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My question is described well in the title.

As you know, $\pi_3(S^2) = {\bf Z}$

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One possible approach is via the long exact sequence of homotopy groups as applied to the Hopf fibration $S^1\rightarrow S^3\rightarrow S^2$. I'm assuming you already know $\pi_k(S^k)\cong \mathbb{Z}$ and $\pi_k(S^1) = 0$ for $k\geq 2$.

The LES looks like $...\rightarrow \pi_k(S^1)\rightarrow \pi_k(S^3)\rightarrow \pi_k (S^2)\rightarrow \pi_{k-1}(S^1)\rightarrow ...$ and applied to any fibration, not just the Hopf fibration.

From this, using the fact that $\pi_k(S^1) = 0$ for $k\geq 2$, we learn that $\pi_k(S^3)\cong \pi_k(S^2)$ for all $k\geq 3$. In particular, $\pi_3(S^2)\cong\pi_3(S^3) \cong \mathbb{Z}$.

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    On second thought, I don't actually know how to make the group structure work out. All I can show using Pontrjagin-Thom is that there is naturally a $\mathbb{Z}$s worth of elements in $\pi_3(S^2)$.2012-11-12