$ and at odd primes $t=1$ when $r=1$.