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

### Chapter 33. SOLVING EQUATIONS BY RADICALS

In this final chapter, Galois theory will be used explicitly to answer practical questions about solving equations.

In the introduction to this book we saw that classical algebra was devoted largely to finding methods for solving polynomial equations. The quadratic formula yields the solutions of every equation of degree 2, and similar formulas have been found for polynomials of degrees 3 and 4. But every attempt to find explicit formulas, of the same kind as the quadratic formula, which would solve a general equation of degree 5 or higher ended in failure. The reason for this was finally discovered by the young Galois, who showed that an equation is solvable by the kind of explicit formula we have in mind if and only if its group of symmetries has certain properties. The group of symmetries is, of course, the Galois group which we have already defined, and the required group properties will be formulated in the next few pages.

Galois showed that the groups of symmetries of all equations of degree ≤4 have the properties needed for solvability, whereas equations of degree 5 or more do not always have them. Thus, not only is the classical quest for radical formulas to solve all equations of degree > 4 shown to be futile, but a criterion is made available to test any equation and determine if it has solutions given by a radical formula. All this will be made clear in the following pages.

Every quadratic equation *ax*^{2} + *bx* + *c* = 0 has its roots given by the formula

Equations of degree 3 and 4 can be solved by similar formulas. For example, the cubic equation *x*^{3} + *ax* + *b* = 0 has a solution given by

Such expressions are built up from the coefficients of the given polynomials by repeated addition, subtraction, multiplication, division, *and taking roots*. Because of their use of radicals, they are called *radical expressions* or *radical formulas*. A polynomial *a*(*x*) is *solvable by radicals* if there is a radical expression giving its roots in terms of its coefficients.

Let us return to the example of *x*^{3} + *ax* + *b* = 0, where *a* and *b* are rational, and look again at __Formula (1)__. We may interpret this formula to assert that if we start with the field of coefficients , adjoin the square root , then adjoin the cube roots , we reach a field in which *x*^{3} + *ax* + *b* = 0 has its roots.

In general, to say that the roots of *a*(*x*) are given by a radical expression is the same as saying that we can extend the field of coefficients of *a*(*x*) by successively adjoining *n*th roots (for various *n*), and in this way obtain a field which contains the roots of *a*(*x*). We will express this notion formally now, in the language of field theory.

*F*(*c*_{1}, …, *c _{n}*) is called a

*radical extension*of

*F*if, for each

*i*, some power of

*c*is in

_{i}*F*(

*c*

_{1}, …,

*c*

_{i}_{ }

_{−}

_{ 1}). In other words,

*F*(

*c*

_{1}, …,

*c*) is an iterated extension of

_{n}*F*obtained by successively adjoining

*n*th roots, for various

*n*. We say that a polynomial

*a*(

*x*) in

*F*[

*x*] is

*solvable by radicals*if there is a radical extension of

*F*containing all the roots of

*a*(

*x*), that is, containing the root field of

*a*(

*x*).

To deal effectively with *n*th roots we must know a little about them. To begin with, the *n*th roots of 1, called *n*th *roots of unity*, are, of course, the solutions of *x ^{n}* − 1 = 0. Thus, for each

*n*, there are exactly

*n n*th roots of unity. As we shall see, everything we need to know about roots will follow from properties of the roots of unity.

In the *n*th roots of unity are obtained by de Moivre’s theorem. They consist of a number *ω* and its first *n* powers: 1 = *ω*^{0}, *ω*, *ω*^{2}, …, *ω ^{n}*

^{ }

^{−}

^{ 1}. We will not review de Moivre’s theorem here because, remarkably, the main facts about roots of unity are true in

*every*field of characteristic zero. Everything we need to know emerges from the following theorem:

**Theorem 1** *Any finite group of nonzero elements in a field is a cyclic group*. (*The operation in the group is the field*’*s multiplication*.)

PROOF: If *F** denotes the set of nonzero elements of *F*, suppose that *G* ⊆ *F**, and that *G*, with the field’s “multiply” operation, is a group of *n* elements. We will compare *G* with * _{n}* and show that

*G*, like

*, has an element of order*

_{n}*n*and is therefore cyclic.

For any integer *k*, let *g*(*k*) be the number of elements of order *k* in *G*, and let *z*(*k*) be the number of elements of order *k* in * _{n}*. For every positive integer

*k*which is a factor of

*n*, the equation

*x*= 1 has

^{k}*at most k*solutions in

*F*; thus,

(*) *G contains at most k elements whose order is a factor of k*.

If *G* has an element *a* of order *k*, then ⟨*a*⟩ = {*e*, *a*, *a*^{2}, …, *a ^{k}*

^{ }

^{−}

^{ 1}) are

*all*the distinct elements of

*G*whose order is a factor of

*k*. [By (*), there cannot be any others.] In

*, the subgroup*

_{n}contains all the elements of * _{n}* whose order is a factor of

*k*.

Since ⟨*a*⟩ and ⟨*n*/*k*⟩ are cyclic groups with the same number of elements, they are isomorphic; thus, the number of elements of order *k* in ⟨*a*⟩ is the same as the number of elements of order *k* in ⟨*n*/*k*⟩. Thus, *g*(*k*) = *z*(*k*).

Let us recapitulate: if *G has* an element of order *k*, then *g*(*k*) = *z*(*k*); but if *G* has *no* such elements, then *g*(*k*) = 0. Thus, for each positive integer *k* which is a factor of *n*, the number of elements of order *k* in *G* is less than (or equal to) the number of elements of order *k* in * _{n}*.

Now, every element of *G* (as well as every element of * _{n}*) has a well-defined order, which is a divisor of

*n*. Imagine the elements of both groups to be partitioned into classes according to their order, and compare the classes in

*G*with the corresponding classes in

*. For each*

_{n}*k*,

*G*has as many

*or fewer*elements of order

*k*than

*does. So if*

_{n}*G*had no elements of order

*n*(while

*have one), this would mean that*

_{n}does*G*has fewer elements than

*, which is false. Thus,*

_{n}*G must*have an element of order

*n*, and therefore

*G*is cyclic. ■

The *n*th roots of unity (which are contained in *F* or a suitable extension of *F*) obviously form a group with respect to multiplication. By __Theorem 1__, it is a cyclic group. Any generator of this group is called a *primitive nth* root of unity. Thus, if *ω* is a primitive *n*th root of unity, the set of all the *n*th roots of unity is

1, *ω*, *ω*^{2}, …, *ω ^{n}*

^{ }

^{−}

^{ 1}

If *ω* is a primitive *n*th root of unity, *F*(*ω*) is an *abelian extension of F* in the sense that *g* ∘ *h* = *h* ∘ *g* for any two *F-*fixing automorphisms *g* and *h* of *F*(*ω*)). Indeed, any automorphism must obviously send *n*th roots of unity to nth roots of unity. So if *g*(*ω*) = *ω ^{r}* and

*h*(

*ω*) =

*ω*, then

^{s}*g*∘

*h*(

*ω*) =

*g*(

*ω*) =

^{s}*ω*and analogously,

^{rs}*h*∘

*g*(

*ω*) =

*ω*. Thus,

^{rs}*g*∘

*h*(

*ω*) =

*h*∘

*g*(

*ω*). Since

*g*and

*h*fix

*F*, and every element of

*F*(

*ω*) is a linear combination of powers of

*ω*with coefficients in

*F*,

*g*∘

*h*=

*h*∘

*g*.

Now, let *F* contain a primitive *n*th root of unity, and therefore all the nth roots of unity. Suppose *a* ∈ *F*, and *a* has an nth root *b* in *F*. It follows, then, that *all the n*th *roots of a are in F*, *for they are b*, *bω*, *bω*^{2}, …, *bω ^{n}*

^{ }

^{−}

^{ 1}. Indeed, if

*c*is any other

*n*th root of

*a*, then clearly

*c*/

*b*is an

*n*th root of 1, say

*ω*hence

^{r};*c*=

*bω*

^{4}. We may infer from the above that

*if F contains a primitive n*th

*root of unity*,

*and b is an n*th

*root of a*,

*then F*(

*b*)

*is the root field of x*−

^{n}*a over F*.

In particular, *F*(*b*) *is an abelian extension of* *F*. Indeed, any *F*-fixing automorphism of *F*(*b*) must send *n*th roots of *a* to *n*th roots of *a*: for if *c* is any *n*th root of *a* and *g* is an *F*-fixing automorphism, then *g*(*c*)* ^{n}* =

*g*(

*c*) =

^{n}*g*(

*a*) =

*a*; hence

*g*(

*c*) is an

*n*th root of

*a*. So if

*g*(

*b*) =

*bω*and

^{r}*h*(

*b*) =

*bω*, then

^{s}*g* ∘ *h*(*b*) = *g*(*bω ^{s}*) =

*bω*=

^{r}ω^{s}*bω*

^{r}^{ + s}

and

*h* ∘ *g*(*b*) = *h*(*bω ^{r}*) =

*bω*=

^{s}ω^{r}*bw*

^{r}^{ + s}

hence *g* ∘ *h*(*b*) = *h* ∘ *g*(*b*). Since *g* and *h* fix *F*, and every element in *F*(*b*) is a linear combination of powers of *b* with coefficients in *F*, it follows that *g* ∘ *h* = *h* ∘ *g*.

If *a*(*x*) is in *F*[*x*], remember that *a*(*x*) is solvable by radicals just as long as there exists *some* radical extension of F containing the roots of *a*(*x*). [*Any* radical extension of F containing the roots of *a*(*x*) will do.] Thus, we may as well assume that any radical extension used here begins by adjoining to F the appropriate roots of unity; *henceforth we will make this assumption*. Thus, if *K* = *F*(*c*_{1}, …, *c _{n}*) is a radical extension of

*F*, then

is a sequence of simple *abelian* extensions. (The extensions are all *abelian* by the comments in the preceding three paragraphs.)

Still, this is not quite enough for our purposes: In order to use the machinery which was set up in the previous chapter, we must be able to say that each field in (2) *is a root field over F*. This may be accomplished as follows: Suppose we have already constructed the extensions *I*_{0} ⊆ *I*_{1} ⊆ ⋯ ⊆ *I _{q}* in (2) so that

*I*is a root field over

_{q}*F*. We must extend

*I*to

_{q}*I*

_{q}_{ + 1}, so

*I*

_{q}_{ + 1}is a root field over

*F*. Also,

*I*

_{q}_{ + 1}must include the element

*c*

_{q}_{ + 1}, which is the

*n*th root of some element

*a*∈

*I*.

_{q}Let *H* = {*h*_{1}, …, *h _{r}*) be the group of all the

*F*-fixing automorphisms of

*I*, and consider the polynomial

_{q}*b*(*x*) = [*x ^{n}* −

*h*

_{1}(

*a*)][

*x*−

^{n}*h*

_{2}(

*a*)] ⋯ [

*x*−

^{n}*h*(

_{r}*a*)]

By the proof of the lemma on page 327, one factor of *b*(*x*) is (*x ^{n}* −

*a*); hence

*c*

_{q}_{ + 1}is a root of

*b*(

*x*). Moreover, by the same lemma, every coefficient of

*b*(

*x*) is in the fixfield of

*H*, that is, in

*F*. We now define

*I*

_{q}_{ + 1}to be the root field of

*b*(

*x*) over

*F*. Since all the roots of

*b*(

*x*) are

*n*th roots of elements in

*I*, it follows that

_{q}*I*

_{q}_{ + 1}is a radical extension of

*I*. The roots may be adjoined one by one, yielding a succession of

_{q}*abelian*extensions, as discussed previously. To conclude, we may assume in (2) that

*K*is a root field over

*F*.

If **G** denotes the Galois group of *K* over *F*, each of these fields *I _{k}* has a fixer which is a subgroup of

**G**. These fixers form a sequence

For each *k*, by __Theorem 4__ of __Chapter 32__, is a normal subgroup of , and ≅ *Gal*(*I _{k}*

_{ + 1}:

*I*which is abelian because is an abelian extension of

_{k}*I*. The following definition was invented precisely to account for this situation.

_{k}*A group G is called solvable if it has a sequence of subgroups* {

*e*} =

*H*

_{0}⊆

*H*

_{1}⊆ ⋯ ⊆

*H*=

_{m}*G such that for each k*,

*G*

_{k}is a normal subgroup of G_{k}_{ + 1}

*and G*

_{k}_{ + 1}/

*G*.

_{k}is abelianWe have shown that *if K is a radical extension of F*, *then Gal*(*K*: *F*) *is a solvable group*. We wish to go further and prove that if *a*(*x*) is any polynomial which is solvable by radicals, its Galois group is solvable. To do so, we must first prove that any homomorphic image of a solvable group is solvable. A key ingredient of our proof is the following simple fact, which was explained on page 152: *G*/*H* is abelian iff *H* contains all the products *xyx*^{−}^{ 1}*y* ^{−}^{ 1} for all *x* and *y* in *G*. (The products *xyx*^{−}^{ 1}*y*^{−}^{ 1} are called “commutators” of *G*.)

**Theorem 2** *Any homomorphic image of a solvable group is a solvable group*.

PROOF: Let *G* be a solvable group, with a sequence of subgroups

{*e*) ⊆ *H*_{1} ⊆ ⋯ ⊆ *H _{m}* =

*G*

as specified in the definition. Let *f*: *G* → *X* be a homomorphism from *G* onto a group *X*. Then *f*(*H*_{0}), *f*(*H*_{1}), …, *f*(*H _{m}*) are subgroups of

*X*, and clearly {

*e*} ⊆

*f*(

*H*

_{0}) ⊆

*f*(

*H*

_{1}) ⊆ ⋯ ⊆

*f*(

*H*) =

_{m}*X*. For each

*i*we have the following: if

*f*(

*a*)∈

*f*(

*H*) and

_{i}*f*(

*x*) ∈

*f*(

*H*

_{i}_{ + 1}), then

*a*∈

*H*and

_{i}*x*∈

*H*

_{i}_{ + 1}; hence

*xax*

^{−}

^{ 1}∈

*H*and therefore

_{i}*f*(

*x*)

*f*(

*a*)

*f*(

*x*)

^{−}

^{ 1}∈

*f*(

*H*). So

_{i}*f*(

*H*) is a normal subgroup of

_{i}*f*(

*H*

_{i}_{ + 1}). Finally, since

*H*

_{i}_{ + 1}/

*H*is abelian, every commutator

_{i}*xyx*

^{−}

^{ 1}

*y*

^{−}

^{ 1}(for all

*x*and

*y*in

*H*

_{i}_{ + 1}) is in

*H*; hence every

_{i}*f*(

*x*)

*f*(

*y*)

*f*(

*x*)

^{−}

^{ 1}

*f*(

*y*)

^{−}

^{ 1}is in

*f*(

*H*). Thus,

_{i}*f*(

*H*

_{i}_{ + 1})/

*f*(

*H*) is abelian. ■

_{i}Now we can prove the main result of this chapter:

**Theorem 3** *Let a*(*x*) *be a polynomial over a field* *F*. *If a*(*x*) *is solvable by radicals, its Galois group is a solvable group*.

PROOF: By definition, if *K* is the root field of *a*(*x*), there is an extension by radicals *F*(*c*_{1}, …, *c _{n}*) such that

*F*⊆

*K*⊆

*F*(

*c*

_{1}, …,

*c*). It follows by

_{n}__Theorem 4__of

__Chapter 32__that

*Gal*(

*F*(

*c*

_{1}, …,

*c*):

_{n}*F*)/

*Gal*(

*F*(

*c*

_{1}, …,

*c*):

_{n}*K*)≅

*Gal*(

*K*:

*F*); hence by that theorem,

*Gal*(

*K*:

*F*) is a homomorphic image of

*Gal*(

*F*(

*c*

_{1}, …,

*c*):

_{n}*F*) which we know to be solvable. Thus, by

__Theorem 2__

*Gal*(

*K*:

*F*) is solvable. ■

Actually, the converse of __Theorem 3__ is true also. All we need to show is that, if *K* is an extension of *F* whose Galois group over *F* is solvable, then *K* may be further extended to a radical extension of *F*. The details are not too difficult and are assigned as __Exercise E__ at the end of this chapter.

__Theorem 3__ together with its converse say that *a polynomial a*(*x*) *is solvable by radicals iff its Galois group is solvable*.

We bring this chapter to a close by showing that there exist groups which are not solvable, and there exist polynomials having such groups as their Galois group. In other words, *there are unsolvable polynomials*. First, here is an unsolvable group:

**Theorem 4** *The symmetric group S*_{5} *is not a solvable group*.

PROOF: Suppose *S*_{5} has a sequence of subgroups

{*e*} = *H*_{0} ⊆ *H*_{1} ⊆ … ⊆ *H _{m}* =

*S*

_{5}

as in the definition of solvable group. Consider the subset of *S*_{5} containing all the cycles (*ijk*) of length 3. We will show that if *H _{i}* contains all the cycles of length 3, so does the next smaller group

*H*

_{i}_{ }

_{−}

_{ 1}. It would follow in

*m*steps that

*H*

_{0}= {

*e*} contains all the cycles of length 3, which is absurd.

So let *H _{i}* contain all the cycles of length 3 in

*S*

_{5}. Remember that if

*α*and

*β*are in

*H*, then their commutator

_{i}*αβα*

^{−}

^{1}

*β*

^{−}

^{1}is in

*H*

_{i}_{ }

_{−}

_{ 1}. But any cycle (

*ijk*) is equal to the commutator

(*ilj*)(*jkm*)(*ilj*)^{−}^{1}(*jkm*)^{−}^{1} = (*ilj*)(*jkm*)(*jli*)(*mkj*) = (*ijk*)

hence every (*ijk*) is in *H _{i}*

_{ }

_{−}

_{ 1}, as claimed. ■

Before drawing our argument toward a close, we need to know one more fact about groups; it is contained in the following classical result of group theory:

**Cauchy’s theorem** *Let G be a finite group of n elements*. *If p is any prime number which divides n*, *then G has an element of order p*.

For example, if *G* is a group of 30 elements, it has elements of orders 2, 3, and 5. To give our proof a trimmer appearance, we will prove Cauchy’s theorem specifically for *p* = 5 (the only case we will use here, anyway). However, the same argument works for any value of *p*.

PROOF: Consider all possible 5-tuples (*a*, *b*, *c*, *d*, *k*) of elements of *G* whose product *abcdk* = *e*. How many distinct 5-tuples of this kind are there? Well, if we select *a*, *b*, *c*, and *d* at random, there is a unique *k* = *d*^{−}^{1}*c*^{−}^{1}*b*^{−}^{1}*a*^{−}^{1} in *G*making *abcdk* = *e*. Thus, there are *n*^{4} such 5-tuples.

Call two 5-tuples *equivalent* if one is merely a cyclic permutation of the other. Thus, (*a*, *b*, *c*, *d*, *k*) is equivalent to exactly five distinct 5-tuples, namely, (*a*, *b*, *c*, *d*, *k*), (*b*, *c*, *d*, *k*, *a*), (*c*, *d*, *k*, *a*, *b*), (*d*, *k*, *a*, *b*, *c*) and (*k*, *a*, *b*, *c*, *d*). The only exception occurs when a 5-tuple is of the form (*a*, *a*, *a*, *a*, *a*) with all its components equal; it is equivalent only to itself. Thus, the equivalence class of any 5-tuple of the form (*a*, *a*, *a*, *a*, *a*) has a single member, while all the other equivalence classes have five members.

Are there any equivalence classes, other than {(*e*, *e*, *e*, *e*, *e*)}, with a single member? *If not*, then 5|(*n*^{4} − 1) [for there are *n*^{4} 5-tuples under consideration, less (*e*, *e*, *e*, *e*, *e*)]; hence *n*^{4} ≡ 1 (mod 5). But we are assuming that 5|*n*; hence *n*^{4} ≡ 0 (mod 5), which is a contradiction.

This contradiction shows that there must be a 5-tuple (*a*, *a*, *a*, *a*, *a*) ≠ (*e*, *e*, *e*, *e*, *e*) such that *aaaaa* = *a*^{5} = *e*. Thus, there is an element *a* ∈ *G* of order 5. ■

We will now exhibit a polynomial in [*x*] having *S*_{5} as its Galois group (remember that *S*_{5} is *not* a solvable group).

Let *a*(*x*) = *x*^{5} − 5*x* − 2. By Eisenstein’s criterion, *a*(*x* + 2) is irreducible over ; hence *a*(*x*) also is irreducible over . By elementary calculus, *a*{*x*) has a single maximum at (−1, 2), a single minimum at (1, −6), and a single point of inflection at (0, −2). Thus (see figure), its graph intersects the *x* axis exactly three times. This means that *a*(*x*) has three real roots, *r*_{1}, *r*_{2}, and *r*_{3}, and therefore two complex roots, *r*_{4} and *r*_{5}, which must be complex conjugates of each other.

Let *K* denote the root field of *a*(*x*) over , and **G** the Galois group of *a*(*x*). As we have already noted, every element of **G** may be identified with a permutation of the roots *r*_{1}, *r*_{2}, *r*_{3}, *r*_{4}, *r*_{5} of *a*(*x*), so **G** may be viewed as a subgroup of *S*_{5}. We will show that **G** is all of *S*_{5}.

Now, [(*r*_{1}): ] = 5 because *r*_{1} is a root of an irreducible polynomial of degree 5 over . Since [*K*: ] = [*K*: (*r*_{1})][(*r*_{1}): ], it follows that 5 is a factor of [*K*: ]. Then, by Cauchy’s theorem, **G** contains an element of order 5. This element must be a cycle of length 5: for by __Chapter 8__, every other element of *S*_{5} is a product of two or more disjoint cycles of length <5, and such products cannot have order 5. (Try the typical cases.) Thus, **G** *contains a cycle of length* 5.

Furthermore, **G** *contains a transposition* because complex conjugation *a* + *bi* → *a* − *bi* is obviously a -fixing automorphism of *K*; it interchanges the complex roots *r*_{4} and *r*_{5} while leaving *r*_{1}, *r*_{2}, and *r*_{3} fixed.

Any subgroup *G* ⊆ *S*_{5} which contains a transposition *τ* and a cycle *σ* of length 5 necessarily contains all the transpositions. (They are *στσ*^{−}^{1}, *σ*^{2}*τσ*^{−}^{2}, *σ*^{3}*τσ*^{−}^{3}, *σ*^{4}*τσ*^{−}^{4}, and their products; check this by direct computation!) Finally, if *G* contains all the transpositions, it contains everything else: for every member of *S*_{5} is a product of transpositions. Thus, **G** = *S*_{5}.

We have just given an example of a polynomial *a*(*x*) of degree 5 over whose Galois group *S*_{5} is not solvable. Thus, *a*(*x*) is an example of a polynomial of degree 5 which cannot be solved by radicals. In particular, there cannot be a radical formula (on the model of the quadratic formula) to solve all polynomial equations of degree 5, since this would imply that every polynomial equation of degree 5 has a radical solution, and we have just exhibited one which does not have such a solution.

In __Exercise B__ it is shown that *S*_{3}, *S*_{4}, and all their subgroups are solvable; hence every polynomial of degree ≤4 is solvable by radicals.

**EXERCISES**

**A. Finding Radical Extensions**

**1** Find radical extensions of containing the following complex numbers:

(*a*)

(*b*)

(*c*)

**2** Show that the following polynomials in [*x*] are not solvable by radicals:

# (* a*)2

*x*

^{5}− 5

*x*

^{4}+ 5

(*b*)x^{5} − 4*x*^{2} + 2

(*c*)x^{5} − 4*x*^{4} + 2*x* + 2

**3** Show that *a*(*x*) = *x*^{5} − 10*x*^{4} + 40*x*^{3} − 80*x*^{2} + 79*x* − 30 is solvable by radicals over , and give its root field. [HINT: Compute (*x* − 2)^{5} − (*x* − 2).]

**4** Show that *ax*^{8} + *bx*^{6} + *cx*^{4} + *dx*^{2} + *e* is solvable by radicals over any field. (HINT: Let *y* = *x*; use the fact that every fourth-degree polynomial is solvable by radicals.)

**5** Explain why parts 3 and 4 do not contradict the principal finding of this chapter: that polynomial equations of degree *n* ≥ 5 do not have a general solution by radicals.

**† B. Solvable Groups**

Let *G* be a group. The symbol *H* *G* is commonly used as an abbreviation of “*H* is a *normal* subgroup of *G*.” A *normal series* of *G* is a finite sequence *H*_{0}, *H*_{1}, …,*H _{n}* of subgroups of

*G*such that

{*e*} = *H*_{0} *H*_{1} ⋯ *H _{n}* =

*G*

Such a series is called a *solvable series* if each quotient group *H _{i}*

_{ + 1}/

*H*is abelian.

_{i}*G*is called a

*solvable group*if it has a solvable series.

**1** Explain why every abelian group is, trivially, a solvable group.

**2** Let *G* be a solvable group, with a solvable series *H*_{0}, …, *H _{n}*. Let

*K*be a subgroup of

*G*. Show that

*J*

_{0}=

*K*∩

*H*

_{0}, …,

*J*=

_{n}*K*∩

*H*is a normal series of

_{n}*K*.

# ** 3** Use the remark immediately preceding

__Theorem 2__to prove that

*J*

_{0}, …,

*J*is a solvable series of

_{n}*K*.

**4** Use parts 2 and 3 to prove: Every subgroup of a solvable group is solvable.

**5** Verify that {*ε*} ⊆ {*ε*, *β*, *δ*} ⊆ *S*_{3} is a solvable series for *S*_{3}. Conclude that *S*_{3}, and all of its subgroups, are solvable.

**6** In *S*_{4}, let *A*_{4} be the group of all the even permutations, and let

*B* = {*ε*, (12)(34), (13)(24), (14)(23)}

Show that {*ε*} ⊆ *B* ⊆ *A*_{4} ⊆ *S*_{4} is a solvable series for *S*_{4}. Conclude that *S*_{4} and all its subgroups are solvable.

The next three sets of exercises are devoted to proving the converse of __Theorem 3__: If the Galois group of *a*(*x*) is solvable, then *a*(*x*) is solvable by radicals.

**† C. pth Roots of Elements in a Field**

Let *p* be a prime number, and *ω* a primitive *p*th root of unity in the field *F*.

**1** If *d* is any root of *x ^{p}* −

*a*∈

*F*[

*x*], show that

*F*(

*ω*,

*d*) is a root field of

*x*−

^{p}*a*. Suppose

*x*−

^{p}*a*is

*not*irreducible in

*F*[

*x*].

**2** Explain why *x ^{p}* −

*a*factors in

*F*[

*x*] as

*x*−

^{p}*a*=

*p*(

*x*)

*f*(

*x*), where both factors have degree ≤2.

# ** 3** If deg

*p*(

*x*) =

*m*, explain why the constant term of

*p*(

*x*) (let us call it

*b*) is equal to the product of

*m*pth roots of

*a*. Conclude that

*b*=

*ω*for some

^{k}d^{m}*k*.

**4** Use part 3 to prove that *b ^{p}* =

*a*.

^{m}**5** Explain why *m* and *p* are relatively prime. Explain why it follows that there are integers *s* and *t* such that *sm* + *tp* = 1.

**6** Explain why *b ^{sp}* =

*a*. Use this to show that (

^{sm}*b*)

^{s}a^{t}*=*

^{p}*a*.

**7** Conclude: If *x ^{p}* −

*a*is

*not*irreducible in

*F*[

*x*], it has a root (namely,

*b*) in

^{s}a^{t}*F*.

We have proved: *x ^{p}* −

*a either has a root in*.

*F or is irreducible over F*.

**† D. Another Way of Defining Solvable Groups**

Let *G* be a group. The symbol *H* *G* should be read, “*H* is a normal subgroup of *G*.” A *maximal* normal subgroup of *G* is an *H* *G* such that, if *H* *J* *G*, then necessarily *J* = *H* or *J* = *G*. Prove the following:

**1** If *G* is a *finite* group, every normal subgroup of *G* is contained in a maximal normal subgroup.

**2** Let *f*: *G* → *H* be a homomorphism. If *J* *H*, then *f*^{−}^{1}(*J*) < *G*.

# ** 3** Let

*K*

*G*. If is a subgroup of

*G*/

*K*, let denote the union of all the cosets which are members of . If

*G*/

*K*, then

*G*. (Use part 2.)

**4** If *K* is a maximal normal subgroup of *G*, then *G*/*K* has no nontrivial normal subgroups. (Use part 3.)

**5** If an abelian group *G* has no nontrivial subgroups, *G* must be a cyclic group of prime order. (Otherwise, choose some *a* ∈ *G* such that ⟨*a*⟩ is a proper subgroup of *G*.)

**6** If *H* *K* *G*, then *G*/*K* is a homomorphic image of *G*/*H*.

# ** 7** Let

*H*

*G*, where

*G*/

*H*is abelian. Then

*G*has subgroups

*H*

_{0}, …,

*H*such that

_{q}*H*=

*H*

_{0}

*H*

_{1}⋯

*H*=

_{q}*G*, where each quotient group

*H*

_{i}_{ + 1}/

*H*is cyclic of prime order.

_{i}It follows from part 7 that if *G* is a solvable group, then, by “filling in gaps, ” *G* has a normal series in which all the quotient groups are cyclic of prime order. Solvable groups are often defined in this fashion.

**E. If Gal(K: F) Is Solvable, K Is a Radical Extension of F**

Let *K* be a finite extension of *F*, where *K* is a root field over *F*, with *G* = *Gal*(*K*: *F*) a solvable group. As remarked in the text, we will assume that *F* contains the required roots of unity. By __Exercise D__, let *H*_{0}, …, *H _{n}* be a solvable series for

*G*in which every quotient

*H*

_{i}_{ + 1}/

*H*is cyclic of prime order. For any

_{i}*i*= 1, …,

*n*, let

*F*. and

_{i}*F*

_{i}_{ + 1}be the fixfields of

*H*and

_{i}*H*

_{i}_{ + 1}.

**1** Prove: *F _{i}* is a normal extension of

*F*

_{i}_{ + 1}, and [

*F*:

_{i}*F*

_{i}_{ + 1}] is a prime

*p*.

**2** Let *π* be a generator of *Gal*(*F _{i}*:

*F*

_{i}_{ + 1}),

*ω*a pth root of unity in

*F*

_{i}_{ + 1}, and

*b*∈

*F*. Set

_{i}*c* = *b* + *ωπ*^{−}^{1}(*b*) + *ω*^{2}*π*^{−}^{2}(*b*) + ⋯ + *ω ^{p}*

^{ }

^{−}

^{ 1}

*π*

^{−}

^{(p }

^{−}

^{ 1)}(

*b*)

Show that *π*(*c*) = *ωc*.

**3** Use part 2 to prove that *π ^{k}*(

*c*) =

^{p}*c*for every

^{p}*k*, and deduce from this that

*c*∈

^{p}*F*

_{i}_{ + 1}

**4** Prove that *F _{i}* is the root field of

*x*−

^{p}*c*over

^{p}*F*

_{i}_{ + 1}.

**5** Conclude that *K* is a radical extension of *F*.