Order of the suspension of the Hopf map I am working on the following problem from an old qualifying exam in algebraic topology:

The Hopf map $\eta:S^3 \rightarrow S^2$ can be defined as $\eta(z_0,z_1) = z_0/z_1$, with $S^3$ the unit sphere in $\mathbb{C}^2$ and $S^2 = \mathbb{C} \cup \infty$. (a) Show that $k \eta \simeq \eta k^2$, where $k$ and $k^2$ denote maps of degree $k$ and $k^2$, respectively.
(b) Using the fact that, after suspension, the compositions here are commutative, show that $\Sigma \eta$ has (additive) order at most $2$.
(c) Show that $\Sigma \eta$ has (additive) order exactly $2$.

I was able to complete part (a), but I'm struggling with completing parts (b) and (c).
In part (b), I know that suspension is a covariant (homotopy) functor, so from the homotopy we showed in part (a), we have $\Sigma(k \eta) \simeq \Sigma (\eta k^2)$ $\Rightarrow$ $\Sigma k \circ \Sigma \eta \simeq \Sigma \eta \circ \Sigma k^2$. I'm not quite sure how to proceed from here to show that $\Sigma \eta$ has additive order at most $2$, though -- for example, where does the commutative property of suspensions mentioned come in handy here?
For part (c), I found something related in the answer to this post, where the user comments that the Freudenthal suspension theorem (which I'm most familiar with of all the ideas expressed there) can be used here. Indeed, by the Freudenthal suspension theorem, the suspension map $\pi_3(S^2) \rightarrow \pi_4(S^3)$ is surjective. I fail to see why this gives that $\Sigma \eta$ has additive order exactly $2$. Firstly, is it easy to see why $\pi_4(S^3) \cong \mathbb{Z}_2$? Then, how does surjectivity of the suspension map $\pi_3(S^2) \rightarrow \pi_4(S^3)$ give that $\Sigma \eta$ is a generator of $\pi_4(S^3) \cong \mathbb{Z}_2$?
Any thoughts would be appreciated.
Thanks!
 A: The point is that $\eta\circ k=\eta+\eta+\dots+\eta$ is the $k$-fold sum of $\eta$ in $\pi_3S^2$. In particular $\eta\circ(-1)=-\eta$. Now,
$$k\circ\eta=\eta\circ k^2$$
and
$$\Sigma(k\circ\eta)=\Sigma k\circ\Sigma\eta=\Sigma\eta\circ\Sigma k=\Sigma(\eta\circ k)$$
by the commutativity of these elements after suspension. Hence
$$\Sigma(\eta\circ k)=\Sigma(\eta\circ k^2).$$
With $k=-1$ we have
$$\Sigma(\eta\circ(-1))=\Sigma(-\eta)=-\Sigma\eta$$
on the left, and
$$\Sigma(\eta\circ(-1)^2)=\Sigma(\eta\circ 1)=\Sigma\eta$$
on the right. In particular
$$\Sigma\eta=-\Sigma\eta,$$
showing that this element has order no greater than $2$.
Thus to show that $\Sigma\eta$ has order $2$ it will suffice to show that it is nontrivial. One way to accomplish this is by noting that the mapping cone construction commutes with suspension. In particular, the mapping cone of $\Sigma\eta$ is $\Sigma\mathbb{C}P^2$. If $\Sigma\eta$ were null-homotopic, then $\Sigma\mathbb{C}P^2$ would be homotopy equivalent to $S^3\vee S^5$ (which is the mapping cone of the constant map $S^4\rightarrow S^3$). That $\Sigma\mathbb{C}P^2$ does not split as a wedge of spheres is a consequence of the non-trivial Steenrod square
$$Sq^2:H^3(\Sigma\mathbb{C}P^2;\mathbb{Z}/2)\rightarrow H^5(\Sigma\mathbb{C}P^2;\mathbb{Z}/2).$$
If this is unfamiliar, then let us stick to the EHP sequence. We know that the suspension $\Sigma:\pi_3S^2\rightarrow\pi_4S^3$ is surjective by Freudenthal's Theorem. Since $\eta$ generates $\pi_3S^2\cong\mathbb{Z}$, and $\Sigma\eta$ has order $\leq2$, the group $\pi_4S^3$ is either $\mathbb{Z}/2$ or is trivial. Both of these groups are devoid of odd torsion, so to answer our problem it will suffice to work 2-locally.
Thus introduce the EHP sequence,
$$\dots\rightarrow\pi_5S^5\xrightarrow{P}\pi_3S^2\xrightarrow\Sigma\pi_4S^3\rightarrow 0$$
which becomes exact after localisation at $2$. By exactness, the right-hand group is trivial if and only if $P$ is surjective. But it is well-known that
$$P(\iota_5)=[\iota_2,\iota_2],$$
where $[\iota_2,\iota_2]$ is the Whitehead product of $\iota_2$ with itself. I calculated the valued of this Whitehead product here (at least up to sign), showing that
$$[\iota_2,\iota_2]=\pm2\eta$$
(the exact sign will depend on your particular conventions).
The point is that $P$ is not surjective, so it must be that $\pi_4S^3\cong\mathbb{Z}/2$, generated by $\Sigma\eta$.
