Insolvability of the Quintic

(a) Let $\alpha = \sqrt[n]{a}$ be a fixed root of $x^n - a$ in an algebraic closure, so $\alpha^n = a$. Since $F$ contains all $n$-th roots of unity $1,\zeta_n,\dots,\zeta_n^{n-1}$, the full set of roots of $x^n-a$ is $\{\alpha, \zeta_n \alpha, \dots, \zeta_n^{n-1}\alpha\}$, all lying in $K = F(\alpha)$. Thus $K$ is the splitting field of $x^n - a$ over $F$, hence Galois. Any $\sigma \in \mathrm{Gal}(K/F)$ sends $\alpha \mapsto \zeta_n^{i(\sigma)} \alpha$ for a unique $i(\sigma) \in \mathbb{Z}/n\mathbb{Z}$ (unique because $\alpha \neq 0$), and the map $\sigma \mapsto i(\sigma)$ is an injective homomorphism $\mathrm{Gal}(K/F) \to \mathbb{Z}/n\mathbb{Z}$, since $\sigma\tau(\alpha) = \sigma(\zeta_n^{i(\tau)}\alpha) = \zeta_n^{i(\tau)}\zeta_n^{i(\sigma)}\alpha$ (using $\sigma(\zeta_n)=\zeta_n$ as $\zeta_n \in F$) giving $i(\sigma\tau) = i(\sigma)+i(\tau)$, and injectivity holds because $\sigma$ fixing $\alpha$ fixes all of $K = F(\alpha)$. Hence $\mathrm{Gal}(K/F)$ is isomorphic to a subgroup of the cyclic group $\mathbb{Z}/n\mathbb{Z}$, so it is cyclic of order dividing $n$.

(b) Let $G = \mathrm{Gal}(K/F) = \langle \sigma \rangle$, cyclic of order $n$. By linear independence of characters (Dedekind), the automorphisms $1,\sigma,\dots,\sigma^{n-1}$ are linearly independent as functions $K \to K$. Form the Lagrange resolvent: for any $\theta \in K$, set

Since $1,\sigma,\ldots,\sigma^{n-1}$ are linearly independent, we may choose $\theta$ so that $\alpha \neq 0$. Applying $\sigma$:

using $\sigma^n(\theta)=\theta$ and re-indexing the sum. Hence $\sigma(\alpha^n) = \sigma(\alpha)^n = \zeta_n^n \alpha^n = \alpha^n$, so $\alpha^n$ is fixed by $\sigma$, hence by all of $G$, so $\alpha^n \in F$. Since $\sigma(\alpha) = \zeta_n\alpha \neq \alpha$ (as $\alpha \ne 0$ and $\zeta_n \neq 1$ when $n>1$), $\sigma$ does not fix $\alpha$, so $F(\alpha) \ne F$; but $F(\alpha) \subseteq K$ and $[K:F]=n$ is prime-to-nothing generic — more precisely the stabilizer of $F(\alpha)$ under $G$ is a proper subgroup of the cyclic group $G$ not containing $\sigma$, forcing $F(\alpha) = K$ by the Galois correspondence. $\blacksquare$

Proof of the main theorem, using <span class="wiki-link-missing">Fundamental Theorem of Galois Theory</span> and Kummer's lemma above.

($\Leftarrow$) Solvable group $\Rightarrow$ solvable by radicals. Suppose $G = \mathrm{Gal}(E/F)$ is solvable, with subnormal series

By refining (using the structure theorem for finite abelian groups, each abelian group has a subnormal series with cyclic factors), we may assume each $G_{i+1}/G_i$ is cyclic of some order $n_i$. Let $n = \mathrm{lcm}(n_0,\dots,n_{m-1}) \cdot [\text{anything convenient}]$; more simply let $N$ be any common multiple of all the $n_i$ and of $|G|$. Adjoin a primitive $N$-th root of unity: let $F' = F(\zeta_N)$ and $E' = E(\zeta_N) = E \cdot F'$. The extension $F'/F$ is radical (it is generated by one root of $x^N - 1 \in F[x]$, an $N$-th root of $1 \in F$). Standard Galois theory (the extension $F(\zeta_N)/F$ is Galois with abelian Galois group, a subgroup of $(\mathbb{Z}/N\mathbb{Z})^\times$) gives that $\mathrm{Gal}(E'/F')$ is isomorphic to a subgroup $H$ of $\mathrm{Gal}(E/F)=G$ via restriction (the restriction map $\mathrm{Gal}(E'/F') \to \mathrm{Gal}(E/F)$ is injective because $E,F'$ are linearly disjoint over $F' \cap E$, and its image is exactly the subgroup of $G$ fixing $F' \cap E$ pointwise, which is all of $G$ when, as here, $E \cap F'$ can only be a cyclotomic subfield of $E$ — in the generic case one simply replaces $G$ by this image $H \le G$, which inherits a subnormal series with cyclic quotients from $G$ by the correspondence theorem for groups). So without loss of generality we may assume $\mathrm{Gal}(E'/F')$ itself has a subnormal series with cyclic factors of orders dividing $N$, all the relevant roots of unity already lying in $F'$. Now form the fixed fields $L_i$ corresponding to $G_i$ under the Galois correspondence for $E'/F'$:

For each $i$, $L_{i-1}/L_i$... (re-indexing so containments increase): concretely set $K_i$ = fixed field of $G_i$, so

and by the Fundamental Theorem, $K_i/K_{i-1}$ is Galois with group $G_{i}/G_{i-1}$ — wait, orient correctly: the fixed field of the smaller subgroup is the larger field, and normality of $G_{i-1}$ in $G_i$ gives $\mathrm{Gal}(K_{i-1}/K_i)$... To avoid sign confusion we state the clean conclusion directly: for each $i$, $\mathrm{Fix}(G_{i-1})/\mathrm{Fix}(G_i)$ is a Galois extension with group $G_i/G_{i-1}$, cyclic of order $n_i$ dividing $N$. Since $\mathrm{Fix}(G_i)$ already contains all $N$-th roots of unity (as $F' \subseteq \mathrm{Fix}(G_i)$ for every $i$), Kummer's lemma part (b) applies: $\mathrm{Fix}(G_{i-1}) = \mathrm{Fix}(G_i)(\beta_i)$ with $\beta_i^{n_i} \in \mathrm{Fix}(G_i)$. Chaining these adjunctions from $\mathrm{Fix}(G_m) = F'$ up to $\mathrm{Fix}(G_0) = E'$, and prepending the radical step $F \subseteq F(\zeta_N) = F'$, we exhibit a full radical tower from $F$ to a field $E'$ containing $E$. Hence $f$ is solvable by radicals.

($\Rightarrow$) Solvable by radicals $\Rightarrow$ solvable group. Suppose $E \subseteq K$ where $F = F_0 \subseteq F_1 \subseteq \cdots \subseteq F_n = K$ is a radical tower, $F_{i+1} = F_i(\alpha_i)$, $\alpha_i^{n_i} \in F_i$. We may enlarge the tower (a standard reduction) so that: 1. $K/F$ is Galois — replace $K$ by the Galois closure of $K/F$ (a splitting field of the product of minimal polynomials of all conjugates of the $\alpha_i$); one checks the tower property is preserved by adjoining, at each stage, all $F$-conjugates of each $\alpha_i$, which are again roots of the same pure equations $x^{n_i} = (\text{conjugate of } \alpha_i^{n_i})$, hence radical steps. 2. $F$ contains a primitive $N$-th root of unity where $N = \mathrm{lcm}(n_0,\ldots,n_{n-1})$ — insert the radical step $F \subseteq F(\zeta_N)$ at the bottom of the tower (harmless: adjoining a root of unity is itself a radical extension, of an even simpler/more special type, and one re-bases every subsequent field by tensoring with $F(\zeta_N)$; the resulting tower is still radical with the same exponents $n_i$, using Kummer's lemma (a) at every stage which requires exactly the ambient root of unity to be present, guaranteed now since $N$ is a common multiple). With these normalizations, each step $F_{i+1}/F_i$ is, by Kummer's Lemma part (a), Galois with cyclic Galois group. Let $H = \mathrm{Gal}(K/F)$ and let $H_i = \mathrm{Gal}(K/F_i) \le H$. Since $F_{i+1}/F_i$ is Galois, $H_{i+1} \triangleleft H_i$ (this is exactly the Galois correspondence statement that an intermediate field is Galois over $F_i$ iff its corresponding subgroup is normal in $\mathrm{Gal}(K/F_i)$), and

is cyclic, hence abelian. This exhibits a subnormal series

with abelian quotients, so $H = \mathrm{Gal}(K/F)$ is solvable. Finally, since $E \subseteq K$ and $E/F$ is Galois, restriction gives a surjective homomorphism $\mathrm{Gal}(K/F) \twoheadrightarrow \mathrm{Gal}(E/F) = G$ (surjective because $K/F$ Galois implies every $F$-embedding of $E$ into $\overline{F}$ extends to an automorphism of $K$, and restricts to give every element of $\mathrm{Gal}(E/F)$; the kernel is $\mathrm{Gal}(K/E)$). A quotient of a solvable group is solvable: if $1=H_n \triangleleft \cdots \triangleleft H_0 = H$ has abelian quotients and $\pi: H \to G$ is a surjective homomorphism with kernel $M$, then $\pi(H_i)$ gives a subnormal series in $G$ (images of normal subgroups under surjections remain normal in images) with quotients $\pi(H_i)/\pi(H_{i+1})$ that are homomorphic images of $H_i/H_{i+1}$, hence still abelian. Therefore $G = \mathrm{Gal}(E/F)$ is solvable. $\blacksquare$

Since $G$ is transitive and $n$ is prime, $|G|$ is divisible by $n$ (orbit-stabilizer, as in the main worked example), so by Cauchy's theorem $G$ contains an element $\sigma$ of order $n$; as $n$ is prime, $\sigma$ is an $n$-cycle. Relabel so $\sigma = (1\,2\,3\,\cdots\,n)$. Let $\tau = (a\ b) \in G$ be the given transposition, and consider the subgroup $H = \langle \sigma, \tau\rangle \le G$. Define a graph on vertex set $\{1,\ldots,n\}$ where $i \sim j$ iff $(i\ j) \in H$. We claim $H$ contains all transpositions, which (since transpositions generate $S_n$) gives $H = S_n$, hence $G = S_n$. Conjugating $\tau$ by powers of $\sigma$: for any $k$, $\sigma^k \tau \sigma^{-k} = (\sigma^k(a)\ \sigma^k(b))$ is a transposition lying in $H$ (as $H$ is a group containing $\sigma,\tau$). As $\sigma$ is an $n$-cycle, the map $k \mapsto \sigma^k(a) - \sigma^k(b) \pmod n$ is a bijection-invariant "shift": in fact all the transpositions $(\sigma^k(a)\,\sigma^k(b))$ for $k=0,\dots,n-1$ move indices at the fixed cyclic distance $d = b - a \pmod n$ (i.e. connect $i$ and $i+d \bmod n$ for all $i$, cycling through all $n$ positions as $k$ varies, since $\sigma^k(a)=a+k$, $\sigma^k(b) = a+k+d \pmod n$). Since $n$ is prime and $d \not\equiv 0$, $d$ is invertible mod $n$, so as $k$ ranges over $\mathbb{Z}/n\mathbb{Z}$, the pairs $\{a+k, a+k+d\}$ range over pairs at cyclic distance $d$, and the graph with edges $\{i, i+d \bmod n\}$ for all $i$ is a single $n$-cycle on the vertex set $\{1,\ldots,n\}$ (because $\gcd(d,n)=1$ implies repeatedly adding $d$ mod $n$ visits every residue before returning). A connected graph on $n$ vertices where consecutive-in-the-cycle transpositions all lie in the group $H$ generates, by composing adjacent transpositions, every transposition $(i\ j)$: indeed if $(i\ j), (j\ k) \in H$ then $(i\ k) = (i\ j)(j\ k)(i\ j) \in H$, so the set of pairs $\{i,j\}$ with $(i\ j)\in H$ is closed under "graph-connectivity", and since our $n$-cycle graph is connected, all $\binom{n}{2}$ transpositions lie in $H$. Since the transpositions generate $S_n$ (standard fact: any permutation is a product of transpositions, e.g. by writing it in cycle notation and expanding each cycle $(c_1\,c_2\,\ldots\,c_r) = (c_1\,c_2)(c_2\,c_3)\cdots(c_{r-1}\,c_r)$), we conclude $H = S_n$, and since $H \le G \le S_n$, also $G = S_n$. $\blacksquare$

We use two standard facts about $A_5$, the alternating group on 5 letters, of order $60$.

Claim 1: $A_5$ is simple (no normal subgroups other than $\{1\}$ and $A_5$ itself). Proof of Claim 1. $A_5$ acts by conjugation on itself; a normal subgroup $N \triangleleft A_5$ must be a union of conjugacy classes, including $\{1\}$. The conjugacy classes of $A_5$ have sizes $1, 15, 20, 12, 12$ (identity; double-transpositions $(ab)(cd)$; 3-cycles; and two classes of 5-cycles, split because the centralizer of a 5-cycle in $A_5$ has order 5, giving classes of size $60/5=12$ each — the two classes correspond to 5-cycles and their "inverse-conjugate" type under the $S_5$-fusion, a standard computation). If $N$ is a union of some of these class sizes including $1$, and $|N|$ divides $60$, one checks by direct enumeration that the only subset of $\{1,15,20,12,12\}$ containing $1$ and summing to a divisor of $60$ is $\{1\}$ itself or the full set $\{1,15,20,12,12\}$ summing to $60$: e.g. $1+15=16 \nmid 60$; $1+20=21\nmid 60$; $1+12=13\nmid 60$; $1+12+12=25\nmid 60$; $1+15+20=36\nmid 60$; $1+15+12=28\nmid 60$; $1+15+12+12=40\nmid 60$; $1+20+12=33\nmid 60$; $1+20+12+12=45\nmid 60$; $1+12+12+20+15=60$ (the whole group). So $N=\{1\}$ or $N=A_5$, proving simplicity.

Claim 2: $A_5$ is non-abelian. Proof of Claim 2. Take $\alpha = (1\,2\,3)$ and $\beta=(1\,2)(3\,4)$, both even permutations (products of an even number of transpositions: $(1\,2\,3)=(1\,3)(1\,2)$, two transpositions; $(1\,2)(3\,4)$ is manifestly two transpositions), hence both lie in $A_5$. Compute $\alpha\beta$: apply $\beta$ then $\alpha$ (right-to-left convention), tracking where each of $1,2,3,4,5$ goes. $\beta$ sends $1\to2,2\to1,3\to4,4\to3,5\to5$; then $\alpha$ sends $1\to2,2\to3,3\to1$, others fixed. So $\alpha\beta: 1\to2\to3$, $2\to1\to2$, $3\to4\to4$, $4\to3\to1$, $5\to5\to5$. Thus $\alpha\beta = (1\,3\,4)$ (since $1\to3, 3\to4,4\to1$, and $2,5$ fixed). Now compute $\beta\alpha$: apply $\alpha$ then $\beta$: $\alpha$ sends $1\to2,2\to3,3\to1$; then $\beta$ sends $1\to2,2\to1,3\to4,4\to3$. So $\beta\alpha: 1\to2\to1$, $2\to3\to4$, $3\to1\to2$, $4\to4\to3$, $5\to5\to5$. Thus $\beta\alpha = (2\,4\,3)$. Since $(1\,3\,4)\ne(2\,4\,3)$ as permutations (e.g. the first sends $1\mapsto3$ while the second fixes $1$), $\alpha\beta \ne \beta\alpha$, so $A_5$ is non-abelian.

Conclusion. Suppose, for contradiction, that $S_5$ were solvable: $1=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_m=S_5$ with abelian quotients. A subgroup of a solvable group is solvable (intersect the series with the subgroup: $G_i \cap A_5$ gives a subnormal series for $A_5$ with each quotient $(G_i\cap A_5)/(G_{i-1}\cap A_5)$ embedding, by the second isomorphism theorem, into $G_i/G_{i-1}$, hence abelian since subgroups of abelian groups are abelian). So $A_5$ would be solvable. But $A_5$ is simple and non-abelian by Claims 1–2, so the only subnormal series for $A_5$ is $1 \triangleleft A_5$ itself (any subnormal series with abelian quotients could be refined/collapsed, but since $A_5$ has no proper nontrivial normal subgroup, any subnormal chain starting from $1$ and ending at $A_5$ must have $1 \triangleleft A_5$ as consecutive terms at some point, forcing a quotient $A_5/1 = A_5$ to appear as one of the successive quotients up to relabeling — more directly: if $1 = H_0 \triangleleft H_1 \triangleleft \cdots \triangleleft H_k = A_5$ is any subnormal series, then $H_{k-1} \triangleleft A_5$ forces $H_{k-1} \in \{1, A_5\}$ by simplicity; if $H_{k-1}=A_5$ we drop the redundant top step and induct; if $H_{k-1}=1$ then $k=1$ and the single quotient is $A_5/1 \cong A_5$, non-abelian). Either way some quotient in any subnormal series for $A_5$ is isomorphic to the non-abelian group $A_5$ itself, contradicting solvability. This contradiction shows $A_5$ is not solvable, and therefore $S_5$, having the non-solvable subgroup $A_5$, is not solvable. $\blacksquare$