Galois Theory 2: The General Quintic is Dead and Évariste Killed Him

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Introduction

In this post, we continue our analysis of Galois Theory and dive further into the proof.

Why the preface about symmetric polynomials? It should be clear by now to readers of the previous post that demonstrating symmetry of polynomial roots and understanding what properties come with the various symmetries will be invaluable for further exploration of general polynomial solubility (or lack thereof).

Recall that $f(x) = x^2 - 2 \implies \alpha = \pm \sqrt{2}$. $\pm \sqrt{2}$ are symmetric roots of $f$ because $\alpha^2 - 2 = f$ regardless of which root we use, and this is true for any polynomial with rational coefficients involving radicals, addition, and multiplication.

The Galois Theory states that if the Galois group of a polynomial $f$ is solvable, then $f$ itself is solvable in radicals. We can use the solvability of Galois groups of a polynomial to show that polynomials of degree $n \geq 5$ are not always solvable, whereas the Galois groups of polynomials of lower orders are always solvable.

Definitions

In order to understand what the hell a Galois group of a polynomial is, let alone what it means to "solve" a group, we need to recap some algebraic and group theoretic definitions.

When a polynomial $f$ with rational coefficients in $\mathbb Q$ can be solved using only rational numbers and the operations $+, -, \div, \times$ and $n$-th roots $\sqrt[n]{x}$, we say that "$f(x)$ is soluble by radicals"

Groups

A group is a collection of objects $G$ together with some operation $\star$ which satisfy the following axioms:

1. For any $x, y \in G$, $x \star y \in G$. That is, the result of the operation between two elements in our group is another element which is also in our collection; a group is closed under its operation.

2. $\exists \epsilon \in G \; s.t. \; \forall x \in G, \epsilon \star x = x = x \star \epsilon$. Every group must have a "no-op" element which is called the identity of the group $G$. The identity of group formed by the integers with addition $G(\Z, +)$ is $0$.

3. The group operation $\star$ is associative. That is, $\forall x,y,z \in G, (x \star y) \star z = x \star (y \star z)$ which allows us to write $x \star y \star z$ unambiguously

4. Every element in $G$ has an inverse: $\forall x \in G \exists x^{-1} \in G \; s.t. \; x \star x^{-1} = \epsilon = x^{-1} \star x$

For example, $(\Z, +)$ form a group, but $(\Z, \times)$ does not, because $\not\exists x \in Z \; s.t. \; 2x = 1$

Abelian Groups

A group $G$ is said to be abelian if its operation is commutative:

$\forall x, y \in G(\star);\; x \star y = y \star x$

Cyclic Groups

$C_p$ is a cyclic group of order $p$ with elements $\{1, x, x^2, ..., x^{p-1} \}$ with some operation $\cdot$ which behaves similar to arithmetic multiplication: $x^n \times x^m = x^{m+n}$ as well as $x^p = 1$.

For example, in $C_5$ the expression $x^2 \cdot x^4$ evaluates to $x^6 = x^5 \cdot x^1 = x$. Hence, the inverse of $x^k$ is $x^{p-k}$ for integer values of $p, k$.

Symmetric Groups

Symmetric groups, denoted $S_n$ and read "the symmetric group on $n$ elements" have objects which are themselves the permutations of the $n$ elements, with the binary operation of composition $\circ$. So, $\sigma \in S_3$ might be the permutation which maps the indices of a triple $(1, 2, 3) \rightarrow (2, 3, 1)$, and another element $\tau \in S_3$ could be $(1, 2, 3) \rightarrow (3, 2, 1)$. Their composition would then be:

$\sigma \circ \tau = (1, 2, 3) \rightarrow (3, 2, 1) \rightarrow (1, 3, 2)$

Alternating Groups and parity

Group parity: A permutation is even if it can be written as a product of an even number of transpositions, and odd if it can be written as an odd number of transpositions.

E.g., a permutation on a non-trivial group which is equivalent to the identity on that group can be expressed: $\sigma = (1 \; 2)( 1 \; 2)$, that is: we send the first element to the 2nd position, and the 2nd element to the first position, then repeat that same cycle, so for some tuple $t = (\color{red}a\color{black} , \color{blue}b\color{black})$:

\begin{aligned} \sigma(t) = \begin{pmatrix} \color{red}a & \color{blue}b \\ \downarrow & \downarrow \\ \color{blue}b & \color{red}a \end{pmatrix} \begin{pmatrix} \color{blue}b & \color{red}a \\ \downarrow & \downarrow \\ \color{red}a & \color{blue}b \end{pmatrix} = (\color{red}a\color{black} , \color{blue}b\color{black}) = \epsilon(t) \end{aligned}

Like the set of real numbers, the parity of a group is multiplicative. That is, the product of an even permutation and an odd permutation is odd, the product of two even permutations or two odd permutations as even. The inverse of an even permutation is even, the inverse of an odd is odd, so we can define a subgroup of of the symmetric group consisting of all even permutations called the Alternating Group: $A_n \subset S_n$:

$A_n = \{ \sigma | \epsilon(\sigma) = 1, \sigma \in S_n \}$

For more on the potential applications of alternating groups, check out this other post about solving Rubik's cubes with group theory.

Fields

A field is like a group $\mathbb F, +$ with an additional operation $\cdot$ which also has identitive elements for both operations, such as $0$ for the "additive" operation, and $1$ for the "multiplicative" operation. Additionally, we specify the following constraints:

1. $(\mathbb F, +)$ is a group, and $(\mathbb F \backslash \{ 0\}, \cdot)$ is a group (division by zero is verboten)
2. The operation $\cdot$ is distributive over $+$ s.t. $(x + y) \cdot z = x\cdot z + y \cdot z$
3. $0 \cdot x = x \cdot 0, x \cdot y = y \cdot x, x + y = y + x$, the multiplicative operation has an identity, and both operations are commutative (which is not always true for a group)

E.g. $\mathbb {Q, R}$ are both fields.

Field Extensions

A field extension is a gerund insofar as we extend fields (verb), and the resultant extension is itself another field (noun). Extensions are accomplished by adjoining additional values to a base field. We write $\mathbb Q(\sqrt{2})$ to denote field extension of the rationals with root two, implicitly adding all other numbers that can be formed by combining rational elements of the base field and $\sqrt{2}$ via the operations of the base field. That is, all numbers that can be expressed in terms of $a + b\sqrt{2}$ are in the field extension $\mathbb Q(\sqrt{2})$ .

The only non-obvious axiom we might want to justify to ourselves is that there exists a multiplicative inverse of a number

$x = \frac{1}{a + b \sqrt{2}}$

which can still be expressed in the form $c + d\sqrt{2}$. We can demonstrate that it exists by multiplying $x$ by its conjugate form of $1$ and simplifying:

\begin{aligned} x &= \frac{1 }{(a + b\sqrt{2})} \cdot \frac{(a - b\sqrt{2})}{(a - b\sqrt{2})} \\ &= \frac{a - b\sqrt{2}}{(a^2 - b^2)} \\ &= \frac{a}{(a^2 - b^2)} - \frac{b\sqrt{2}}{(a^2 - b^2)} \\ \end{aligned}

leaving us with an expression of the form $c + d\sqrt{2}$ where

$c = \frac{a}{(a^2 - b^2)},\quad d = - \frac{b}{(a^2 - b^2)} \\$

As mentioned before, a field extension of $F$ is itself a field $K$ which contains $F$. We write $F \subseteq K$, or more concretely, $\mathbb Q \subseteq \mathbb R$, or $\mathbb Q \subseteq \mathbb Q(\sqrt{2})$. (We didn't show that $\mathbb Q(\sqrt{2})$ is a field, but it's plain to see that $\forall a \in \mathbb Q, a + 0\sqrt{2} \in \mathbb Q(\sqrt{2}))$

Because the rationals are a field $\mathbb Q$, but $\mathbb Q$ is not closed under the $n$-th root operation, oftentimes we will want to adjoin our roots to the rationals.

Abel-Ruffini Theorem

The Abel-Ruffini theorem –which predates Galois as hinted at in the previous post– states that some quintics have roots that are not expressible in terms of radicals.

Even if our root is some horrendous mess such as:

$\alpha = \sqrt[5]{\frac{3}{7}} + \frac{1}{5}\sqrt[23]{2 - \frac{3}{7}\sqrt{89}}$

we can still feasibly construct a splitting field by adjoining

$\mathbb Q \subseteq \mathbb Q(\sqrt{89}) \subseteq \mathbb Q(\sqrt{89}, \sqrt[23]{2 - \sqrt{89}}) \subseteq \cdots$

and so on.

Splitting Fields

Given any polynomial $f(x)$, the splitting field of $f$ is the smallest field extension of $\mathbb Q$ that contains all the roots of $f$.

For example, the splitting field of $f(x) = x^2 - 2$ is $\mathbb Q(\sqrt{2})$. This is obviously the minimal field since we can't remove any elements from it without either violating axioms of it being a field (closed under addition and multiplication) or removing one of the roots.

Another example of a splitting field is

\begin{aligned} f(x) &= x^4 - 5x^2 + 6 \\ &= (x^2 - 3)(x^2 - 2) \\ &\implies \{ \alpha \} = \{ \pm \sqrt{3}, \pm \sqrt{2}, \} \end{aligned}

Therefore the splitting field of $f$ is $\mathbb Q(\sqrt{3}, \sqrt{2})$.

Automorphisms

An automorphism is an invertible, one-to-one, structure-preserving map of a set (or a field) onto itself: $\sigma: F \rightarrow F$ which satisfies the following axioms for all $x,y \in F$:

1. $\sigma(x + y) = \sigma(x) + \sigma(y)$
2. $\sigma(xy) = \sigma(x)\sigma(y)$
3. $\sigma(1/x) = 1/\sigma(x)$

That is, applying the automorphism to a compound expression constructed from the operations $\{+, -, \times, \div \}$ is equivalent to to applying it to the individual terms of that expression.

For example, conjugation is an automorphism in $\mathbb Q(\sqrt{2})$ given by:

\begin{aligned} \sigma : \mathbb Q(\sqrt{2}) &\rightarrow \mathbb Q(\sqrt{2}) \\ \sigma(a + b\sqrt{2}) &= a - b\sqrt{2} \\ \end{aligned}

$F$-Automorphisms

An $F$-automorphism is an automorphism $\sigma$ over a field extension $K/F$ s.t.

$\forall x \in F; \quad \sigma(x) = x$

That is, $\sigma$ leaves elements of the base field unchanged. We can imagine how this might be useful when we want to describe permutations of just the roots of a polynomial which leaves all other rational elements of our field extension $\mathbb Q(\alpha)/\mathbb Q$ unchanged.

As it turns out, conjugation is the only non-trivial $\mathbb Q$-automorphism for radical extensions $\mathbb Q(\sqrt{k})$.

Formally, we describe this relationship for any polynomial $f$ with rational coefficients, the field extension $K/\mathbb Q$, and a $\mathbb Q$-automorphism $\sigma$ of $K$ as:

$\sigma(f(x)) = f(\sigma(x))$

This is useful since it shows that a $\mathbb Q$-automorphism of the splitting field $K$ of $f$ permutes just the roots:

\begin{aligned} f(x) = 0 &\implies f(\sigma(x)) = \sigma(f(x)) = 0 \\ &\implies \sigma(\alpha) = 0 \end{aligned}

Knowing how a single $\mathbb Q$-automorphism $\sigma$ of a splitting field $\mathbb Q \subseteq K$ rearranges the $\alpha$ roots of a polynomial $f$ is sufficient to derive how $\sigma$ acts on all elements of $K$. For example, for $f(x) = x^4 - 5x + 6$, with the splitting field $K = \mathbb Q(\sqrt{2},\sqrt{3})$, and the roots $\{\alpha \} = \{ \pm \sqrt{2}, \pm \sqrt{3}, \}$, there is no $\mathbb Q$-automorphism $\sigma$ of $K$ where $\sigma(\sqrt{2}) = \sqrt{3}$.

We can prove this by contradiction, supposing that there does exist such a permutation where $\sigma(\sqrt{2}) = \sqrt{3}$. Then,

\begin{aligned} \sigma(\sqrt{2})^2 &= \sigma(\sqrt{2}^2) = \sigma(2) = 2 \end{aligned}

because $\sigma$ must preserve multiplicative structure of $K$, and $\sigma(x) = x$ for all $x$ in the base field $\mathbb Q$. But if $\sigma(\sqrt{2}) = \sqrt{3}$, then $\sigma(\sqrt{2})^2 = \sqrt{3}^2$, or $2 = 3$, which is not true.

Galois Groups

A Galois group of a field extension $K/\mathbb Q$ is the collection of all $\mathbb Q$-automorphisms where the group operation of composition $\circ$ on permutations means:

$\sigma \circ \tau \implies (\sigma \circ \tau)(x) = \sigma(\tau(x))$

Galois groups can be denoted $Gal(K/\mathbb Q)$ or just $Gal(f)$ if $K$ is a splitting field of a polynomial $f$.

For example, $f(x) = x^2 - 2$ has the Galois group $Gal(f) = (\sigma, \tau)$ with $\sigma$ being the conjugate we've seen before defined by $\sigma(a + b\sqrt{2}) = a - b\sqrt{2}$ and $\tau$ is the identity: $\tau(x) = x = \sigma \circ \sigma$.

This Galois group is the same as the cyclic group of order 2: $C_2$ (which is also the symmetric group on 2 elements: $S_2$). It's cyclic because it takes two applications of the non-identitive element to reach the identity on the group: $\sigma^2 = \tau$.

Begin the Problem

Galois Theory builds on the Abel-Ruffini Theorem claiming that if an equation $f$ is soluble by radicals, then we should be able to construct a sort of "tower" of radical extensions of the rationals with the roots.

Galois theory simplifies the problem of general radical solutions of higher order polynomials by translating it from the infinite domain of fields (ring theory) to the domain of groups (specifically, groups of finite permutations of the roots) making the problem infinitely easier. Specifically, we consider the special case of automorphisms acting on the splitting field of a polynomial.

Formally, for a polynomial $f$ of the form:

$f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$

with rational coefficients $a_i \in \mathbb Q$, we consider the field extension containing all the roots of $f$ and denote $L = \mathbb Q(\alpha_1, \alpha_2, ..., \alpha_n)$.

By definition of a root, $f(\alpha_k) = 0$ for all values of $\alpha_k$ we've adjoined to the rationals (I'll omit the subscript on $\alpha$ going forward since we cease caring about specific roots, but rather whole permutations of roots). But what happens when we apply a $\mathbb Q$-automorphism to the polynomial?

\begin{aligned} f(x) &= a_{n}x^{n} + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \\ f(\alpha) &= a_{n}\alpha^{n} + a_{n-1}\alpha^{n-1} + \cdots + a_1\alpha + a_0 = 0 \\ \\ \sigma(0) &= \sigma(a_{n}\alpha^{n} + a_{n-1}\alpha^{n-1} + \cdots + a_1\alpha + a_0) \\ &= \sigma(a_{n}\alpha^{n}) + \sigma(a_{n-1}\alpha^{n-1}) + \cdots + \sigma(a_1\alpha) + \sigma(a_0) \\ &= \sigma(a_{n})\sigma(\alpha^{n}) + \cdots \\ &= \sigma(a_{n})\sigma(\alpha)^n + \cdots \\ &= a_{n}\sigma(\alpha)^n + a_{n-1}\sigma(\alpha)^{n-} \cdots a_1\sigma(\alpha) + a_0 \\ \end{aligned}

We're left with just the roots being expressed in terms of $\sigma$, since by definition of $\sigma$ a $\mathbb Q$-automorphism of the splitting field of $f$, it leaves elements of $\mathbb Q$ unchanged.

If we fairly assume all roots of our polynomial have single multiplicity,1 then according to our properties of automorphisms, $\sigma$ just permutes the roots leaving the rest of the expression unchanged.

We call this minimal extension $\mathbb Q(\alpha)$ a Galois Extension. The collection of all automorphisms of this extension (sometimes denoted $Aut(L/\mathbb Q)$) is the Galois group of $f$.

More Extensions

Two additional types of field extensions omitted from the front-loaded glossary are cyclotomic extensions and Kummer extensions.

Cyclotomic Extensions

Consider all polynomials of the form $f(x) = x^n - 1$. For $n=5$ We can plot the roots of unity2 on the complex plane which are evenly distributed along the unit circle:

Where, in general, $\{ \alpha_1, \alpha_2, ..., \alpha_n\}$ are $1$ and the powers of the $n$-th roots of unity $\zeta$.

To analyze the Galois Group of $f$, we extend the rationals to include our roots of unity: $\mathbb Q(\zeta)$.

Again, it is sufficient to adjoin just $\zeta$ since we know the other roots are its powers up to $\zeta^{n-1}$. So, what automorphisms are in this Galois Group? $1$ needs to remain unchanged as it is in the base field $\mathbb Q$, but starting with the first root of unity $\zeta$, we can consider any automorphism sending this to any other root.

Let's consider the automorphism $\sigma: \mathbb Q(\zeta) \rightarrow \mathbb Q(\zeta)$ which permutes $\zeta$ by sending it to $\zeta^3$. As mentioned earlier, we can derive the rest of $\sigma$ from just this relation. Moving to the next root of unity $\zeta^2$, we apply the definition of $\sigma$:

\begin{aligned} \sigma(\zeta^2) = \sigma(\zeta)^2 &= \sigma(\zeta)\sigma(\zeta) \\ &= \zeta^3 \zeta^3 = \zeta^6 \\ &= \zeta^5 \zeta = 1 \cdot \zeta \\ &= \zeta \end{aligned}

Similarly for the other roots we get

\begin{aligned} \sigma(\zeta^3) = \sigma(\zeta)^3 &= \sigma(\zeta)\sigma(\zeta)\sigma(\zeta) \\ &= \zeta^3 \zeta^3 \zeta^3 = \zeta^9 \\ &= \zeta^5 \zeta^4 = 1 \cdot \zeta^4 \\ &= \zeta^4 \\ \sigma(\zeta^4) = \sigma(\zeta)^4 &= \sigma(\zeta)\sigma(\zeta)\sigma(\zeta)\sigma(\zeta) \\ &= \zeta^3 \zeta^3 \zeta^3 \zeta^3 = \zeta^{12} \\ &= \zeta^5 \zeta^5 \zeta^2 = 1 \cdot 1 \cdot \zeta^2 \\ &= \zeta^2 \\ \end{aligned}

The complete permutation derived from $\zeta \rightarrow \zeta^3$ is then:

We can denote this unique mapping as $\sigma_3$ since it sends the first non-fixed element (since $\sigma(1) = 1$) $\zeta$ to $\zeta^3$. In general $\sigma_l = \zeta^{l}$. Note that for cyclotomic automorphisms we have the following property:

$\sigma_l\sigma_m(\zeta) = \sigma_l(\zeta^m) = (\sigma_l(\zeta))^m = \zeta^{lm}$

Furthermore, this also implies commutativity of such permutations: $\sigma_l\sigma_m = \sigma_m\sigma_l = \zeta^{lm}$. Only some Galois groups will have this abelian property.

Recall that A group $G$ is said to be abelian if its operation is commutative. Here, the elements of our Galois group are permutations $\sigma, \tau$, and the operation is composition (implicitly, $\circ$):

$\forall \sigma, \tau \in G;\; \sigma \tau = \tau \sigma$

So, the Galois group of $f(x) = x^n - 1$ is abelian.

Kummer Extensions

Consider the generalization of the simple polynomial from the previous example $f(x) = x^n - \theta$. Again, the $n$-th roots of $\theta$ fall along a circle on the complex plane (just not necessarily the unit circle).

Our first root becomes $\alpha$, rather than $1$, and the rest of the roots take the form $\alpha\zeta, \alpha\zeta^2, ..., \alpha\zeta^{n}$. To find the Galois group, our base field is no longer simply the rationals, but instead it's some field extension $K$. We'll define $K$ to be the field containing $\zeta$, though by default it doesn't matter since we can simply adjoin $\zeta$ to an existing field and then call that sumbitch $K$ without loss of generality. We also have $\theta \in K$, and next we adjoin the roots of $f$ to $K$ yielding the splitting field over $f$ denoted $K(\alpha)$.

In this case, unlike the Cyclotomic extension, the first element of our root $\alpha$ is not constrained by the $K$-automorphism over $K(\alpha)$, so $\sigma(\alpha)$ is not fixed like $\sigma(1) = 1$ was. We can permute $\alpha$ anywhere, so lets choose the automorphism $\sigma_2$ which sends $\alpha \rightarrow \alpha\zeta^2$.

Continuing derivation of the next element in our group $\alpha\zeta$, according to the preservation of algebraic structure, we get:

\begin{aligned} \sigma_2(\alpha\zeta) &= \sigma_2(\alpha)\sigma_2(\zeta) \\ &= \sigma_2(\alpha)\zeta \\ &= \alpha\zeta^2\zeta \\ &= \alpha\zeta^3 \end{aligned}

Note that since $\zeta$ is a member of the base group $K$, $\sigma$ leaves it unchanged and instead effectively only permutes $\alpha$. Similarly, we get

\begin{aligned} \sigma_2(\alpha\zeta^2) &= \sigma_2(\alpha)\sigma_2(\zeta^2) \\ &= \sigma_2(\alpha)\zeta^2 \\ &= \alpha\zeta^2\zeta^2 \\ &= \alpha\zeta^4\\ \\ \sigma_2(\alpha\zeta^3) &= \sigma_2(\alpha)\sigma_2(\zeta^3) \\ &= \sigma_2(\alpha)\zeta^3 \\ &= \alpha\zeta^2\zeta^3 \\ &= \alpha\zeta^5 \\ &= \alpha\zeta \\ \\ \sigma_2(\alpha\zeta^4) &= \sigma_2(\alpha)\sigma_2(\zeta^4) \\ &= \sigma_2(\alpha)\zeta^4 \\ &= \alpha\zeta^2\zeta^4 \\ &= \alpha\zeta^6 \\ &= \alpha\zeta^2 \end{aligned}

So, all together, $\sigma_2$ on $Gal(f)$ permutes the roots in this starshape:

And, similar to the unitary expansion, we can label an automorphism $\sigma_l(\alpha) = \alpha\zeta^l$ according to where $\sigma$ sends the first root $\alpha$, and we once again can try to determine whether these groups of automorphisms are abelian:

$\sigma_l\sigma_m(\alpha) = \sigma_l(\alpha\zeta^m) = \alpha\zeta^{l + m}$

Though the rules of composition result in summing the powers rather than multiplying them, these powers still commute:

$\sigma_l\sigma_m = \sigma_m\sigma_l = \alpha\zeta^{l+m}$

so the group is still abelian. Note, however, that since the automorphisms over this kind of field extension are all determined by where $\alpha$ goes, it eliminates several possibilities of permutations. The only automorphism sending $\alpha \rightarrow \alpha\zeta^2$ permutes the roots in the starshape illustrated above, meaning that a permutation

$\alpha \rightarrow \alpha\zeta^2 \rightarrow \alpha\zeta^4 \rightarrow \alpha\zeta \rightarrow \alpha\zeta^3 \rightarrow \alpha$

like this is not possible:

In this sense, the Galois group, or rather the size of the Galois group is a measure of how related the roots of a polynomial are. In our example, the roots differ by powers of $\zeta$, so there really aren't that many automorphisms.

Naturally, we'll want to compose these extensions for some polynomial of the form $f = x^n - \theta$ with roots $\alpha$.

But what about a direct extension from $\mathbb Q$ to the largest field extension $\mathbb Q(\zeta, \alpha)$?

The result would still be a splitting field over the same polynomial $f$ with roots in $\alpha$ (assuming $\theta \in \mathbb Q$) to begin with. So it would still be a Galois extension and we can then consider the Galois group of this extension.

Towers of Galois

Now we can begin to show how analysis of the Galois groups of field extensions can tell us meaningful things about the general solubility of polynomials. Ultimately, we want to devise a sequence of field extensions resembling the following:

where the $\beta$ terms are whatever unholy perversions like those

$\sqrt[23]{2 - \sqrt{89}}$

terms we ran into earlier. First, we'll consider a generic, simple case involving 3 field extensions to formalize the problem of Group Solubility.

We're most interested in the Galois group of the Galois extension $M/K$ containing the $K$-automorphisms of $M$ as well as the $L$-automorphisms of $M$. Note that because

$K \subseteq L \subseteq M$

The $L$-automorphisms of $M$, in addition to leaving elements of $L$ unchanged by definition, also leave $K$ unchanged!

On the other hand, the Galois group $Aut(M/K)$ containing automorphisms of the field extension from $K$ straight to $M$ implies the existence of a broader Galois group $Aut(M/K)$ which also contains $Aut(M/L)$ since they both leave elements of $K$ unchanged, and every $\sigma \in Aut(M/L)$ necessarily leaves elements of $K$ unchanged too, so we can also say:

$Aut(M/L) \subseteq Aut(M/K)$

Furthermore, because $L/K$ is a Galois extension, we can say something more. Our prior constraint of automorphisms –that they are transformations $\sigma: L \rightarrow L$ where $L$ is the splitting field of $f$– can be loosened. The only requirement for this tower we've constructed is that $\alpha \in L$ which needs to be true so that we can apply $\sigma(\alpha)$. In fact, $L$ can be a bigger field so long as it still contains the splitting field over $F$.

Returning to our tower, we fix $L$ to in fact be the splitting field of $f$ over $K$. $Aut(M/K$) then will still contain permutations of the roots of $f$. So, even though we're only fixing our smaller field $K$, permutations $\sigma_L: \alpha_i \rightarrow \alpha_j$ are still transformations entirely within $L$, that is:

$L = K(\alpha_1, \alpha_2, ..., \alpha_n)$

Next we consider a new pair of automorphisms: $\sigma \in Aut(M/K)$ is a $K$-automorphism of $M$ insofar as it leaves all elements of $K$ unchanged. Additionally, we can consider an $L$-automorphism of $M$ denoted $\tau \in Aut(M/L)$, and observe that the composition of these two automorphisms on some element $\ell \in L$:

$\sigma\tau\sigma^{-1}(\ell)$

And, without knowing anything else about $\sigma, \tau$ or $\ell$ other than these properties of groups, we can infer that:

• $\sigma^{-1}(\ell) = \ell' \in L$ because $\ell \in L$
• $\tau(\ell') = \ell'' \in L$ leaves $\ell$ unchanged, so it is also in $L$
• $\sigma(\ell'')$ is effectively just $\sigma\sigma^{-1} = \ell''$

So the whole chain of permutations is itself an automorphism in $Aut(M/L)$. This kind of relationship allows us to say that $Aut(M/L)$ is a normal subgroup of $Aut(M/K)$.

In general, we say that $N$ is a normal subgroup of $G$ (denoted $N \vartriangleleft G$) iff:

\begin{aligned} \gamma\nu\gamma^{-1}(x) = (x) \\ \nu \in N, \gamma \in G \end{aligned}

Like the characters from Lost and the island, we return to the tower once again we can label the Galois group over the field extension from our base group $K$ to $M$ as $G = Aut(M/K)$ and the Galois group of the extension from $L$ to $M$ as $N = Aut(M/L)$ which we recognize to be a normal subgroup of $G$.

Considering finally the third Galois group of the lowest order field extension $L/K$, elements of this group $H$ are the $K$-automorphisms over $L$. And the last insight we'll draw from this simplest-form of a Galois tower is the consistent behavior of automorphisms of a subgroup $H$ within a broader group $N$.

To illustrate, we consider automorphisms $\nu \in N$ again as one-to-one mappings:

Straightforwardly, $\nu$ sends elements of $L$ to itself. $\nu$ on $L$ behaves like an automorphism of $L$.

It's also possible to have another automorphism $\nu' \in N$ which preserves the transformations on elements of $L$ while doing something completely different to the elements in $M$:

The composition $\nu \circ \nu'$ of any two $L$-automorphisms $\nu, \nu' \in Aut(M/L)$ on the subfield $L$ is effectively the same as $\sigma \circ \sigma^{-1}$.

Automorphic Difference and Quotient Groups

We say that the result of $\nu_1^{-1}\nu_2$ (or sometimes this composite permutation itself) is the difference between automorphisms. If $\nu_1^{-1}, \nu_2$ both do the same thing to the elements of the smaller field $L$, then their composition leaves $L$ fixed ($\nu_2$ sends an element of $L$ from left to right, and $\nu_1^{-1}$ sends an element of $L$ back from right to left).

This observation is restricted to elements in $L$, we can make no inferences about the nature of $\nu$ on $M$. This difference itself though is another $L$-automorphism of the group $Aut(M/L)$ leaving $L$ unchanged. When $\nu_1^{-1}, \nu_2 \in N$ we say that $\nu_1^{-1}\nu_2$ is a quotient group $G/N$. Two automorphisms of $G$ are in $G/N$ if they differ by an element of $N$.

So, all together, the relationships between the field extensions and their respective groups of our tower are as follows:

At the moment, it may seem like we're no closer to knowing anything about quintics though...

If we reference the initial target tower:

Additionally, if we simply further adjoin the roots of unity $\zeta$ with $\mathbb Q(\sqrt[3]{2})$ yielding $\mathbb Q(\sqrt[3]{2}, \zeta)$ this would not be a radical extension.3 Instead, we would want to first adjoin the unities with the rationals which (a familiar cyclotomic extension), then adjoin the cube roots of $\theta = 2$ which is also a valid Kummer extension:

We know that Cyclotomic and Kummer extensions are both Galois extensions, so now we're kosher again. This tells us that, when we're constructing our tower, we want to first add all the roots via Galois extension, then add just the roots at each stage ensure that these are radical extensions.

We construct it this way so that we can reuse the results of previous constructions, namely that Galois groups of both types of extensions are Abelian.

To construct the target tower, we add all $n_i$-th roots of unity: $\beta_i = \sqrt[n_i]{\theta_i}$ where $n_i$ are the orders of radicals $\beta_i$ making all these radical extensions either Cyclotomic or Kummer, and thus all Galois.

For the big, direct extension from $\mathbb Q \rightarrow \mathbb Q(\beta_1, \beta_2, ..., \beta_r)$, there is a tedious and recursive construction algorithm:

1. Let $K_0 = \mathbb Q$
2. Before adjoining $\beta_i = \sqrt[n_i]{\theta_i}$ to $K_{i-1}$, first adjoin those $n_i$-th roots of unity.
3. After adjoining $\beta_i$, then add all numbers of the form $\beta_i(\sigma) = \sqrt[n_i]{\sigma(\theta_i)}$ where $\sigma \in Aut(K_{i-1}/\mathbb Q)$

This works via the fact that the only thing fixed by all automorphisms in $Aut(K_{i-1}/\mathbb Q)$ must already be in $\mathbb Q$ to begin with.

This big Galois extension is the splitting field of some polynomial $f$ with rational coefficients in $\mathbb Q$. Since $\mathbb Q$ is our base field, the coefficients are necessarily also in all our intermediate extensions, so all extensions from an intermediate field to the biggest target field are also Galois – including the splitting field of the quintic $\mathbb Q(\alpha_1, \alpha_2, ..., \alpha_n)$. So we have:

Which has the same form as our pedagogical tower:

We can label all the extension's Galois groups of the target tower $G, G_1, ..., N, H$ and prove that the target tower is "reducible" to the trivial tower.

An Aside about Group Solubility

We've now introduced enough concepts about groups (namely, normal subgroups and quotient groups) to define group solubility.

There is a very deep theorem in finite group theory which is known as the Feit-Thompson theorem. It states that if $|G|$ is odd, then $G$ is a solvable group. The paper giving this proof, entitled ”Solvability of Groups of Odd Order,” was published in 1963 and is more than 250 pages long.4

Suppose that $G$ is a finite group. We say that $G$ is soluble if the following statements about it are true:

• If $|G| > 1$, then $G$ has a normal subgroup $H_1 \vartriangleleft G$ such that the quotient group $G/H_1$ is abelian
• If $|H_1| > 1$ then $H_1$ also has a normal subgroup $H_2 \vartriangleleft H_1$ such that the quotient group $H_1/H_2$ is abelian
• Etc. till the base case of $|H_k| = 1$ (with the possibility that $|G| = 1$)

Alternatively, if we denote $G = H_0$, $G$ is soluble if there exists a sequence of subgroups $H_1, ..., H_k$ s.t. for each $j, 1 \leq j \leq k$:

• $H_j$ is a normal subgroup of its predecessor $H_j \vartriangleleft H_{j-1}$
• the quotient group $H_{j-1}/H_j$ is abelian
• and $H_k = \{ \epsilon_G\}$, that is, the final subgroup contains merely the identity on $H_0$

If $G$ is a solvable group, then every subgroup of $G$ is a solvable group, and every quotient group of $G$ is also a solvable group. Suppose that $G$ is a group and that $N$ is a normal subgroup of G: $N \vartriangleleft G$, then it can be proved that $G$ is a solvable group iff both $G/N$ and $N$ are solvable.

To reduce the target tower to an instance of the trivial tower, first we focus on $G, G_1$, and the Cyclotomic extension.

We know that these are related as subgroups where $G_1 \vartriangleleft G$, and with the Cyclotomic extension having the Galois quotient group $G/G_1$ which means that it's abelian, so any two automorphisms will commute.

Similarly, $G_2 \vartriangleleft G_1$ and $G_1/G_2$ is abelian. We can repeat this relation till we reach the last radical extension with Galois group $G_r$ which itself also needs to be abelian, giving us a big ol' chian:

$G \vartriangleright G_1 \vartriangleright G_2 \vartriangleright \cdots \vartriangleright G_r$

where the constituent quotient groups $G_i/G_{i+1}$. This chain of groups is said to be soluble, and the Galois theory states that if there exists abelian subgroups $G_i$ to build this chain, then the biggest, direct group $G$ is solvable. If the tower $\mathbb Q \rightarrow \cdots \rightarrow \mathbb Q(\beta_1, \beta_2,...\beta_r)$ is solvable, what about the right side of the tower from $\mathbb Q \rightarrow \mathbb Q(\alpha_1, \alpha_2, ..., \alpha_n) \rightarrow \mathbb Q(\beta_1, \beta_2,...\beta_r)$?

We automatically have $H = G/N$ as solvable since it's determined by our quintic $f$ and this setup we've constructed implies that if $f$ is soluble by radicals, then the Galois group generated by our radical extensions will be soluble. So all that's left to do is find some $f$ whose Galois group is not soluble.

Again, the comprehensive proof strategy is a bit tedious but I'll outline it below before taking a shortcuit to intuition which leverages the in-solubility of Symmetric Groups shared by most polynomials of degree five and above.

Rigorous Proof Strategy & $S_5$ and Above

• Claim 1: $A_n$ is generated by 3-cycles
• Claim 2: Any normal subgroup of $B_n \vartriangleleft A_n$ contains a 3-cycle
• Case 1: $B_n$ contains an $n$-cycle for $n \geq 4$
• Case 2: $B_n$ contains at least 2 disjoint 3-cycles
• Case 3: $B_n$ contains one 3-cycle with the remaining permutations being transpositions
• Case 4: $B_n$ consists solely of transpositions

Intuitively, though, what we've done is layer so many constraints that it becomes impossibles to satisfy all of the for polynomials of degree five or higher in the general case.

Recall that a subgroup $N \vartriangleleft G$ means that $gng^{-1} \in N$ where $N$ is a group s.t. $n_1n_2, n^{-1}, \epsilon = nn^{-1} \in N$ and. These conditions on $N$ end up being so restrictive that there's really only a few possible groups which satisfy this criteria e.g. $A_5 \vartriangleleft S_5$.

This group we consider now contains all possible permutations of the five roots of a quintic polynomial and it's called the symmetric group of order five, denoted $S_5$. Consider the conjugal5 property of a subgroup $N$ and how $gng^{-1} \in N$ acts on an element:

And consider also the permutation $g$ on a 5-cycle in $S_5$:

$s_1 \rightarrow s_2 \rightarrow s_3 \rightarrow s_4 \rightarrow s_5 \rightarrow s_1$

If we fix $g$ to permute the first two elements of the set that $S_5$ is defined over, $gng^{-1}$ is still a 5-cycle:

In general, any 5-cycle can be obtained by $g$ in our Galois group. So, if any 5-cycle is in $N$, all 5-cycles must be in $N$, per the conjugation property. And also, the group property of composition of automorphisms $\nu_1 \circ \nu_2 \in N$. So, if the composition of two 5-cycles gives a 3 cycle, then all 3-cycles are in $N$ as well...

The point being that the existence of even a single 5-cycle quickly expands/reveals $N$ to be the whole group it's supposed to be normal to. And if we recall that for a group $G$ to be solvable, every quotient subgroup $G/G_{i+1}$ as well as the final group $G_r$ must all be abelian, so it should be relatively rare that XYZ, considering all these constraints. It's no surprise then that $S_5$ is not solvable in general.

$S_2, S_3, S_4$ on the other hand are very solvable, completely solvable in general. Because $gng^{-1} \in N$, and $N \vartriangleleft G$ requires fewer conjugations to admit all permutations, so there's fewer constraints.

E.g. for $S_4$, ff $G$ is abelian, then $G$ is a solvable group understood as:

• $H_0 = S_4$
• $H_1 = A_4$
• $H_2 = \{\epsilon, (1 \; 2)(3 \; 4), (1 \; 3)(2 \; 4), (1 \; 4)(2 \; 3) \}$
• $H_3 = \{ \epsilon \}$

Note that the quotient group $H_0/H_1$ is of order 2, $H_1/H_2$ of order 3, and $H_2/H_3 ≅ H_2$ is a group of order 4, and all of these quotient groups are abelian. Every subgroup $H_k \subset G$ is a solvable group, and every quotient group of $G$ is also a solvable group, therefore $S_4$ is solvable.

More on Insolvency of $S_5$

$S_5$ is not solvable because it has a composition series $\{\epsilon, A_5, S_5 \}$, giving factor groups isomorphic to $A_5$ and $C_2$, and $A_5$ is not abelian! $A_5$ is a simple, non-trivial (has more than one element), proper (not equal to the whole group) normal subgroup and is non-abelian.

A normal subgroup $S_5$ must be a union of conjugacy classes, that is – a set of element that are connected by conjugation: $a, b$ are conjugates of each other if $\exists g \in G$ s.t. $a = gbg^{-1}$.

Conjugacy classes partition the elements of a group into disjoint subsets, which are the orbits of the group acting on itself by conjugation. Conjugacy classes of a group can be used to classify groups; they can be used to show that two groups are not isomorphic, or to discover properties of an isomorphism between groups if it exists. In general, the sizes of conjugacy classes in a group give information about its structure.

The orbit of an element $x \in X$ is the set of elements $\{y | y\in X: g \cdot x = y, g \in G\}$, the elements that $x$ can become by combining it with $y \in X$ via an operation $G$. Orbits are to groups as characteristics are to fields.

The conjugacy classes in $S_5$ consist of permutations having the same cycle-decompositions. $S_5$ has seven such decompositions shapes with the following cardinalities (number of possible distinct permutations in $S_5$ with that shape):

$\sigma$Cardinality
$\epsilon$1
$(a \; b)$10
$(a \; b \; c)$20
$(a \; b \; c \; d)$30
$(a \; b \; c \; d \; e)$24
$(a \; b)(c \; d)$15
$(a \; b \; c)(d \; e)$20

Note that the sum total of these cardinalities, denoted $S_5$ = 120 = 5!\$ which checks out since that's the number of ways to permute 5 distinct elements

Additionally, any subgroup of $S_5$ must contain the identity element and also have order dividing 120. Note that all of these conjugacy classes are disjoint. These criteria indicate that the only possible normal subgroups of $S_5$ will have orders: 1, 40, 60, 120. $S_5$ only has one subgroup of order 60, which is $A_5$ formed by the union of conjugacy classes with the following shapes: $(a \; b \; c) \cup (a \; b)(c \; d) \cup (a \; b \; c \; d \; e)$ as well as the identity permutation $\epsilon$ of course.

For order 40, there is one choice of conjugacy classes with total cardinality equal to 40 consisting of the unique conjugacy classes of sizes 1, 24 and 15:

$\epsilon \cup (a \; b \; c \; d \; e) \cup (a \; b) (c \; d)$

but the union is not a subgroup because $(a \; b)(c \; d)$ and $(a \; b)(c \; e)$ are in the conjugacy class of size 15 and their product $(a \; b)(c \; d)(a \; b)(c \; e) = (c \; d \; e)$, which is one of the conjugacy classes of size 20. Similarly, the only possibility for n = 60 is the union of the unique conjugacy classes of size 1, 24 and 15 together with the conjugacy class of size 20 consisting of the 3-cycles.

Hence, the only possible choice of the proper and normal subgroup $H_1$ is $A_5$ since if $H_1 = \{ \epsilon \}$, then $G/H_1$ is nonabelian. $A_5$'s conjugacy classes and their cardinalities are:

$\sigma$Cardinality
$\epsilon$1
$(a \; b \; c)$20
$(a \; b \; c \; d \; e)$2 * 12 = 24
$(a \; b)(c \; d)$15

Considering all possible unions of these sets, the total cardinality cannot divide 60 excepts for $\{ \epsilon \}$ which is trivial and $A_5$ itself which is not proper. Therefore, there is no possible choice of a proper, normal subgroup $H_2$ of $H_1 = A_5$ if we require (which we do) that $H_1/H_2$ be abelian. Since $A_5$ is not solvable, $S_5$ is not solvable.

Which $f$ have $Gal(f) = S_5$

So, we've found a Galois group that's insoluble: $S_5$, and if we can now find a quintic polynomial with Galois Group $S_5$ that is still soluble by radicals, then we've proved by contradiction the claim.

Turns out, most quintics chosen at random will have Galois group $S_5$ which we can verify with some Mathematica incantations6

(* The group p: *)
p = PermutationGroup[
{Cycles[{{2, 3, 5, 4}}],
Cycles[{{1, 2, 3, 4, 5}}]}
];

(* has elements g: *)
g = GroupElements[p]

(*
g =	[   {Cycles[{}], Cycles[{{2, 3, 5, 4}}],
Cycles[{{2, 4, 5, 3}}],  Cycles[{{2, 5}, {3, 4}}],
Cycles[{{1, 2}, {3, 5}}], Cycles[{{1, 2, 3, 4, 5}}],
Cycles[{{1, 2, 4, 3}}], Cycles[{{1, 2, 5, 4}}],
Cycles[{{1, 3, 4, 2}}], Cycles[{{1, 3}, {4, 5}}],
Cycles[{{1, 3, 5, 2, 4}}], Cycles[{{1, 3, 2, 5}}],
Cycles[{{1, 4, 5, 2}}],  Cycles[{{1, 4, 3, 5}}],
Cycles[{{1, 4}, {2, 3}}], Cycles[{{1, 4, 2, 5, 3}}],
Cycles[{{1, 5, 4, 3, 2}}], Cycles[{{1, 5, 3, 4}}],
Cycles[{{1, 5, 2, 3}}], Cycles[{{1, 5}, {2, 4}}]}
]

where the order of g given by Length[g] = 20
*)

(* We can see that C_5 is a subgroup of p: *)
c5 = GroupElements[CyclicGroup[5]];
in = Intersection[g, c5]

(* and furthermore, that C_5 is a normal subgroup of p: *)
conj[x_] := PermutationProduct[#, x, InversePermutation[#]] & /@ g
Flatten[conj /@ c5] // Union

(* which yields:
[
{Cycles[{}],
Cycles[{{1, 2, 3, 4, 5}}], Cycles[{{1, 3, 5, 2, 4}}],
Cycles[{{1, 4, 2, 5, 3}}], Cycles[{{1, 5, 4, 3, 2}}]}
]

establishing that gHg' = H, where H = C_5.

And furthermore, we can show that the extension p/C_5 is abelian
since the quotient group elements commute:
*)

cos = {a, b, c, d} =
Union[RightCosetRepresentative[CyclicGroup[5], #] & /@ g]
su = Subsets[cos, {2}]
sur = Reverse /@ su
PermutationProduct @@@ su
PermutationProduct @@@ sur

(* yielding:
[
{Cycles[{{2, 3, 5, 4}}], Cycles[{{2, 4, 5, 3}}],
Cycles[{{2, 5}, {3, 4}}], Cycles[{}],
Cycles[{{2, 4, 5, 3}}], Cycles[{{2, 3, 5, 4}}]}

{Cycles[{{2, 3, 5, 4}}], Cycles[{{2, 4, 5, 3}}],
Cycles[{{2, 5}, {3, 4}}], Cycles[{}],
Cycles[{{2, 4, 5, 3}}], Cycles[{{2, 3, 5, 4}}]}
]

As C_5 <| p, and p/C_5, C_5/{e} are abelian, p is solvable
*)

Example problems

Finally, a few example problems to further illustrate how to take all this lofty theory and apply it to some easy inquiries about splitting fields and Galois groups of various polynomials

Splitting Field of a Quartic

Find the splitting field of $f(x) = x^4 + x^2 + 1$ over $\mathbb Q$.

Without doing any actual number crunching, we can recognize that $f$ will factor into:

\begin{aligned} f(x) &= (x^2 + ax + b)(x^2 + cx + d) \\ &= (a + c)x^3 + (ac + 2)x^2 \end{aligned}

for some values of $a,b,c,d$. We can deduce that:

• $b = d = 1$ because the constant term in $f$ is 1,
• $(ac + 2) = 1$ corresponds to the coefficient of the middle $x^2$ term in our non-factored polynomial,
• $0$ is the coefficient of our $x^3$ term.

These observations are sufficient to tell us what $a$ and $c$ are:

\begin{aligned} a + c = 0 &\implies c = - a \\ ac + 2 = 1 &\implies -c^2 = - 1 \\ &\implies c = \pm 1 \\ \end{aligned}

So $f$ will factor to:

$f(x) = \underbrace{(x^2 + x + 1)}_{\Phi_3(x)}\underbrace{(x^2 - x + 1)}_{6th \text{ roots of unity}}$

From here, we can make two claims / observations.

The first being that the first factor with a positive linear $x$ term is the third cyclic polynomial in $x$ whose roots are $e^{2 \pi i/3}, e^{-2 \pi i/3}$.

The second claim is that the roots of the second factor with a negative linear $x$ term are $e^{ \pi i/3}, e^{-\pi i/3}$.

We can verify these are roots by converting the roots of unity into rectangular coordinates and plugging them into their respective polynomial factors e.g.:

$e^{\pi i/3} = \frac{1}{2} \pm \frac{\sqrt{3}}{2}i$

and showing that the expression evaluates to zero. This gives us four roots of our quartic polynomial, all of which are powers of the third roots of unity $e^{ \pi i/3}$, so we can obtain our splitting field by adjoining this third root of unity to the radicals: $\mathbb Q(e^{ \pi i/3})$. We might equivalently express this field extension in terms of the rectangular coordinates as $\mathbb Q(\sqrt{3} i)$.

And finally, we might inquire about the degree of our extension, which is equal to the degree of the minimal polynomial of $\sqrt{3}i$. We can simply square this term $(\sqrt{3}i)^2$ to get $-3$ which is a root of $x^2+3$ which is irreducible,7 so the degree of our extension is two.

$\;$

Finding the Galois Group of a Quartic

For some polynomial $f(x) = x^4 -20x + 80$, we might inquire about the Galois group. To get started, we find the roots of $f$ which can be thought of as a quadratic equation in $x^2$. That is, we can solve for the roots by using $x^2$ as our variable, and then plug it into ye ole quadratic equation:

\begin{aligned} x^2 &= \frac{20\pm \sqrt{400-320}}{2} \\ &\implies x^2 = 10 \pm 2\sqrt{5} \\ &\implies x = \pm \sqrt{10 \pm 2\sqrt{5}} \\ \end{aligned}

We know that $10 \pm 2\sqrt{5}$ is always non-negative, so we get all four roots as the various permutations of plus and minus signs of within our $x$ term. We'll alias $\alpha, \beta$ to be partitions of the roots divided by the sign of the inner root:

\begin{aligned} \alpha = \pm \sqrt{10 + 2\sqrt{5}} \\ \beta = \pm \sqrt{10 - 2\sqrt{5}} \\ \end{aligned}

And we know that our splitting field, then, is $K = \mathbb Q(\alpha, \beta)$.

Next, to determine the contents of the Galois group of $f$, we might consider what kinds of permutations exist between these roots. We can observe that $f$ is irreducible per Eisenstein's Criterion which states that a polynomial of the form

$f(x) = a*nx^n + a*{n-1}x^{n-1} + \cdots + a_1x^{1} + a_0$

is irreducible iff there is a prime $p$ s.t.:

• $p$ does not divide the coefficient of the highest order term: $p \not{|} \; a_n$
• $p$ does divide the coefficients of the rest of the term: $p \; | \; a_{n-1}, ... , p \; | \; a_{0}$
• $p^2$ does not divide the constant term: $p^2 \not{|}\; a_0$

So, our polynomial $f(x) = x^4 -20x + 80$ is irreducible for $p = 5$ since:

• $5$ does not divide 1
• 5 does divide 20 and 80
• 25 does not divide 80

This tells us that the minimal polynomial for $\alpha$ over $\mathbb Q$, denoted $irr(\alpha; \; \mathbb Q)$ is actually $f$ since it has $\alpha$ as a root, and is irreducible. This tells us that the degree of this extension $[K : \mathbb Q]$ is at least 4 since the degree of the minimal polynomial of $\mathbb Q(\alpha) \subset K$ is 4. But, $K$ also contains $\beta$, which could possibly push the lower bound higher.

We can use this to determine possible elements of $Gal(f)$. Consider the automorphisms of $K$:

\begin{aligned} \psi : K &\rightarrow K \\ \psi(-\alpha) &= -\psi(\alpha) \\ \psi(-\beta) &= -\psi(\beta) \\ \end{aligned}

And while this would be true of any automorphism because it's linear w.r.t. multiplication, but it's particularly special because our roots are $\pm \alpha, \pm \beta$, so any permutation of the roots our Galois group is singularly determined by where it sends $\alpha$ and where it sends $\beta$.

Using that lil tidbit, it's straightforward to tabulate all possible elements of our Galois group in terms of where they send each positive root respectively:

$\alpha \mapsto$$\alpha$$\beta$$-\alpha$$-\beta$$\beta$$-\alpha$$-\beta$$\alpha$
$\beta \mapsto$$\beta$$-\alpha$$-\beta$$\alpha$$\alpha$$\beta$$-\alpha$$-\beta$

Note that these are the possible combinations of places we can send the roots, which do not necessarily fix automorphisms. And though this ordering seems entirely arbitrary, of course I'm cheating since I already know what $Gal(f)$ is, so I structured it accordingly.

$\epsilon$$\sigma$$\sigma^2$$\sigma^3$$\tau$$\sigma\tau$$\sigma^2\tau$$\sigma^3\tau$
$\alpha \mapsto$$\alpha$$\beta$$-\alpha$$-\beta$$\beta$$-\alpha$$-\beta$$\alpha$
$\beta \mapsto$$\beta$$-\alpha$$-\beta$$\alpha$$\alpha$$\beta$$-\alpha$$-\beta$

We can retcon an intuitive derivation of these labels starting with $\sigma$ which sends $\alpha \mapsto \beta$. We can convince ourselves that $\sigma^2$ would send $\alpha \mapsto -\alpha$ per:

$\alpha \underset{\sigma}\longmapsto \beta \underset{\sigma}\longmapsto -\alpha$

So these labels are consistent with the laws of composition, what a coincidence! Students of Group theory will recognize this table to be the elements of the dihedral group with eight elements $D_4$ which has one element of order 4 which is $\sigma$ and another element with order 2 which is $\tau$, the product of which gives this dihedral group its name.

Because of the way the roots of this particular polynomial work, the possible automorphisms are contrived to be the elements of $D_4$. However, we don't know a priori that each of these elements is an automorphism, so we do a little bit more massaging to narrow down $D_4$ to the elements of our Galois group. We do this by considering a special element $q \in Q \subset K$ and apply each permutation $\psi \in D_4$ to see if it fixes $q$ since we know that $\mathbb Q$-automorphisms of $K$ leave elements of $\mathbb Q$ unchanged. We set $q$ to some expression evaluates to a rational number, but will also be sufficiently greebled by $\psi$ such as:

\begin{aligned} q &= \alpha\beta(\alpha^2 - \beta^2) \\ &=\Big(\sqrt{10 + 2\sqrt{5}}\Big)\Big(\sqrt{10 - 2\sqrt{5}}\Big)\Bigg(\sqrt{10 + 2\sqrt{5}}^2 - \sqrt{10 - 2\sqrt{5}}^2\Bigg) \\ &= \sqrt{80}\sqrt{80} \\ &= 80 \in \mathbb Q \\ \end{aligned}