Unsolvability of a Quintic and its link with "Simplicity" of $A_{5}$

See Thereom 1 from "The Topological Proof of Abel–Ruffini Theorem," Henryk Żolądek (link), which I repeat below:

Theorem 1.

The groups S(n), n ≥ 5, are not solvable.

Proof

(We follow the book of J. Browkin [2]). Because the alternating group $A(n)$ is normal subgroup of $S(n)$ with two-element quotient group, it is enough to show that $A(n)$ is not solvable. But this follows from the following observation. If the cycles $\sigma = (123)$ and $\tau = (345)$ (with one common element) belong to a subgroup $H \subset A(n)$, then the elements $[\sigma, \tau] = (\sigma(3)\sigma(4)\sigma(5)) \cdot \tau^{−1} = (145) \cdot (354) = (143)$ and $[\sigma^{−1}, \tau^{−1}] = (253)$ belong to the commutator $H^{(1)}$. The latter are also cycles with one common element.

Repeating this argument we see that all the derivative groups $A(n)^{(j)}$ contain two cycles with one common element. Therefore none of them can be trivial. 􏰆

The paper uses the definition of solvable where $G$ is solvable if $G^{(r)}=\{e\}$ for some finite $r$, where $G^{(k + 1)} = \left(G^{(k)}\right)^{(1)}$ (the derivative groups), and $G^{(1)} = [G, G]$ (the commutator group).

The proof obviously doesn't work for $n < 5$ because you can't find two three-cycles in $A(n)$ with one common element in those cases.