Chapter 21

Rigorous

Galois Theory — Solvability by Radicals

00 · Symbol Glossary

$G^{(i)}$i-th derived subgroup (informal)

Shorthand sometimes used for the $i$ -th term of a chain witnessing solvability — a group that can be broken down into a finite series of abelian pieces, made precise in Section 01.

$G_0\trianglelefteq G_1\trianglelefteq\cdots$Subnormal series

A chain of subgroups, each normal in the next one (not necessarily normal in $G$ itself) — the structure a solvable group's decomposition takes.

$\sqrt[n]{a}$Radical (nth root)

An element $\beta$ with $\beta^n=a$ . A radical extension adjoins a chain of such roots, one at a time, to build up a tower of fields.

01 · Solvable Groups

Definition — Solvable Group

A group $G$ is solvable if there is a chain of subgroups $\{e\}=G_0\trianglelefteq G_1\trianglelefteq\cdots\trianglelefteq G_n=G$ (each $G_i$ normal in $G_{i+1}$ , though not necessarily in $G$ ) such that every quotient $G_{i+1}/G_i$ is abelian. In the full Galois story, each abelian step corresponds to adjoining roots of unity and $n$ th roots along a tower of extensions — the reason solvable groups eventually line up with polynomials solvable by radicals.

Example — Every abelian group is solvable

Take the chain $\{e\}\trianglelefteq G$ directly: $G/\{e\}\cong G$ is abelian by hypothesis. One step suffices.

02 · Solvability Is Inherited

Theorem 21.1 — Subgroups of Solvable Groups Are Solvable

If $G$ is solvable and $H\le G$ , then $H$ is solvable.

Proof

Let $\{e\}=G_0\trianglelefteq\cdots\trianglelefteq G_n=G$ witness solvability. Set $H_i := H\cap G_i$ . Each $H_i\le H$ , and $H_i\trianglelefteq H_{i+1}$ : for $h\in H_{i+1}$ and $a\in H_i$ , $hah^{-1}\in G_i$ (since $a\in G_i$ and $G_i\trianglelefteq G_{i+1}$ , with $h\in G_{i+1}$ too) and $hah^{-1}\in H$ (closure in $H$ ), so $hah^{-1}\in H\cap G_i=H_i$ . The map $H_{i+1}/H_i \to G_{i+1}/G_i$ , $hH_i\mapsto hG_i$ , is a well-defined injective homomorphism (well-defined and injective by the same style of coset-membership argument as Theorem 4.1; a homomorphism directly from the quotient operation), so $H_{i+1}/H_i$ is isomorphic to a subgroup of the abelian group $G_{i+1}/G_i$ — hence abelian itself (subgroups of abelian groups are abelian). The chain $\{e\}=H_0\trianglelefteq\cdots\trianglelefteq H_n=H$ witnesses $H$ 's solvability.

Theorem 21.2 — Quotients of Solvable Groups Are Solvable

If $G$ is solvable and $N\trianglelefteq G$ , then $G/N$ is solvable.

Proof

Let $\{e\}=G_0\trianglelefteq\cdots\trianglelefteq G_n=G$ witness solvability. Let $\pi: G\to G/N$ be the quotient map, and set $\bar G_i := \pi(G_i) = (G_iN)/N$ . Since $\pi$ is surjective and $G_i\trianglelefteq G_{i+1}$ , $\bar G_i\trianglelefteq\bar G_{i+1}$ (homomorphic images of normal-in preserve normal-in, by the same argument as Theorem 6.2 combined with surjectivity). The map $G_{i+1}/G_i \to \bar G_{i+1}/\bar G_i$ induced by $\pi$ is surjective (by construction), so $\bar G_{i+1}/\bar G_i$ is a homomorphic image of the abelian group $G_{i+1}/G_i$ — hence abelian (images of abelian groups under homomorphisms are abelian, directly from the definition). The chain $\{\bar e\}=\bar G_0\trianglelefteq\cdots\trianglelefteq \bar G_n=G/N$ witnesses $G/N$ 's solvability.

Theorem 21.3 — Extensions of Solvable Groups Are Solvable

If $N\trianglelefteq G$ with $N$ and $G/N$ both solvable, then $G$ is solvable.

Proof

Let $\{e\}=N_0\trianglelefteq\cdots\trianglelefteq N_m=N$ witness $N$ 's solvability, and $\{\bar e\}=\bar H_0\trianglelefteq\cdots\trianglelefteq \bar H_k = G/N$ witness $G/N$ 's solvability. Let $H_i \le G$ be the preimage of $\bar H_i$ under $\pi: G\to G/N$ (a subgroup, preimages of subgroups under homomorphisms). Since $\bar H_i\trianglelefteq\bar H_{i+1}$ , $H_i\trianglelefteq H_{i+1}$ (preimages preserve normality-in, the ideal-theoretic analog of which appeared already in Theorem 12.4's style of argument), and $H_{i+1}/H_i \cong \bar H_{i+1}/\bar H_i$ (a standard correspondence between subgroups containing $N$ and subgroups of $G/N$ , mirroring Theorem 6.7's Third Isomorphism Theorem machinery), abelian.

Concatenate: $\{e\}=N_0\trianglelefteq\cdots\trianglelefteq N_m=N=H_0\trianglelefteq H_1\trianglelefteq\cdots\trianglelefteq H_k=G$ , with every consecutive quotient abelian (the first stretch by $N$ 's solvability, the second by the correspondence just shown). This witnesses $G$ 's solvability.

03 · Solvable Examples

Example — S_3 is solvable

$\{e\}\trianglelefteq A_3\trianglelefteq S_3$ : $A_3\cong\mathbb{Z}_3$ (order 3, prime, cyclic by Corollary 4.6), and $S_3/A_3\cong\mathbb{Z}_2$ (Chapter 06's example). Both quotients abelian. Solvable.

Example — S_4 is solvable

Let $V=\{e,(1\,2)(3\,4),(1\,3)(2\,4),(1\,4)(2\,3)\}$ (the Klein four group, closed under composition: any two of the nonidentity elements compose to the third, checkable directly). Every element of $V$ is even, so $V\le A_4$ . Moreover $V\trianglelefteq A_4$ for a clean reason: conjugation preserves cycle type, and $V$ consists of the identity together with all three elements of $S_4$ of cycle type $(2,2)$ — so conjugating any element of $V$ by anything lands on an element of the same cycle type, which is necessarily back inside $V$ . $V$ is abelian (isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ ). $A_4/V$ has order $12/4=3$ , cyclic (Corollary 4.6), abelian. $S_4/A_4\cong\mathbb{Z}_2$ (index 2, Chapter 06 style). Chain: $\{e\}\trianglelefteq V\trianglelefteq A_4\trianglelefteq S_4$ , all quotients abelian. Solvable.

04 · A_5 Is Simple

Definition — Simple Group

A group is simple if its only normal subgroups are $\{e\}$ and itself.

Theorem 21.4 — A_5 Is Simple

$A_5$ (order 60) is simple.

Proof

Step 1: Conjugacy class sizes in $A_5$ . Using the Orbit-Stabilizer Theorem (Theorem 8.4) applied to conjugation, each class size is $[A_5:C_{A_5}(x)]=60/\lvert C_{A_5}(x)\rvert$ .

Double transpositions, e.g. $(1\,2)(3\,4)$ : its centralizer in $S_5$ has order 8 (generated by $(1\,2)$ , $(3\,4)$ , and the swap $(1\,3)(2\,4)$ exchanging the two pairs), of which the even elements form a subgroup of order 4. Class size in $A_5$ : $60/4=15$ .

3-cycles, e.g. $(1\,2\,3)$ : centralizer in $S_5$ has order 6 (powers of the 3-cycle, times permutations of the 2 fixed points); the even elements form a subgroup of order 3 (the 3-cycle's powers only — the transposition of fixed points is odd). Class size in $A_5$ : $60/3=20$ .

5-cycles, e.g. $(1\,2\,3\,4\,5)$ : centralizer in $S_5$ has order 5 (only its own powers), entirely even — so this centralizer already sits inside $A_5$ , meaning the single $S_5$ -class of size $120/5=24$ splits into two $A_5$ -classes of size $24/2=12$ each (a class splits into two equal halves exactly when its $S_5$ -centralizer contains no odd permutation).

Tally: $1+15+20+12+12=60$ . ✓

Step 2: No proper nontrivial normal subgroup exists. A normal subgroup $N\trianglelefteq A_5$ is a union of entire conjugacy classes (Theorem 8.2-style: conjugation orbits are exactly the classes, and normality means $N$ is a union of orbits), always including $\{e\}$ , with $\lvert N\rvert$ dividing 60 (Lagrange). Checking every possible sum of $1$ plus a nonempty sub-collection of $\{15,20,12,12\}$ against divisibility by 60:

Classes included (besides $\{e\}$ )	$\lvert N\rvert$	Divides 60?
none	1	trivial case
$15$	16	no
$20$	21	no
$12$	13	no
$12+12$	25	no
$15+20$	36	no
$15+12$	28	no
$15+12+12$	40	no
$20+12$	33	no
$20+12+12$	45	no
$15+20+12$	48	no
$15+20+12+12$	60	all of $A_5$

No proper nonempty combination gives a divisor of 60 — only the trivial subgroup and all of $A_5$ survive. So $A_5$ 's only normal subgroups are $\{e\}$ and $A_5$ : simple.

05 · S_n Is Not Solvable for n ≥ 5

Theorem 21.5 — S_n Is Not Solvable, n ≥ 5

$S_n$ is not solvable for any $n\geq5$ .

Proof

$A_5$ is not solvable. Since $A_5$ is simple (Theorem 21.4) and nonabelian (it contains noncommuting elements, e.g. two different 3-cycles), any subnormal series for $A_5$ can only have the trivial jump $\{e\}\trianglelefteq A_5$ (no intermediate normal subgroups exist to insert), giving the single quotient $A_5/\{e\}\cong A_5$ — which is not abelian. So $A_5$ admits no series with all-abelian quotients: not solvable.

$S_5$ is not solvable. If $S_5$ were solvable, Theorem 21.1 would force every subgroup, including $A_5\le S_5$ , to be solvable too — contradicting the previous paragraph. So $S_5$ is not solvable.

$S_n$ is not solvable for $n\geq5$ . $S_5$ embeds as a subgroup of $S_n$ for $n\geq5$ (permutations of $\{1,\ldots,5\}$ extended to fix $6,\ldots,n$ ). If $S_n$ were solvable, Theorem 21.1 would force this copy of $S_5$ to be solvable — contradiction.

06 · Radical Extensions and Galois's Criterion

Definition — Radical Extension

$K/F$ is a radical extension if there is a tower $F=K_0\subseteq K_1\subseteq\cdots\subseteq K_r=K$ with each $K_{i+1}=K_i(\beta_i)$ for some $\beta_i$ satisfying $\beta_i^{n_i}\in K_i$ for some positive integer $n_i$ (each step adjoins an $n_i$ -th root of a previously available element).

Example — The quadratic formula as a radical tower

For $f=x^2+bx+c\in\mathbb{Q}[x]$ , the roots $\frac{-b\pm\sqrt{b^2-4c}}{2}$ live in $K_1=\mathbb{Q}(\sqrt{b^2-4c})$ , a single radical step ( $\beta_1=\sqrt{b^2-4c}$ with $\beta_1^2=b^2-4c\in K_0=\mathbb{Q}$ ). Solvability by radicals for a polynomial means its splitting field sits inside some radical extension of the base field — exactly the classical notion of "a formula using $+,-,\times,\div$ , and $n$ -th roots."

Theorem 21.6 — Galois's Solvability Criterion

Let $F$ have characteristic $0$ and $f\in F[x]$ . The roots of $f$ can be expressed by radicals over $F$ (equivalently, $f$ 's splitting field is contained in a radical extension of $F$ ) if and only if $\mathrm{Gal}(f) := \mathrm{Gal}(K/F)$ (for $K$ the splitting field of $f$ ) is a solvable group.

A genuinely deep theorem, taken as given

Both directions of Theorem 21.6 require machinery beyond this course: adjoining roots of unity (so that $K_i(\beta_i)/K_i$ becomes a genuine Galois extension with a cyclic Galois group — the "Kummer theory" needed to match radical steps with abelian quotients precisely), and careful bookkeeping about how solvability transfers between the radical tower and the Galois correspondence's subgroup chain. We take the statement as given, in the same spirit as the Fundamental Theorem of Algebra and Artin's Lemma, and spend this chapter's remaining effort on its most famous consequence — which needs only the group theory already built.

07 · The Insolvability of the General Quintic

Example — x⁵ - 4x + 2 has Galois group $S_5$

Consider $f=x^5-4x+2\in\mathbb{Q}[x]$ .

Irreducible over $\mathbb{Q}$ : by Eisenstein's criterion with the prime $2$ (each non-leading coefficient — $0,0,0,-4,2$ — is divisible by 2, the constant term $2$ is not divisible by $2^2=4$ , and the leading coefficient $1$ is not divisible by $2$ ).

Exactly 3 real roots: a calculus fact (checking $f'(x)=5x^4-4$ has exactly two real critical points, and evaluating $f$ there shows one local max is positive and the other local min is negative, giving three sign changes — taken as given here, since it is a computation in analysis rather than algebra).

The Galois group contains a 5-cycle: since $f$ is irreducible of degree 5, $\mathrm{Gal}(f)$ acts transitively on the 5 roots (Corollary 20.3's embedding is transitive whenever the polynomial is irreducible — any root can be sent to any other via some automorphism, since all roots share the same minimal polynomial, Theorem 18.4). A transitive subgroup of $S_5$ has order divisible by 5 (Orbit-Stabilizer: the orbit of size 5 forces $5\mid\lvert\mathrm{Gal}(f)\rvert$ ), so by Cauchy's Theorem (Theorem 8.6), $\mathrm{Gal}(f)$ contains an element of order 5 — necessarily a single 5-cycle (the only order-5 elements of $S_5$ ).

The Galois group contains a transposition: with exactly 3 real roots and 2 nonreal (complex conjugate) roots, complex conjugation restricts to an automorphism of the splitting field $K\subset\mathbb{C}$ fixing $\mathbb{Q}$ (conjugation fixes every rational number and is a field automorphism of $\mathbb{C}$ , hence of any subfield it preserves — and it preserves $K$ since $K$ is generated by roots of $f$ , which conjugation permutes among themselves by Theorem 20.2, since $f$ has real coefficients). This automorphism fixes the 3 real roots and swaps the 2 complex-conjugate roots: a single transposition, viewed inside $\mathrm{Gal}(f)\le S_5$ .

A 5-cycle and a transposition generate all of $S_5$ (a classical fact about permutation groups of prime degree — omitted here, but intuitively: repeatedly conjugating the transposition by powers of the 5-cycle produces enough transpositions to generate every transposition, and transpositions alone generate $S_5$ ). So $\mathrm{Gal}(f)=S_5$ .

By Theorem 21.5, $S_5$ is not solvable. By Theorem 21.6 (Galois's criterion), $f$ 's roots cannot be expressed by radicals — a specific, explicit quintic with no radical formula for its roots.

The classical conclusion

Since some quintic has no radical formula, no general formula (built from $+,-,\times,\div$ , and $n$ -th roots, applying uniformly to every degree-5 polynomial's coefficients) can exist — such a formula would apply to $x^5-4x+2$ in particular, contradicting what was just shown. This is the Abel–Ruffini Theorem, and the entire argument — from group actions in Chapter 08, through Sylow-style counting, through the Galois correspondence in Chapter 20 — converges on this single explicit polynomial as its final witness.

08 · Exercises

EXERCISE 21.1

Apply Theorem 21.1: if $S_5$ contains a non-solvable subgroup, what does that say about $S_5$ itself?

$S_6$ contains $S_5$ as a subgroup (permutations fixing the 6th point). Since $A_5\le S_5\le S_6$ and $A_5$ is not solvable (Theorem 21.5's proof), Theorem 21.1 (contrapositive) shows $S_6$ cannot be solvable — if it were, its subgroup $A_5$ would have to be solvable too.

Using Theorem 21.1, explain why $S_6$ is not solvable, without repeating the full argument of Theorem 21.5.

EXERCISE 21.2

Check whether $D_4$ (order 8) could possibly be nonsolvable — recall that all groups of prime-power order have special structure (Chapter 09).

$D_4$ has order $8=2^3$ , a $p$ -group. By Theorem 9.1, $Z(D_4)\neq\{e\}$ ; in fact $Z(D_4)=\{e,r^2\}$ (order 2). $D_4/Z(D_4)$ has order 4, abelian by Corollary 9.2. Chain: $\{e\}\trianglelefteq Z(D_4)\trianglelefteq D_4$ , with $Z(D_4)$ abelian (order 2) and $D_4/Z(D_4)$ abelian (order 4). $D_4$ is solvable. (In fact every finite $p$ -group is solvable, by this same style of argument iterated using Theorem 9.1 repeatedly.)

Show that $D_4$ (order 8) is solvable, using Chapter 09's fact that nontrivial $p$ -groups have nontrivial centers.

EXERCISE 21.3

Apply Theorem 21.6 directly, given the Galois group is already identified.

From Chapter 20, $\mathrm{Gal}(x^3-2/\mathbb{Q}) \cong S_3$ . $S_3$ is solvable (Section 03's example). By Theorem 21.6 (Galois's criterion), the roots of $x^3-2$ can be expressed by radicals over $\mathbb{Q}$ — consistent with the classical fact that cubic equations have a radical formula (Cardano's formula), unlike the quintic example of Section 07.

Using Theorem 21.6 and Chapter 20's computation that $\mathrm{Gal}(x^3-2)\cong S_3$ , determine whether the roots of $x^3-2$ can be expressed by radicals.

09 · Chapter Summary

Concept	Statement
Solvable group	Subnormal series $\{e\}=G_0\trianglelefteq\cdots\trianglelefteq G_n=G$ with abelian quotients
Subgroups, quotients, extensions	All preserve solvability (Thm 21.1, 21.2, 21.3)
$S_3$ , $S_4$	Solvable, via explicit chains through $A_3$ / $V_4,A_4$
$A_5$ is simple	Only normal subgroups are $\{e\}$ and $A_5$ (Thm 21.4, via conjugacy class counting)
$S_n$ not solvable, $n\geq5$	Since $A_5\le S_n$ is nonabelian and simple (Thm 21.5)
Radical extension	Tower of $n_i$ -th root adjunctions
Galois's Criterion	Solvable by radicals $\iff \mathrm{Gal}(f)$ is a solvable group (Thm 21.6)
$x^5-4x+2$	Galois group $S_5$ , not solvable — no radical formula exists

Next: Chapter 22 — Modules over Rings & Structure Theorems generalizes vector spaces by replacing the field of scalars with an arbitrary ring, unifying abelian groups, vector spaces, and ideals under a single framework and closing the course with the structure theorem for finitely generated modules over a PID.