[proofplan] We show that every polynomial of degree at most four over a characteristic-zero field is solvable by radicals. The strategy is to invoke the [Solvability by Radicals and Solvable Galois Groups](/theorems/1321) criterion, which reduces the problem to showing that the Galois group of such a polynomial is solvable. The Galois group of a degree-$n$ polynomial embeds as a transitive subgroup of $S_n$, so it suffices to show that every subgroup of $S_n$ is solvable for $n \le 4$. We establish this by exhibiting an explicit composition series for $S_4$ with abelian successive quotients, which implies $S_4$ is solvable; since every subgroup of a solvable group is solvable, the result follows. [/proofplan] [step:Identify the Galois group as a subgroup of $S_n$ for $n = \deg f \le 4$] Let $f \in K[t]$ with $n := \deg f \le 4$. We may assume $n \ge 1$ (constant polynomials have no roots to solve for). Let $L$ be a splitting field of $f$ over $K$, and let $\alpha_1, \ldots, \alpha_n$ be the roots of $f$ in $L$ (counted without multiplicity in the splitting field). Every $\sigma \in \operatorname{Gal}(L/K)$ permutes $\{\alpha_1, \ldots, \alpha_n\}$ because $\sigma$ fixes the coefficients of $f$. The map \begin{align*} \operatorname{Gal}(L/K) &\to S_n \\ \sigma &\mapsto \text{the permutation } \pi_\sigma \text{ defined by } \sigma(\alpha_i) = \alpha_{\pi_\sigma(i)} \end{align*} is an injective group homomorphism (injectivity follows because $L = K(\alpha_1, \ldots, \alpha_n)$ and a $K$-automorphism is determined by its action on the generators). Therefore $\operatorname{Gal}(L/K)$ is isomorphic to a subgroup of $S_n$ with $n \le 4$. [guided] Why does the Galois group embed into $S_n$? The splitting field $L$ is generated over $K$ by the roots $\alpha_1, \ldots, \alpha_n$ of $f$. Every $K$-automorphism $\sigma$ of $L$ must send each root of $f$ to another root of $f$: if $f(\alpha_i) = 0$, then $f(\sigma(\alpha_i)) = \sigma(f(\alpha_i)) = \sigma(0) = 0$ because $\sigma$ fixes the coefficients of $f$ in $K$. So $\sigma$ permutes $\{\alpha_1, \ldots, \alpha_n\}$, inducing a permutation $\pi_\sigma \in S_n$. The map $\sigma \mapsto \pi_\sigma$ is a group homomorphism: $(\sigma \circ \tau)(\alpha_i) = \sigma(\alpha_{\pi_\tau(i)}) = \alpha_{\pi_\sigma(\pi_\tau(i))}$, so $\pi_{\sigma \circ \tau} = \pi_\sigma \circ \pi_\tau$. It is injective: if $\pi_\sigma$ is the identity permutation, then $\sigma(\alpha_i) = \alpha_i$ for all $i$. Since $L = K(\alpha_1, \ldots, \alpha_n)$ and $\sigma$ fixes both $K$ and every $\alpha_i$, the automorphism $\sigma$ fixes every element of $L$, so $\sigma = \operatorname{id}_L$. Thus $\operatorname{Gal}(L/K) \cong \operatorname{im}(\sigma \mapsto \pi_\sigma) \le S_n$ with $n \le 4$. The proof will be complete once we show that every subgroup of $S_4$ is solvable. [/guided] [/step] [step:Exhibit a subnormal series for $S_4$ with abelian quotients] We construct a subnormal series for $S_4$. Define the Klein four-group \begin{align*} V_4 := \{e, (12)(34), (13)(24), (14)(23)\} \subset A_4. \end{align*} Consider the chain \begin{align*} S_4 \rhd A_4 \rhd V_4 \rhd \langle (12)(34) \rangle \rhd \{e\}. \end{align*} We verify that each inclusion is normal in the preceding group and compute the successive quotients. **$A_4 \lhd S_4$:** The alternating group $A_4$ is the kernel of the sign homomorphism $\operatorname{sgn} \colon S_4 \to \{+1, -1\}$, so $A_4$ is a normal subgroup of $S_4$. The quotient is $S_4 / A_4 \cong \mathbb{Z}/2\mathbb{Z}$, which is abelian. **$V_4 \lhd A_4$:** We verify that $V_4$ is normal in $A_4$ by observing that the three non-identity elements of $V_4$ are exactly the three elements of $S_4$ that have cycle type $(2,2)$ (products of two disjoint transpositions). Conjugation in $S_4$ preserves cycle type, so conjugation in $A_4 \subset S_4$ also preserves cycle type. Since $V_4 = \{e\} \cup \{\text{all elements of cycle type } (2,2)\}$, $V_4$ is closed under conjugation by elements of $A_4$, hence $V_4 \lhd A_4$. The quotient has order $|A_4|/|V_4| = 12/4 = 3$, so $A_4/V_4 \cong \mathbb{Z}/3\mathbb{Z}$, which is abelian. **$\langle (12)(34) \rangle \lhd V_4$:** The group $V_4$ is abelian (every element has order at most $2$: $e^2 = e$ and $\sigma^2 = e$ for each non-identity $\sigma \in V_4$ since each is a product of two disjoint transpositions). In an abelian group, every subgroup is normal, so $\langle (12)(34) \rangle \lhd V_4$. The quotient $V_4 / \langle (12)(34) \rangle$ has order $4/2 = 2$, so $V_4 / \langle (12)(34) \rangle \cong \mathbb{Z}/2\mathbb{Z}$, which is abelian. **$\{e\} \lhd \langle (12)(34) \rangle$:** The trivial subgroup is normal in every group. The quotient is $\langle (12)(34) \rangle / \{e\} \cong \langle (12)(34) \rangle \cong \mathbb{Z}/2\mathbb{Z}$, which is abelian. The successive quotients are $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/3\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}$, all of which are cyclic and hence abelian. Therefore $S_4$ is solvable. [guided] A group $G$ is solvable if and only if it admits a subnormal series \begin{align*} G = G_0 \rhd G_1 \rhd \cdots \rhd G_m = \{e\} \end{align*} in which each quotient $G_i / G_{i+1}$ is abelian. For $S_4$, we produce the chain $S_4 \rhd A_4 \rhd V_4 \rhd \langle (12)(34) \rangle \rhd \{e\}$. Let us verify each normality claim and quotient carefully. **First link: $A_4 \lhd S_4$.** The alternating group is the kernel of the sign homomorphism, which maps each permutation to $+1$ or $-1$ according to the parity of its decomposition into transpositions. Kernels of homomorphisms are always normal subgroups. The quotient $S_4/A_4$ has order $|S_4|/|A_4| = 24/12 = 2$, so it is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. **Second link: $V_4 \lhd A_4$.** The Klein four-group $V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$ consists of the identity and all three permutations of cycle type $(2,2)$ in $S_4$. To verify normality, we need $\sigma V_4 \sigma^{-1} = V_4$ for every $\sigma \in A_4$. Conjugation by any permutation preserves cycle type: if $\tau$ has cycle type $(2,2)$, then $\sigma\tau\sigma^{-1}$ also has cycle type $(2,2)$. Since $V_4$ contains *all* elements of cycle type $(2,2)$ (together with the identity), the set $V_4$ is closed under conjugation by any element of $S_4$, and in particular by elements of $A_4$. Hence $V_4 \lhd A_4$. The quotient $A_4/V_4$ has order $12/4 = 3$. A group of prime order is cyclic, so $A_4/V_4 \cong \mathbb{Z}/3\mathbb{Z}$. **Third link: $\langle (12)(34) \rangle \lhd V_4$.** Every element of $V_4$ has order $1$ or $2$, so $V_4 \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. This group is abelian, so every subgroup is normal. The subgroup $\langle (12)(34) \rangle = \{e, (12)(34)\}$ has order $2$, and the quotient $V_4/\langle (12)(34) \rangle$ has order $4/2 = 2$, hence is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. **Fourth link: $\{e\} \lhd \langle (12)(34) \rangle$.** The trivial group is normal in any group. The quotient $\langle (12)(34) \rangle / \{e\} \cong \mathbb{Z}/2\mathbb{Z}$. All four quotients -- $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/3\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}$ -- are cyclic of prime order, hence abelian. The existence of this subnormal series with abelian quotients certifies that $S_4$ is solvable. Why does this series work while no such series exists for $S_5$? The key obstruction for $S_5$ is that $A_5$ is simple (its only normal subgroups are $\{e\}$ and $A_5$ itself), and $A_5$ is non-abelian. So the series $S_5 \rhd A_5 \rhd \{e\}$ has $A_5/\{e\} \cong A_5$, which is non-abelian, and no further refinement is possible. By contrast, $A_4$ is not simple: it contains the normal subgroup $V_4$, which allows the chain to continue. [/guided] [/step] [step:Deduce solvability of $\operatorname{Gal}(f)$ from solvability of $S_4$] From the first step, $\operatorname{Gal}(L/K)$ is isomorphic to a subgroup $H$ of $S_n$ with $n \le 4$, and hence $H \le S_n \le S_4$ (since $S_n$ embeds naturally into $S_4$ for $n \le 4$ by acting on $\{1, \ldots, n\}$ and fixing the remaining elements). Every subgroup of a solvable group is solvable. We verify this: if $G = G_0 \rhd G_1 \rhd \cdots \rhd G_m = \{e\}$ is a subnormal series with abelian quotients and $H \le G$, then $H = H \cap G_0 \rhd H \cap G_1 \rhd \cdots \rhd H \cap G_m = \{e\}$ is a subnormal series for $H$, and the map $H \cap G_i \to G_i/G_{i+1}$ given by $h \mapsto hG_{i+1}$ has kernel $H \cap G_{i+1}$, so $(H \cap G_i)/(H \cap G_{i+1})$ embeds into $G_i/G_{i+1}$. A subgroup of an abelian group is abelian, so each quotient $(H \cap G_i)/(H \cap G_{i+1})$ is abelian. Since $S_4$ is solvable (by the previous step), the subgroup $H \cong \operatorname{Gal}(L/K)$ is solvable. [guided] The principle that subgroups of solvable groups are solvable is a standard result in group theory. Let us spell out the argument in our setting. We showed that $S_4$ has the subnormal series $S_4 \rhd A_4 \rhd V_4 \rhd \langle (12)(34) \rangle \rhd \{e\}$ with abelian quotients. Let $H \le S_4$ be any subgroup. Define $H_i := H \cap G_i$ for each term $G_i$ in the series. This gives the chain \begin{align*} H = H \cap S_4 \supset H \cap A_4 \supset H \cap V_4 \supset H \cap \langle (12)(34) \rangle \supset H \cap \{e\} = \{e\}. \end{align*} We claim $H_{i+1} \lhd H_i$ for each $i$. Let $h \in H_i = H \cap G_i$ and $x \in H_{i+1} = H \cap G_{i+1}$. Then $hxh^{-1} \in H$ (since $H$ is a group and $h, x \in H$) and $hxh^{-1} \in G_{i+1}$ (since $G_{i+1} \lhd G_i$ and $h \in G_i$, $x \in G_{i+1}$). Therefore $hxh^{-1} \in H \cap G_{i+1} = H_{i+1}$, confirming normality. For the quotients, consider the map $\varphi_i \colon H_i \to G_i/G_{i+1}$ defined by $h \mapsto hG_{i+1}$. This is a group homomorphism with $\ker(\varphi_i) = \{h \in H_i : h \in G_{i+1}\} = H \cap G_{i+1} = H_{i+1}$. By the first isomorphism theorem, $H_i/H_{i+1} \cong \operatorname{im}(\varphi_i) \le G_i/G_{i+1}$. Since $G_i/G_{i+1}$ is abelian and a subgroup of an abelian group is abelian, $H_i/H_{i+1}$ is abelian. This shows $H$ has a subnormal series with abelian quotients, so $H$ is solvable. In particular, $\operatorname{Gal}(L/K) \cong H$ is solvable. Note that this argument works for $n \le 4$ but fails for $n = 5$: $S_5$ is not solvable (because $A_5$ is simple and non-abelian), so there exist subgroups of $S_5$ that are not solvable -- for instance, $S_5$ itself and $A_5$. [/guided] [/step] [step:Apply the Galois solvability criterion to conclude $f$ is solvable by radicals] Since $\operatorname{char} K = 0$ and $\operatorname{Gal}(L/K)$ is solvable, the [Solvability by Radicals and Solvable Galois Groups](/theorems/1321) criterion (which states that a polynomial over a characteristic-zero field is solvable by radicals if and only if its Galois group is solvable) implies that $f$ is solvable by radicals. This holds for every polynomial $f \in K[t]$ with $\deg f \le 4$, completing the proof. [guided] The [Solvability by Radicals and Solvable Galois Groups](/theorems/1321) theorem provides the bridge between the group-theoretic condition (solvability of the Galois group) and the algebraic condition (solvability by radicals). Its hypotheses require: (i) the base field has characteristic zero, and (ii) the Galois group of the polynomial is solvable. We have verified both: (i) is given in the theorem statement ($\operatorname{char} K = 0$), and (ii) was established in the preceding steps. This result is sharp in the following sense. For $n = 5$, the [Abel-Ruffini-Galois](/theorems/1322) theorem shows that the general polynomial of degree $5$ is *not* solvable by radicals, because $S_5$ is not solvable. The breakdown occurs precisely at degree $5$ because $A_5$ is the smallest non-abelian simple group, preventing the construction of a subnormal series with abelian quotients. [/guided] [/step]