Worked solution: Abel–Ruffini theorem and the Galois solvability criterion (1824)

For the criterion in Step 5 to make the generic quintic unsolvable by radicals, $S_5$ itself must fail to be a solvable group. For $n\ge5$, the symmetric group $S_n$ has exactly one nontrivial proper normal subgroup, the alternating group $A_n$ (the even permutations), so $1\triangleleft A_n\triangleleft S_n$ is the only possible composition series. For $n\ge5$, $A_n$ is simple — it has no normal subgroups at all besides itself and $\{1\}$ — and it is not abelian (for instance in $A_5$, two different $3$-cycles typically fail to commute). A simple non-abelian group can never be replaced in a subnormal chain by anything smaller, so the chain $1\triangleleft A_5\triangleleft S_5$ cannot be refined into one with only abelian quotients: the quotient $A_5/1\cong A_5$ is itself non-abelian. Hence $S_5$ is not a solvable group.