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Show that $\pi_7(S^4)$ contains an element of infinite order.

Now, I know that I should probably use the Hopf bundle here somewhere. I also know that $\pi_3(S^7) = 0$. But I am stuck. Can anyone help me?

Math
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    This is part of a more general theorem of Serre that $\pi_{4n-1}(S^{2n})\otimes\mathbb Q=\mathbb Q$, and in fact these are the only non-finite homotopy groups of spheres except $\pi_n(S^n)$. The easiest way I know to see this is rational homotopy theory. The proof will really depend on your background. – Cheerful Parsnip Sep 03 '15 at 19:05
  • Do you mean $\pi_7(S^3) = 0$? –  Sep 03 '15 at 19:06
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    @GrumpyParsnip: I don't know any better way to prove that very general theorem than rational homotopy theory, but the argument below proves $\pi_3(S^2), \pi_7(S^4)$, and $\pi_{15}(S^8)$ all have elements of infinite order. –  Sep 03 '15 at 19:08
  • @JohnMa: No, that $\pi_3(S^7) = 0$ is relevant. –  Sep 03 '15 at 19:08
  • @MikeMiller: indeed, I was just reading this argument on Wikipedia and was about to sketch it in an answer. You beat me to it though! – Cheerful Parsnip Sep 03 '15 at 19:10

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As you implied, there is a fiber bundle $S^3 \hookrightarrow S^7 \to S^4$ given by thinking of $S^4$ as the quaternionic projective line $\Bbb{HP}^1$. Pass to the long exact sequence of homotopy groups of a fibration to obtain the exactness of $$\pi_7(S^3) \to \pi_7(S^7) \to \pi_7(S^4).$$ Because the inclusion of the fiber $S^3 \to S^7$ is null-homotopic, we have that the map $\pi_7(S^3) \to \pi_7(S^7)$ is zero; so we have the exact sequence $$0 \to \pi_7(S^7) \to \pi_7(S^4).$$ But $\pi_7(S^7) \cong \Bbb Z$ and is generated by the identity map. So it maps to an element of infinite order (the quaternionic Hopf bundle).

Note that we didn't really need the fact that $\pi_3(S^7) = 0$ - the inclusion of the fiber is automatically not surjective, so just invoke the fact that $S^7 \setminus \{pt\}$ is contractible (it's $\Bbb R^7$!).

This exact same argument with the complex numbers or octonions proves $\pi_3(S^2)$ and $\pi_{15}(S^8)$ have an element of infinite order, too.