TITLE: Intuition on finite homotopy groups
QUESTION [16 upvotes]: As I have been studying algebraic topology, something that I found puzzling was the existence of finite homotopy groups. For instance, $\pi_{4}(S^{2})\cong\pi_{5}(S^{4})\cong\mathbb{Z}/2\mathbb{Z}$. I was wondering if there was any kind of intuitive reason for why this might be true, and if there are spaces $X$ such that $\pi_{1}(X)$ is finite. Speaking very roughly, it would seem that a finite, nontrivial fundamental group means that if you repeat a closed path enough times, it can be contracted to a point, something which I find rather hard to visualize. So the question is: Is there any intuitive reason for the existence of finite homotopy groups?

REPLY [3 votes]: Perhaps it isn't necessary to add another answer, but here is one more nevertheless.

Given any discrete group $G$, form the classifying space $BG$. A map of $S^1$ to $BG$
is equivalent to giving a principal $G$-bundle over $S^1$. Such a bundle can formed by
gluing the ends of the trivial bundle $[0,1]\times G$ with a twist given by an element
of $G$. So in this way, we get $\pi_1(BG)=G$.
Of course, invoking $BG$ is perhaps violating the spirit of the question. 
When $G$ is finite, we can approximate
this by something more concrete. Let $V$ be a faithful unitary representation of $G$ with no
trivial summands (e.g. the complement of $\mathbb{C}$ in the regular representation).
After replacing $V$ by $V\oplus V$ if necessary, we can assume that $\dim V>1$. 
Then the unit sphere $S\subset V$ will be simply connected, and $G$ will act on it without
fixed points. So the quotient $S/G$ will have fundamental group $G$.
This is the same construction used (implicitly) in many of the other answers.