Toy example for computing stable homotopy group. I am currently reading Hatcher's (Algebraic Topology) explanation of stable homotopy groups. My understanding may be a bit shaky and I made a sort of toy example.
I am assuming that $i$ is fixed and arbitrary, and that $X$ can be any $CW$ complex. For example, let $i=4$ and $X=S^3$, then we have the following sequence 
$$\pi_4(S^3) \to \pi_5(S^4) \to \pi_6(S^5) \to \pi_7(S^6)$$ 
where we can interpret $S^4 = \Sigma(S^3)$, $S^5 = \Sigma^2(S^3)$. Via Freudenthal suspension theorem we find that $\pi_6(S^5) \cong \pi_7(S^6)$ and for all following elements in the sequence. 
Hatcher notates the stable homotopy group as $\pi_i^s(X)$. 
In this case do we write $\pi_4^2(S^3) = \pi_6(S^5)$? As in, do we say that $\pi_6(S^5)$ is the stable homotopy group corresponding to the choice $i=4$ and $X= S^3$ ?
 A: The Fredenthal Suspension Theorem states that the suspension homomorphism $\pi_{n+k}(S^n) \to \pi_{n+k+1}(S^{n+1})$ is an isomorphism for $n > k + 1$.
Taking $n = 3$ and $k = 1$, we see that the suspension homomorphism $\pi_4(S^3) \to \pi_5(S^4)$ is an isomorphism as $3 > 1 + 1$.
As we continue to take the suspension homomorphism, we maintain the same value of $k$ but $n$ increases, so $n > k + 1$ is still true, and hence the homomorphisms will be isomorphisms. That is, we get
$$\pi_4(S^3) \cong \pi_5(S^4) \cong \pi_6(S^5) \cong \pi_7(S^6) \cong \dots$$
This group, whatever it is, is what we call the stable homotopy group $\pi_1^S = \pi_1^S(S^0)$. In this case, the group is $\mathbb{Z}_2$, so we would write $\pi_1^S \cong \mathbb{Z}_2$.
Note, when $n \leq k + 1$, the groups $\pi_{n+k}(S^n)$ and $\pi_{n+k+1}(S^{n+1})$ may not be isomorphic, and therefore $\pi_k^S$ may not be isomorphic to $\pi_{n+k}(S^n)$ for every $n$. For example, $\pi_3(S^2) \cong \mathbb{Z}$ but $\pi_4(S^3) \cong \mathbb{Z}_2 \cong \pi_1^S$. 
However, $\pi_k^S \cong \pi_{n+k}(S^n)$ for $n > k + 1$, which justifies the notation
$$\pi_k^S \cong \lim_{n\to\infty}\pi_{n+k}(S^n) = \lim_{n\to\infty}\pi_{n+k}(\Sigma^nS^0).$$
In particular, $\pi_k^S \cong \pi_{2k+2}(S^{k+2})$; the case above corresponds to $k = 1$.

In general, for an $(n-1)$-connected CW complex $X$, the suspension homomorphism $\pi_k(X) \to \pi_{k+1}(\Sigma X)$ is an isomorphism for $k < 2n-1$. 
Now suppose $X$ is a CW complex and consider the sequence of suspension homomorphisms
$$\pi_k(X) \to \pi_{k+1}(\Sigma X) \to \pi_{k+2}(\Sigma^2 X) \to \pi_{k+3}(\Sigma^3 X) \to \dots$$
Note that $\Sigma^n X$ is $(n-1)$-connected, so the suspension homomorphism $\pi_{k+n}(\Sigma^n X) \to \pi_{k+n+1}(\Sigma^{n+1} X)$ is an isomorphism for $k + n < 2n - 1$ (i.e. $n > k + 1$). So we see that
$$\pi_{2k+2}(\Sigma^{k+2} X) \cong \pi_{2k+3}(\Sigma^{k+3} X) \cong \pi_{2k+4}(\Sigma^{k+4} X) \cong \pi_{2k+5}(\Sigma^{k+5} X) \cong \dots$$
This group is the stable homotopy group $\pi_k^S(X)$ and one writes
$$\pi_k^S(X) = \lim_{n\to\infty}\pi_{n+k}(\Sigma^n X).$$
Note that $\pi_k^S(X) \cong \pi_{2k+2}(\Sigma^{k+2}X)$. 
From this point of view, the stable homotopy group you were trying to calculate was $\pi_4^S(S^3)$; so you would write $\pi_4^S(S^3) \cong \mathbb{Z}_2$. To see that $\pi_4^S(S^3)$ coincides with $\pi_1^S = \pi_1^S(S^0)$, note that for $i < k$
$$\pi_k^S(\Sigma^i X) = \lim_{n\to\infty}\pi_{n+k}(\Sigma^n\Sigma^i X) = \lim_{n\to\infty}\pi_{n+k}(\Sigma^{n+i} X) = \lim_{n\to\infty}\pi_{(n+i)+(k-i)}(\Sigma^{n+i} X) = \lim_{N\to\infty}\pi_{N + (k-i)}(\Sigma^N X) = \pi_{k-i}^S(X).$$
Therefore, $\pi_4^S(S^3) = \pi_4^S(\Sigma^3 S^0) = \pi_1^S(S^0) = \pi_1^S$.
