## A Book of Abstract Algebra, Second Edition (1982)

### Chapter 32. GALOIS THEORY: THE HEART OF THE MATTER

If *K* is a field and *h* is an isomorphism from *K* to *K*, we call *h* an *automorphism* of *K* (automorphism = “self-isomorphism”).

We begin this chapter by restating __Theorems 5__ and __6__ of __Chapter 31__:

*Let K be the root field of some polynomial over F*; *suppose a* ∈ *K*:

(i)*Any isomorphism with domain K which fixes F is an automorphism of K*.

(ii)*If a and b are roots of an irreducible polynomial p*(*x*) *in F*[*x*], *there is an automorphism of K fixing F and sending a to b*.

Rule (i) is merely a restatement of __Theorem 5__ of __Chapter 31__, using the notion of automorphism. Rule (ii) is a result of combining __Theorem 4__ of __Chapter 31__ [which asserts that there exists an *F*-fixing isomorphism from *L* = *F*(*a*) to *L*′ = *F*(*b*)] with __Theorem 6__ of the same chapter.

Let *K* be the root field of a polynomial *a*(*x*) in *F*[*x*], If *c*_{1}, *c*_{2}, …, *c _{n}* are the roots of

*a*(

*x*), then

*K*=

*F*(

*c*

_{1},

*c*

_{2}, …,

*c*), and, by (*) on page 316, any automorphism

_{n}*h*of

*K*which fixes

*F permutes c*

_{1},

*c*

_{2}, …,

*c*. On the other hand, remember that every element

_{n}*a*in

*F*(

*c*

_{1},

*c*

_{2}, …,

*c*) is a sum of terms of the form

_{n}where the coefficient *k* of each term is in *F*. If *h* is an automorphism which fixes *F*, *h* does not change the coefficients, so *h*(*a*) is completely determined once we know *h*(*c*_{1}), …, *h*(*c _{n}*). Thus,

*every automorphism*

*of K fixing F is completely determined by a permutation of the roots of a*(

*x*).

This is *very* important!

What it means is that we may identify the automorphisms of *K* which fix *F* with permutations of the roots of *a*(*x*).

It must be pointed out here that, just as the symmetries of geometric figures determine their geometric properties, so the symmetries of equations (that is, permutations of their roots) give us all the vital information needed to analyze their solutions. Thus, if *K* is the root field of our polynomial *a*(*x*) over *F*, we will now pay very close attention to the automorphisms of *K* which fix *F*.

To begin with, how many such automorphisms are there? The answer is a classic example of mathematical elegance and simplicity.

**Theorem 1** *Let K be the root field of some polynomial over F. The number of automorphisms of K fixing F is equal to the degree of K over F*.

PROOF: Let [*K* : *F*] = *n*, and let us show that *K* has exactly *n* automorphisms fixing *F*. By __Theorem 2__ of __Chapter 31__, *K* = *F*(*a*) for some *a* ∈ *K*.Let *p*(*x*) be the minimum polynomial of *a* over *F*; if *b* is any root of *p*(*x*), then by (ii) on the previous page, there is an automorphism of *K* fixing *F* and sending *a* to *b*. Since *p*(*x*) has *n* roots, there are exactly *n* choices of *b*, and therefore *n* automorphisms of *K* fixing *F*.

[Remember that every automorphism *h* which fixes *F* permutes the roots of *p*(*x*) and therefore sends *a* to *some* root of *p*(*x*); and *h* is completely determined once we have chosen *h*(*a*).] ■

For example, we have already seen that () is of degree 2 over . () is the root field of *x*^{2} – 2 over because () contains both roots of *x*^{2} – 2, namely ±. By __Theorem 1__, there are exactly two automorphisms of () fixing : one sends to ; it is the identity function. The other sends to –, and is therefore the function *a* + *b* → *a* − *b*.

Similarly, we saw that = (*i*), and is of degree 2 over . The two automorphisms of which fix are the identity function and the function *a* + *bi* → *a* − *bi* which sends every complex number to its complex conjugate.

As a final example, we have seen that (, ) is an extension of degree 4 over , so by __Theorem 1__, there are four automorphisms of (, ) which fix : Now, (, ) is the root field of (*x*^{2} – 2)(*x*^{2} − 3) over for it contains the roots of this polynomial, and any extension of containing the roots of (*x*^{2} – 2)(*x*^{2} − 3) certainly contains and . Thus, by (*) on page 316, each of the four automorphisms which fix sends roots of *x*^{2} – 2 to roots of *x*^{2} − 2, and roots of *x*^{2} − 3 to roots of *x*^{2} – 3. But there are only four possible ways of doing this, namely,

Since every element of (, ) is of the form , these four automorphisms (we shall call them *ε*, *α*, *β*, and *γ*) are the following:

If *K* is an extension of F, the automorphisms of *K* which fix *F form a group*. (The operation, of course, is composition.) This is perfectly obvious: for if *g* and *h* fix *F*, then for every *x* in *F*,

that is, *g* ∘ *h* fixes *F*. Furthermore, if

that is, if *h* fixes *F* so does *h*^{–1}

This fact is perfectly obvious, but nonetheless of great importance, for it means that we can now use all of our accumulated knowledge about groups to help us analyze the solutions of polynomial equations. And that is precisely what Galois theory is all about.

If *K* is the root field of a polynomial *a*(*x*) in *F*[*x*], *the group of all the automorphisms of K which fix F is called the Galois group of a*(*x*). We also call it the *Galois group of K over F*, and designate it by the symbol

*Gal*(*K* : *F*)

In our last example we saw that there are four automorphisms of (, ) which fix . We called them *ε*, *α*, *β*, and *γ*. Thus, the Galois group of (, ) over is *Gal*((, ) : ) = {*ε*, *α*, *β*, *γ*}; the operation is composition, giving us the table

As one can see, this is an abelian group in which every element is its own inverse; almost at a glance one can verify that it is isomorphic to _{2} × _{2}.

Let *K* be the root field of *a*(*x*), where *a*(*x*) is in *F*[*x*]. In our earlier discussion we saw that every automorphism of *K* fixing *F* [that is, every member of the Galois group of *a*(*x*)] may be identified with a permutation of the roots of *a*(*x*). However, it is important to note that *not every permutation of the roots of a*(*x*) *need be in the Galois group of a*(*x*), even when *a*(*x*) is irreducible. For example, we saw that (, ) = ( + ), where + is a root of the irreducible polynomial *x*^{4} – 10*x*^{2} + 1 over . Since *x*^{4} – 10*x*^{2} + 1 has four roots, there are 4! = 24 permutations of its roots, only four of which are in its Galois group. This is because only four of the permutations are genuine symmetries of *x*^{4} – 10*x*^{2} + 1, in the sense that they determine automorphisms of the root field.

In the discussion throughout the remainder of this chapter, let *F* and *K* remain fixed. *F* is an arbitrary field and *K* is the root field of some polynomial *a*(*x*) in *F*[*x*]. The thread of our reasoning will lead us to speak about fields *I*where *F* ⊆ *I* ⊆ *K*, that is, fields “between” *F* and *K*. We will

refer to them as *intermediate fields*. Since *K* is the root field of *a*(*x*) over *F*, it is also the root field of *a*(*x*) over *I* for every intermediate field *I*.

The letter **G** will denote the Galois group of *K* over *F*. With each intermediate field *I*, we associate the group

*I** = *Gal*(*K* : *I*)

that is, the group of all the automorphisms of *K* which fix *I*. It is obviously a subgroup of **G**. We will call *I** the *fixer* of *I*.

Conversely, with each subgroup *H* of **G** we associate the subfield of *K* containing all the *a* in *K* which are not changed by any *π* ∈ *H*. That is,

{*a* ∈ *K* : *π*(*a*) = *a* for every *π* ∈ *H*}

One verifies in a trice that this is a subfield of *K*. It obviously contains *F*, and is therefore one of the intermediate fields. It is called the *fixed field* of *H*. For brevity and euphony we call it the *fixfield* of *H*.

Let us recapitulate: Every subgroup *H* of **G** fixes an intermediate field *I*, called the *fixfield* of *H*. Every intermediate field *I* is fixed by a subgroup *H* of **G**, called the *fixer* of *I*. This suggests very strongly that there is a one-to-one correspondence between the subgroups of **G** and the fields intermediate between *F* and *K*. Indeed, this is correct. This one-to-one correspondence is at the very heart of Galois theory, because it provides the tie-in between properties of field extensions and properties of subgroups.

Just as, in __Chapter 29__, we were able to use vector algebra to prove new things about field extensions, now we will be able to use group theory to explore field extensions. The vector-space connection was a relative lightweight. The connection with group theory, on the other hand, gives us a tool of tremendous power to study field extensions.

We have not yet proved that the connection between subgroups of **G** and intermediate fields is a one-to-one correspondence. The next two theorems will do that.

**Theorem 2** *If H is the fixer of I*, *then I is the fixfield of H*.

PROOF: Let *H* be the fixer of *I*, and *I*′ be the fixfield of *H*. It follows from the definitions of *fixer* and *fixfield* that *I* ⊆ *I*′, so we must now show that *I*′ ⊆ *I*. We will do this by proving that *a* ∉ *I* implies *a* ∉ *I*′. Well, if *a* is an element of *K* which is not in *I*, the minimum polynomial *p*(*x*) of *a* over *I* must have degree ≥2 (for otherwise, *a* ∈ *I*). Thus, *p*(*x*) has another root *b*. By Rule (ii) given at the beginning of this chapter, there is an automorphism of *K*fixing *I* and sending *a* to *b*. This automorphism moves *a*, so *a* ∉ *I*′. ■

**Lemma** *Let H be a subgroup of* **G**, *and I the fixfield of H*. *The number of elements in H is equal to* [*K* : *I*].

PROOF: Let *H* have *r* elements, namely, *h*_{1}, …, *h _{r}*. Let

*K*=

*I*(

*a*). Much of our proof will revolve around the following polynomial:

*b*(*x*) = [*x* – *h*_{1}(*a*)][*x* − *h*_{2}(*a*)] ⋯ [*x* − *h _{r}*(

*a*)]

Since one of the *h _{i}* is the identity function, one factor of

*b*(

*x*) is (

*x*−

*a*), and therefore

*a is a root of b*(

*x*). In the next paragraph we will see that all the coefficients of

*b*(

*x*) are in

*I*, so

*b*(

*x*) ∈

*I*[

*x*]. It follows that

*b*(

*x*) is a multiple of the minimum polynomial of

*a*over

*I*, whose degree is exactly [

*K*:

*I*]. Since

*b*(

*x*) is of degree

*r*, this means that

*r*≥ [

*K*:

*I*], which is half our theorem.

Well, let us show that all the coefficients of *b*(*x*) are in *I*. We saw on page 314 that every isomorphism *h _{i}* :

*K*→

*K*can be extended to an isomorphism

*, :*

_{i}*K*[

*x*] →

*K*[

*x*]. Because

*is an isomorphism of polynomials, we get*

_{i}But *h _{i}* ∘

*h*

_{1},

*h*∘

_{i}*h*

_{2}, …,

*h*∘

_{i}*h*are

_{r}*r distinct*elements of

*H*, and

*H*has exactly

*r*elements, so they are all the elements of

*H*(that is, they are

*h*

_{1}, …,

*h*, possibly in a different order). So the factors of

_{r}*(*

_{i}*b*(

*x*)) are the same as the factors of

*b*(

*x*), merely in a different order, and therefore

*(*

_{i}*b*(

*x*)) =

*b*(

*x*). Since equal polynomials have equal coefficients,

*h*leaves the coefficients of

_{i}*b*(

*x*) invariant. Thus, every coefficient of

*b*(

*x*) is in the fixfield of

*H*, that is, in

*I*.

We have just shown that [*K* : *I*] ≤ *r*. For the opposite inequality, remember that by __Theorem 1__, [*K* : *I*] is equal to the number of *I*-fixing automorphisms of *K*. But there are at least *r* such automorphisms, namely *h*_{1}, …, *h _{r}*. Thus, [

*K*:

*I*] ≥

*r*, and we are done. ■

**Theorem 3** *If I is the fixfield of H*, *then H is the fixer of I*.

PROOF: Let *I* be the fixfield of *H*, and *I** the fixer of *I*. It follows from the definitions of fixer and fixfield that *H* ⊆ *I**. We will prove equality by showing that there are as many elements in *H* as in *I**. By the lemma, the order of *H* is equal to [*K* : *I*]. By __Theorem 2__, *I* is the fixfield of *I**, so by the lemma again, the order of *I** is also equal to [*K* : *I*]. ■

It follows immediately from __Theorems 2__ and __3__ that *there is a one-to-one correspondence between the subgroups of Gal*(*K* : *F*) *and the intermediate fields between K and F*. This correspondence, which matches every subgroup with its fixfield (or, equivalently, matches every intermediate field with its fixer) is called a *Galois correspondence*. It is worth observing that larger subfields correspond to smaller subgroups; that is,

As an example, we have seen that the Galois group of (, ) over is **G** = {*ε*, *α*, *β*, *γ*} with the table given on page 325. This group has exactly five subgroups—namely, {*ε*}, {*ε*, *α*}, {*ε*, *β*}, {*ε*, *γ*}, and the whole group **G**. They may be represented in the “inclusion diagram”:

On the other hand, there are exactly five fields intermediate between and (, ), which may be represented in the inclusion diagram:

If *H* is a subgroup of any galois group, let *H°* designate the fixfield of *H*. The subgroups of **G** in our example have the following fixfields:

(This is obvious by inspection of the way *ε*, *α*, *β*, and *γ* were defined on page 325.) The Galois correspondence, for this example, may therefore be represented as follows:

In order to effectively tie in subgroups of **G** with extensions of the field *F*, we need one more fact, to be presented next.

Suppose *E* ⊆ *I* ⊆ *K*, where *K* is a root field over *E* and *I* is a root field over *E*. (Hence by __Theorem 8__ of __Chapter 31__, *K* is a root field over *I*.) If *h* ∈ *Gal*(*K* : *E*), *h* is an automorphism of *K* fixing *E*. Consider the *restriction of h to I*, that is, *h* restricted to the smaller domain *I*. It is an isomorphism with domain *I* fixing *E*, so by Rule (i) given at the beginning of this chapter, it is an automorphism of *I*, still fixing *E*. We have just shown that if *h* ∈ *Gal*(*K* : *E*), then the restriction of *h* to *I* is in *Gal*(*I* : *E*). This permits us to define a function *μ* : *Gal*(*K* : *E*) → *Gal*(*I* : *E*) by the rule

*μ*(*h*) = the restriction of *h* to *I*

It is very easy to check that *μ* is a homomorphism. *μ* is surjective, because every *E*-fixing automorphism of *I* can be extended to an *E*-fixing automorphism of *K*, by __Theorem 6__ in __Chapter 31__.

Finally, if *h* ∈ *Gal*(*K* : *E*), the restriction of *h* to *I* is the identity function iff *h*(*x*) = *x* for every *x* ∈ *I*, that is, iff *h* fixes *I*. This proves that the kernel of *μ* is *Gal*(*K* : *I*).

To recapitulate: *μ* is a homomorphism from *Gal*(*K* : *E*) *onto Gal*(*I* : *E*) with kernel *Gal*(*K* : *I*). By the FHT, we immediately conclude as follows:

**Theorem 4** *Suppose E* ⊆ *I* ⊆ *K*, *where I is a root field over E and K is a root field over E*. *Then*

*It follows*, *in particular*, *that Gal*(*K* : *I*) *is a normal subgroup of Gal*(*K* : *E*).

**EXERCISES**

**† A. Computing a Galois Group**

**1**Show that (*i*, ) is the root field of (*x*^{2} + 1)(*x*^{2} − 2) over .

**# 2**Find the degree of (

*i*, ) over .

**3**List the elements of *Gal*((*i*, ) : ) and exhibit its table.

**4**Write the inclusion diagram for the subgroups of *Gal*((*i*, ) : ), and the inclusion diagram for the fields intermediate between and (*i*, ). Indicate the Galois correspondence.

**† B. Computing a Galois Group of Eight Elements**

**1**Show that (, , ) is the root field of (*x*^{2} – 2)(*x*^{2} – 3)(*x*^{2} − 5) over .

**2**Show that the degree of (, , ) over is 8.

**3**List the eight elements of **G** = *Gal*((, , ) : ) and write its table.

**4**List the subgroups of **G**. (By Lagrange’s theorem, any proper subgroup of **G** has either two or four elements.)

**5**For each subgroup of **G**, find its fixfield.

**6**Indicate the Galois correspondence by means of a diagram like the one on page 329.

**† C. A Galois Group Equal to S_{3}**

**1**Show that (, *i*) is the root field of *x*^{3} – 2 over , where designates the *real* cube root of 2. (HINT: Compute the complex cube roots of unity.)

**2**Show that [() : ] = 3.

**3**Explain why *x*^{2} + 3 is irreducible over (), then show that [(, *i*): ()] = 2. Conclude that [(, *i*) : ] = 6.

**4**Use part 3 to explain why *Gal*((, ) : ) has six elements. Then use the discussion following Rule (ii) on page 323 to explain why every element of *Gal*((, *i*) : ) may be identifed with a permutation of the three cube roots of 2.

**5**Use part 4 to prove that *Gal*((, *i*) : ) ≅ *S*_{3}.

**† D. A Galois Group Equal to D_{4}**

If *α* = is a real fourth root of 2, then the four fourth roots of 2 are ±*α* and ±*iα*. Explain parts 1–6, briefly but carefully:

# ** 1** (

*α*,

*i*) is the root field of

*x*

^{4}− 2 over .

**2** [(*α*) : ] = 4.

**3** *i* ∉ (*α*); hence [(*α*, *i*) : (*α*)] = 2.

**4** [(*α*, *i*) : ] = 8.

**5** {1, *α*, *α*^{2}, *α*^{3}, *i*, *iα*, *iα*^{2}, *iα*^{3}} is a basis for (*α*, *i*) over .

**6** Any -fixing automorphism *h* of (*α*, *i*) is determined by its effect on the elements in the basis. These, in turn, are determined by *h*(*α*) and *h*(*i*).

**7** Explain: *h*(*α*) must be a fourth root of 2 and *h*(*i*) must be equal to ±*i*. Combining the four possibilities for *h*(*α*) with the two possibilities for *h*(*i*) gives eight possible automorphisms. List them in the format

**8** Compute the table of the group *Gal*((*α*, *i*) : ) and show that it is isomorphic to *D*_{4}, the group of symmetries of the square.

**† E. A Cyclic Galois Group**

**# 1** Describe the root field

*K*of

*x*

^{7}− 1 over . Explain why [

*K*: ] = 6.

**2** Explain: If *α* is a primitive seventh root of unity, any *h* ∈ *Gal*(*K* : ) must send α to a seventh root of unity. In fact, *h* is determined by *h*(*α*).

**3** Use part 2 to list explicitly the six elements of *Gal*(*K* : ). Then write the table of *Gal*(*K* : ) and show that it is cyclic.

**4** List all the subgroups of *Gal*(*K* : ), with their fixfields. Exhibit the Galois correspondence.

**5** Describe the root field *L* of *x*^{6} − 1 over , and show that [*L* : ] = 2. Explain why it follows that there are no intermediate fields between and *L* (except for and *L* themselves).

**# 6** Let

*L*be the root field of

*x*

^{6}– 2 over . List the elements of

*Gal*(

*L*: ) and write its table.

**† F. A Galois Group Isomorphic to S_{5}**

Let *a*(*x*) = *x*^{5} − 4*x*^{4} + 2*x* + 2 ∈ [*x*], and let *r*_{1}, …, *r*_{5} be the roots of *a*(*x*) in . Let *K* (*r*_{1}, …, *r*_{5}) be the root field of *a*(*x*) over .

Prove: parts 1–3:

**1** *a*(*x*) is irreducible in ⌈*x*⌉.

**2** *a*(*x*) has three real and two complex roots. [HINT: Use calculus to sketch the graph of *y* = *a*(*x*), and show that it crosses the *x* axis three times.]

**3** If *r*_{1} denotes a real root of *a*(*x*), [(*r*_{1}) : ] = 5. Use this to prove that [*K* : ] is a multiple of 5.

**4** Use part 3 and Cauchy’s theorem (__Chapter 13__, __Exercise E__) to prove that there is an element *α* of order 5 in *Gal*(*K* : ). Since *α* may be identified with a permutation of {*r*_{1}, …, *r*_{5}}, explain why it must be a cycle of length 5. (HINT: Any product of disjoint cycles on {*r*_{1}, …, *r*_{5}} has order ≠ 5.)

**5** Explain why there is a transposition in *Gal* (*K* : ). [It permutes the conjugate pair of complex roots of *a*(*x*).]

**6** Prove: Any subgroup of *S*_{5} which contains a cycle of length 5 and a transposition must contain all possible transpositions in *S*_{5}, hence all of *S*_{5}. Thus, *Gal*(*K* : ) = *S*_{5}.

**G. Shorter Questions Relating to Automorphisms and Galois Groups**

Let *F* be a field, and *K* a finite extension of *F*. Suppose *a*, *b* ∈ *K*. Prove parts 1–3:

**1** If an automorphism *h* of *K* fixes *F* and *a*, then *h* fixes *F*(*a*).

**2** *F*(*a*, *b*)* = *F*(*a*)* *F*(*b*)*.

**3** Aside from the identity function, there are no -fixing automorphisms of (). [HINT: Note that () contains only real numbers.]

**4** Explain why the conclusion of part 3 does not contradict __Theorem 1__.

In the next three parts, let *ω* be a primitive *p*th root of unity, where *p* is a prime.

**5** Prove: If *h* ∈ *Gal*((*ω*) : ), then *h*(*ω*) = *ω ^{k}* for some

*k*where 1 ≤

*k*≤

*p*− 1.

**6** Use part 5 to prove that *Gal*((*ω*) : ) is an abelian group.

**7** Use part 5 to prove that *Gal*((*ω*) : ) is a cyclic group.

**† H. The Group of Automorphisms of C**

**1** Prove: The only automorphism of is the identity function. [HINT: If *h* is an automorphism, *h*(1) = 1; hence *h*(2) = 2, and so on.]

**2** Prove: Any automorphism of sends squares of numbers to squares of numbers, hence positive numbers to positive numbers.

**3** Using part 2, prove that if *h* is any automorphism of , *a* < *b* implies *h*(*a*) < *h*(*b*).

**# 4** Use parts 1 and 3 to prove that the only automorphism of is the identity function.

**5** List the elements of *Gal*( : ).

**6** Prove that the identity function and the function *a* + *bi* → *a* − *bi* are the only automorphisms of which fix .

**I. Further Questions Relating to Galois Groups**

Throughout this set of questions, let *K* be a root field over *F*, let **G** = *Gal*(*K* : *F*), and let *I* be any intermediate field. Prove the following:

**1** *I** = *Gal*(*K* : *I*) is a subgroup of **G**.

**2** If *H* is a subgroup of **G** and *H*° = {*a* ∈ *K* : *π*(*a*) = *a* for every *π* ∈ *H*}, then *H*° is a subfield of *K*, and *F* ⊆ *H*°.

**3** Let *H* be the fixer of *I*, and *I*′ the fixfield of *H*. Then *I* ⊆ *I*′. Let *I* be the fixfield of *H*, and *I** the fixer of *I*. Then *H* ⊆ *I**.

**# 4** Let

*I*be a normal extension of

*F*(that is, a root field of some polynomial over

*F*). If

**G**is abelian, then

*Gal*(

*K : I*) and

*Gal*(

*I : F*) are abelian. (HINT: Use

__Theorem 4__.)

**5** Let *I* be a normal extension of *F*. If **G** is a cyclic group, then *Gal*(*K* : *I*) and *Gal*(*I* : *F*) are cyclic groups.

**6** If **G** is a cyclic group, there exists exactly one intermediate field *I* of degree *k*, for each integer *k* dividing [*K* : *F*].

**† J. Normal Extensions and Normal Subgroups**

Suppose *F* ⊆ *K*, where *K* is a normal extension of *F*. (This means simply that *K* is the root field of some polynomial in *F*[*x*]: see __Chapter 31__, __Exercise K__.) Let *I*_{1} ⊆ *I*_{2} be intermediate fields.

**1** Deduce from __Theorem 4__ that, if *I*_{2} is a normal extension of *I*_{1}, then is a normal subgroup of .

**2** Prove the following for any intermediate field *I*: Let *h* ∈ *Gal* (*K* : *F*), *g* ∈ *I**, *a* ∈ *I*, and *b* = *h*(*a*). Then [*h* ∘ *g* ∘ *h*^{−}^{ 1}](*b*) = *b*. Conclude that

*hI** *h*^{−}^{ 1} ⊆ *h*(*I*)*

**3** Use part 2 to prove that *hI** *h*^{−}^{ 1} = *h*(*I*)*.

Two intermediate fields *I*_{1} and *I*_{2} are called *conjugate* iff there is an automorphism [i.e., an element *i* ∈ *Gal*(*K* : *F*)] such that *i*(*I*_{1}) = *I*_{2}.

**4** Use part 3 to prove that *I*_{1} and *I*_{2} are conjugate iff and are conjugate subgroups in the Galois group.

**5** Use part 4 to prove that for any intermediate fields *I*_{1} and *I*_{2} : iff is a normal subgroup of , then *I*_{2} is a normal extension of *I*_{1}.

Combining parts 1 and 5 we have: *I*_{2} is a normal extension of *I*_{1} iff is a normal subgroup of . (Historically, this result is the origin of the word “normal” in the term “normal subgroup.”)