KindiakMath

Galois Group Theory

Previously, we have proven the fundamental theorem of Galois theory, which establishes a direct correspondence between field extensions and subgroups of automorphisms. We say that a finite extension \mathbb K \supseteq \mathbb F is Galois if there exists some (unique) subgroup G \leq \mathrm{Aut}(\mathbb K) such that \mathbb K^G = \mathbb F. In this case, we write

G = \mathrm{Gal}(\mathbb K/\mathbb F) \leq \mathrm{Aut}(\mathbb K).

Subsequently, for any intermediate field \mathbb F \subseteq \mathbb L \subseteq \mathbb K, \mathbb K \supseteq \mathbb L is Galois, and the intermediate field \mathbb L is in direct correspondence with the subgroup

\mathrm{Gal}(\mathbb K/\mathbb L) \leq \mathrm{Gal}(\mathbb K/\mathbb F).

Conversely, the subgroup H \leq \mathrm{Gal}(\mathbb K/\mathbb F) corresponds to the intermediate field \mathbb K^H. Finally, the extension \mathbb L \supseteq \mathbb F is Galois if and only if

\mathrm{Gal}(\mathbb L/\mathbb F) \triangleleft \mathrm{Gal}(\mathbb K/\mathbb F).

In this case,

\mathrm{Gal}(\mathbb K/\mathbb F)/\mathrm{Gal}(\mathbb K/\mathbb L) \cong \mathrm{Gal}(\mathbb L/\mathbb F).

Now let \mathbb K \supseteq \mathbb F denote a finite extension.

Lemma 1. Given f \in \mathbb F[x], suppose f can be written as a product of linear factors in \mathbb F(\mathcal Z_f)[x], then \mathbb F(\mathcal Z_f) \supseteq \mathbb F is Galois with Galois group \mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f)).

Proof. We remark that \mathbb F(\mathcal Z_f) \supseteq \mathbb F is always normal. The linear factor criterion implies that \mathbb F(\mathcal Z_f) \supseteq \mathbb F is separable. Since the extension \mathbb F(\mathcal Z_f) \supseteq \mathbb F is finite, we have that \mathbb F(\mathcal Z_f) \supseteq \mathbb F is Galois where, by definition, \mathrm{Gal}(\mathbb F(\mathcal Z_f)/\mathbb F) = \mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f)).

Remark 1. Recall that

\mathcal Z_f := \{ \alpha \in \bar{\mathbb F} : f(\alpha) = 0\},

where \bar{\mathbb F} denotes the algebraic closure. We say that a polynomial f \in \mathbb F[x] is separable if it can be written as a product of linear factors in \mathbb F(\mathcal Z_f)[x]. In particular, f is separable if and only if \mathbb F(\mathcal Z_f) \supseteq \mathbb F is separable.

Theorem 1. For any f \in \mathbb F[x], define the Galois group of f over \mathbb F by \mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f)). Then there exists some n \leq \deg(f) such that \mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f)) is isomorphic to some subgroup of \mathcal S_n (i.e. the group of bijections on \{1,\dots, n\}).

Proof. Suppose \mathcal Z_f = \{\alpha_1,\dots,\alpha_n\} where n \leq \deg(f). By definition, for any \sigma \in \mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f)), \sigma(\mathcal Z_f) = \mathcal Z_f so that \sigma|_{\mathcal Z_f} \in \mathrm{Sym}(\mathcal Z_f). In particular, for any i, there exists a unique j_i such that \sigma(\alpha_i) = \alpha_{j_i}. For each \sigma, define \phi_\sigma \in \mathcal S_n by \phi_{\sigma}(i) = j_i. Then the map \phi : \mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f)) \to \mathcal S_n defined by \phi(\sigma) = \phi_\sigma forms the desired group monomorphism, so that \mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f)) \cong \phi(\mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))) \leq \mathcal S_n.

Corollary 1. If f is separable and irreducible with degree n, then n \mid |\mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))|.

Proof. If f is separable, then \mathbb F(\mathcal Z_f) \supseteq \mathbb F is separable too, and thus Galois with

\mathbb F(\mathcal Z_f)^{\mathrm{Gal}(\mathbb F(\mathcal Z_f) / \mathbb F)} = \mathbb F,\quad \mathrm{Gal}(\mathbb F(\mathcal Z_f) / \mathbb F) = \mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f)).

If f is irreducible, [\mathbb F(\alpha_1) : \mathbb F] = n. By the tower rule,

|\mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))| = [\mathbb F(\mathcal Z_f) : \mathbb F(\alpha_1)] \cdot [\mathbb F(\alpha_1) : \mathbb F] = [\mathbb F(\mathcal Z_f) : \mathbb F(\alpha_1)] \cdot n,

so that n \mid |\mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))|.

Remark 1. By Theorem 1, if f is irreducible and separable, then \mathrm{Gal}(\mathbb F(\mathcal Z_f) / \mathbb F) is isomorphic to some subgroiup of \mathcal S_n, where n = \deg(f).

Example 1. If f \in \mathbb F[x] is irreducible with degree 2, then \mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f)) is isomorphic to either \mathbb Z_2 or the trivial group.

Lemma 1. If G \leq \mathcal S_n contains an n-cycle and a transposition, then G = \mathcal S_n.

Proof. Suppose G contains (1\ 2) and (1\ 2\ \dots\ n) without loss of generality. Recall that

\mathcal S_n = \langle (1\ 2), (1\ 2\ \dots\ n) \rangle.

Hence,

\mathcal S_n \leq \langle (1\ 2), (1\ 2\ \dots\ n) \rangle \leq G \leq \mathcal S_n.

Recall that in a finite group G and a cycle \sigma, we define \sigma^n to be the n-fold composition of \sigma, and |\sigma| := \min\{n \in \mathbb N : \sigma^n = 1\}.

Theorem 2. Let p be prime and f \in \mathbb Q[x] be irreducible with degree p. If f has exactly two non-real roots in \mathbb C, then \mathrm{Aut}_{\mathbb Q}(\mathbb Q(\mathcal Z_f)) \cong \mathcal S_p.

Proof. Denote \mathbb L := \mathbb Q(\mathcal Z_f) and G := \mathrm{Aut}_{\mathbb Q}(\mathbb L) for brevity. We remark that \mathbb L \supseteq \mathbb F is Galois if and only if \mathbb L^G = \mathbb Q. By the fundamental theorem of algebra, \mathcal Z_f \subseteq \mathbb C. By hypothesis,

\mathcal Z_f \backslash \mathbb R = \{\alpha_1,\alpha_2\},\quad \mathcal Z_f \cap \mathbb R = \{\alpha_3,\dots,\alpha_p\}.

By Theorem 1, we can regard G \leq \mathcal S_p by passing to an isomorphism. Since f is irreducible, by Corollary 1, p \mid |G|. Therefore, G contains a permutation \sigma with |\sigma| = p. Write \sigma = \sigma_r \circ \cdots \circ \sigma_1 for disjoint cycles \sigma_i. For r \geq 2, |\sigma_i| < p implies that

|\sigma| = \mathrm{lcm}( |\sigma_1|,\dots, |\sigma_r|) \nmid p,

a contradiction. Therefore, we must have \sigma = \sigma_1, that is, \sigma is a p-cycle.

On the other hand, since \alpha_1,\alpha_2 are non-real, by the conjugate root theorem, \alpha_2 = \alpha_1^*, where (\cdot)^* denotes complex conjugation. Therefore, G contains the transposition \tau := \phi_{12}. By Lemma 1, G \cong \mathcal S_p, as required.

In particular, consider a general quintic polynomial f with degree n \geq 5. If f is reducible, then we can write it in the form f = f_k \cdot f_m, where \deg(f_k) = k \geq 1 and k+m = n. Then f has solutions in radicals if and only if f_k and f_m do. This is certainly the case for n = 5.

However, the more interesting case arises if f is irreducible. Then Theorem 2 tells us that \mathrm{Aut}_{\mathbb Q}(\mathbb Q(\mathcal Z_f)) \cong \mathcal S_5. Surprisingly, the closely related alternating group \mathcal A_5 that consists of the even permutations of \mathcal S_5 is a simple group. This simplicity turns out to be the key to the non-solvability of the quintic.

Theorem 3. Call a Galois extension \mathbb K \supseteq \mathbb F Abelian over \mathbb F if \mathrm{Gal}(\mathbb K /\mathbb F) is Abelian. Given \mathbb K \supseteq \mathbb L \supseteq \mathbb F, if \mathbb K \supseteq \mathbb F is Abelian, then \mathbb K \supseteq \mathbb L and \mathbb L \supseteq \mathbb F are Abelian Galois extensions.

Proof. By hypothesis, \mathrm{Gal}(\mathbb K / \mathbb F) is Abelian. Therefore, \mathrm{Aut}_{\mathbb F}(\mathbb L) \leq \mathrm{Gal}(\mathbb K / \mathbb F) is Abelian, and thus normal. Therefore, \mathbb L \supseteq \mathbb F is Galois with

\mathrm{Gal}(\mathbb L/\mathbb F) = \mathrm{Aut}_{\mathbb F}(\mathbb L)

being Abelian. Furthermore, since \mathbb K \supseteq \mathbb L is Galois,

\displaystyle \mathrm{Gal}(\mathbb L/\mathbb F) \cong \frac{\mathrm{Gal}(\mathbb K/\mathbb F) }{\mathrm{Gal}(\mathbb K/\mathbb L) }

implies that \mathrm{Gal}(\mathbb K/\mathbb L) must be Abelian, so that \mathbb K \supseteq \mathbb L is Abelian too.

Remark 1. Replacing ‘Abelian’ with ‘cyclic’ in Theorem 3 mutatis mutandis still renders the statement true.

It seems like there is a close connection between Galois extensions and sequences of normal subgroups. We formalise this intuition by defining the subnormal series.

Definition 1. A subnormal series of a group G is a finite sequence H_0, H_1,\dots, H_n of normal subgroups of G such that

\{1\} = H_0 < H_1 < \cdots < H_n = G.

Here, we denote the group identity by 1 and denote H_s < H_t to mean that H_s \triangleleft H_t and H_s \neq H_t. We call this series normal if H_i < G for all i.

Example 1. The series \{1\} < \mathcal A_3 < \mathcal S_3 is normal.

Example 2. The series

\{1\} < \{(1), (1\ 2)\} < \{ (1), (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3) \} < \mathcal A_4 < \mathcal S_4

is subnormal but not normal.

Definition 2. We say that a group G is solvable if it has a subnormal series \{H_i : i \in I\} such that all the factors H_{i+1}/H_i are Abelian.

Example 3. Any Abelian group is solvable.

Theorem 4. For n \geq 5, \mathcal A_n and \mathcal S_n are not solvable.

Proof. Recalling that \mathcal A_n is simple, the only subnormal series of $ltaex \mathcal S_n$ is

\{1\} < \mathcal A_n < \mathcal S_n.

Furthermore, \mathcal A_n/\{1\} \cong \mathcal A_n is not Abelian. Even more so, the factor \mathcal S_n/\mathcal A_n is congruent to the subgroup of transpositions, which is clearly not Abelian. Therefore, \mathcal S_n cannot be solvable.

The naming of “solvable” should already strongly hint at its connection to the non-existence of the quintic formula. But we will only properly discuss this idea the next time, through connecting it to radical extensions.

—Joel Kindiak, 25 Apr 26, 1340H

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