A group is solvable if you can break it down into cyclic pieces, step by step. This algebraic property is the exact condition for a polynomial to be solvable by radicals — the deepest theorem in Galois theory.

Composition series

A composition series for a group $G$ is a chain of subgroups:

$${e} = G_0 \trianglelefteq G_1 \trianglelefteq G_2 \trianglelefteq \cdots \trianglelefteq G_n = G$$

where each $G_i$ is normal in $G_{i+1}$, and each quotient $G_{i+1}/G_i$ is simple (has no further normal subgroups).

The group $G$ is solvable if every quotient in some composition series is cyclic (isomorphic to $\mathbb{Z}_p$ for some prime $p$).

The chains for degrees 2 through 5

$S_2$: $S_2 \supset E$. Quotient: $\mathbb{Z}_2$ (cyclic). Solvable ✓

$S_3$: $S_3 \supset A_3 \supset E$. Quotients: $\mathbb{Z}_2$, $\mathbb{Z}_3$ (both cyclic). Solvable ✓

$S_4$: $S_4 \supset A_4 \supset K \supset \mathbb{Z}_2 \supset E$. Quotients: $\mathbb{Z}_2$, $\mathbb{Z}_3$, $\mathbb{Z}_2$, $\mathbb{Z}_2$. Solvable ✓

$S_5$: $S_5 \supset A_5 \supset E$. The quotient $S_5/A_5 \cong \mathbb{Z}_2$ is cyclic. But $A_5/E \cong A_5$ is simple of order 60 — and not cyclic. Not solvable ✗

The chain for $S_5$ can’t be refined further because $A_5$ has no normal subgroups. It’s simple, it’s big (60 elements), and it’s not cyclic. The chain gets stuck.

Why the connection works

Each step in the Galois correspondence matches:

Adjoining a $p$-th root creates a field extension of degree $p$ whose Galois group is cyclic of order $p$. So solving by radicals means descending through cyclic quotients — exactly the definition of solvability.

The fundamental theorem

Galois’s theorem: A polynomial is solvable by radicals if and only if its Galois group is a solvable group.

For degree 2, 3, and 4: the symmetric groups $S_2$, $S_3$, $S_4$ are all solvable. So the general quadratic, cubic, and quartic all have radical solutions.

For degree 5: the symmetric group $S_5$ is not solvable. So the general quintic does not have a radical solution.

This doesn’t mean specific quintics can’t be solved — $x^5 - 1 = 0$ has the solution $x = 1$ and the fifth roots of unity. It means there’s no general formula involving only radicals that works for all quintics.