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

### ANSWERS TO SELECTED EXERCISES

**CHAPTER 2**

** A3**This is not an operation on , since

*a**

*b*is not uniquely defined for any

*a*,

*b*∈ ,

*a*≠ 0, and

*b*≠ 0. If

*a*≠ 0 and

*b*≠ 0, then the equation

*x*

^{2}−

*a*

^{2}

*b*

^{2}= 0 has two roots—namely,

*x*=

*a**

*b*= ±

*ab*.

__B7__

(ii)*Associative law*:

(iii)*Identity element*: To find an identity element (if one exists), we try to solve the equation *x* * *e* = *x* for *e*:

Cross multiplying gives *xe* = *x*^{2} + *xe* + *x*, and thus 0 = *x*(*x* + 1). Consequently, this equation can be solved only for *x* = 0 and *x* = – 1. For positive values of *x* there is no solution.

** C3**We shall examine only one of the operations—namely, the one whose table is

To say that the operation is associative is to say that the equation

*x* * (*y* * *z*) = (*x* * *y*) * *z*

is true for all possible choices of *x*, of *y*, and of *z*. Each of the variables may be equal to either *a* or *b*, yielding eight cases to be checked:

1.*a* * (*a* * *a*) = *a* * *b* = *a*, (*a* * *a*) * *a* = *b* * *a* = *a*

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

3.*a* * (*b* * *a*) = *a* * *a* = *b*, (*a* * *b*) * *a* = *a* * *a* = *b*

4.*a* * (*b* * *b*) = *a* * *a* = *b*, (*a* * *b*) * *b* = *a* * *b* = *a*

5.*b* * (*a* * *a*) = *b* * *b* = *a*, (*b* * *a*) * *a* = *a* * *a* = *b*

6.*b* * (*a* * *b*) = *b* * *a* = *a*, (*b* * *a*) * *b* = *a* * *b* = *a*

7.*b* * (*b* * *a*) = *b* * *a* = *a*, (*b* * *b*) * *a* = *a* * *a* = *b*

8.*b* * (*b* * *b*) = *b* * *a* = *a*, (*b* * *b*) * *b* = *a* * *b* = *a*

Since *a* * (*a* * *b*) ≠ (*a* * *a*) * *b*, * is *not* associative.

By the way, this operation is commutative, since *a* * *b* = *a* = *b* * *a*. Using commutativity, we need not check all eight cases above; in fact, equality in cases 1, 3, 6, and 8 follows using commutativity. For example, in case 3, it follows from commutativity that *a* * (*b* * *a*) = (*b* * *a*) * *a* = (*a* * *b*) * *a*. Thus, for commutative operations, only four of the eight cases need to be checked. If you are able to show that it is sufficient to check only cases 2 and 4 for commutative operations, your work will be further reduced.

We have checked only one of the 16 operations. Check the remaining 15.

**CHAPTER 3**

** A4**(ii)

*Associative law*:

(iii) *Identity element*: Solve *x* * *e* = *x* for *e*.

If

then *e*(1 − *x*^{2}) = 0, so *e* = 0. Now check that *x* * 0 = 0 * *x* = *x*.

(iv) *Inverses*: Solve *x* * *x*′ = 0 for *x*′. (You should get *x*′ = − *x*.) Then check that *x* * (− *x*) = 0 = (− *x*) * *x*.

** B2**(ii)

*Associative law*:

(*a*, *b*) * [(*c*, *d*) * (*e*, *f*)] = (*a*, *b*) * (*ce*, *de* + *f*) = (*ace*, *bce* + *de* + *f*)

[(*a*, *b*) * (*c*, *d*)] * *f*) = (*ac*, *bc* + *d*) * (*e*, *f*) = (*ace*, *bce* + *de* + *f*)

(iii) *Identity element*: Solve (*a*, *b*) * (*e*_{1}, *e*_{2}) = (*a*, *b*) for (*e*_{1}, *e*_{2}). Suppose

(*a*, *b*) * (*e*_{1}, *e*_{2}) = (*ae*, *be*_{1} + *e*_{2}) = (*a*, *b*)

This implies that *ae*_{1} = *a* and *be*_{1} + *e*_{2} = *b*, so *e*_{1} = 1 and *e*_{2} = 0. Thus, (*e*_{1}, *e*_{2}) = (l, 0). Now check that (*a*, *b*) * (1, 0) = (1, 0) * (*a*, *b*) = (*a*, *b*).

(iv) *Inverses*: Solve (*a*, *b*) * (*a*′, *b*′) = (1, 0) for (*x*′, *b*′). You should get *a*′ = 1/*a* and *b*′ = − *b*/*a*. Then check.

** G1** The equation

*a*

_{4}=

*a*

_{1}+

*a*

_{3}means that the fourth digit of every codeword is equal to the sum of the first and third digits. We check this fact for the eight codewords of

*C*

_{1}in turn: 0 = 0 + 0, 1 = 0 + 1, 0 = 0 + 0, 1 = 0 + 1, 1 = 1 + 0, 0=1 + 1, 1 = 1 + 0, and 0 = 1 + 1.

** G2**(

*a*) As stated, the first three positions are the information positions. Therefore there are eight codewords; omitting the numbers in the last three positions (the redundancy positions), the codewords are 000, 001, 010, 011, 100, 101, 110, and 111. The numbers in the redundancy positions are specified by the parity-check equations given in the exercise. Thus, the complete codewords are 000000, 001001, …. (Complete the list.)

** G6**Let

*a*,

_{k}*b*,

_{k}*x*denote the digits in the

_{k}*k*th position of

**a**,

**b**, and

**x**, respectively. Note that if

*x*≠

_{k}*a*and

_{k}*x*≠

_{k}*b*, then

_{k}*a*=

_{k}*b*. But if

_{k}*x*≠

_{k}*a*and

_{k}*x*=

_{k}*b*, then

_{k}*a*≠

_{k}*b*. And if

_{k}*x*=

_{k}*a*and

_{k}*x*≠

_{k}*b*, then

_{k}*a*≠

_{k}*b*. Finally, if

_{k}*x*=

_{k}*a*and

_{k}*x*=

_{k}*b*, then

_{k}*a*=

_{k}*b*. Thus,

_{k}**a**differs from

**b**in only those positions where a differs from

**x**or

**b**differs from

**x**. Since a differs from

**x**in

*t*or fewer positions, and

**b**differs from

**x**in

*t*or fewer positions,

**a**cannot differ from

**b**in more than 2

*t*positions.

**CHAPTER 4**

** A3**Take the first equation,

*x*

^{2}

*a*=

*bxc*

^{−}

^{1}, and multiply on the right by

*c*:

*x*^{2}*ac* = (*bxc*^{−}^{ 1})*c* = *bx*(*c*^{−}^{ 1}*c*) = *bxe* = *bx*

From the second equation, *x*^{2}*ae* = *x*(*xac*) = *xacx*. Thus,

*xaca* = *bx*

By the cancellation law (cancel *x* on the right),

*xac* = *b*

Now multiply on the right by (*ac*)^{−}^{ 1} to complete the problem.

** C1**From

__Theorem 3__,

*a*

^{−}

^{ 1}

*b*

^{−}

^{ 1}= (

*ba*)

^{−}

^{ 1}and

*b*

^{−}

^{ 1}

*a*

^{−}

^{ 1}= (

*ab*)

^{−}

^{ 1}.

** D2**From

__Theorem 2__, if (

*ab*)

*c*=

*e*, then

*c*is the inverse of

*ab*.

** G3**Let

*G*and

*H*be abelian groups. To prove that

*G*×

*H*is abelian, let (

*a*,

*b*) and (

*c*,

*d*) be any two elements of

*G*×

*H*(so that

*a*∈

*G*,

*c*∈

*G*,

*b*∈

*H*, and

*d*∈

*H*) and show that (

*a*,

*b*) · (

*c*,

*d*) = (

*c*,

*d*) · (

*a*,

*b*).

PROOF:

**CHAPTER 5**

** B5**Suppose

*f*,

*g*∈

*H*. Then

*df*/

*dx*=

*a*and

*dg*/

*dx*=

*b*for constants

*a*and

*b*. From calculus,

*d*(

*f*+

*g*)

*dx*=

*df*/

*dx*+

*dg*/

*dx*=

*a*+

*b*, which is constant. Thus,

*f*+

*g*∈

*H*.

** C5**PROOF: (i)

*K*is closed under products. Let

*x*,

*y*∈

*K*. Then there are integers

*n*,

*m*> 0 such that

*x*∈

^{n}*H*and

*y*∈

^{m}*H*. Since

*H*is a subgroup of

*G*, it is closed under products—and hence under exponentiation (which is repeated multiplication of an element by itself). Thus (

*x*)

^{n}*∈*

^{m}*H*and (

*x*)

^{m}*∈*

^{n}*H*. Set

*q*=

*mn*.

*Since G is abelian*, (

*xy*)

*=*

^{q}*x*=

^{q}y^{q}*x*= (

^{mn}y^{mn}*x*)

^{n}*(*

^{m}*y*)

^{m}*∈*

^{n}*H*since both (

*x*)

^{n}*and (*

^{m}*y*)

^{m}*are in*

^{n}*H*. Complete the problem.

__D5__*S* = {*a*_{1} …, *a _{n}*} has

*n*elements. The

*n*products

*a*

_{1}

*a*

_{1},

*a*

_{1}

*a*

_{2}, …,

*a*

_{1}

*a*are elements of

_{n}*S*(why?) and no two of them can be equal (why?). Hence every element of

*S*is equal to one of these products. In particular,

*a*

_{1}=

*a*

_{1}

*a*for some

_{k}*k*. Thus,

*a*

_{1}

*e*=

*a*

_{1}

*a*, and hence

_{k}*e*=

*a*. This shows that

_{k}*e*∈

*S*. Now complete the problem.

** D7**(

*a*) Suppose

*x*∈

*K*. Then

(i) if *a* ∈ *H*, then *xax*^{−}^{ 1} ∈ *H*, and

(ii) if *xbx*^{−}^{ 1} ∈ *H*, then *b* ∈ *H*.

We shall prove that *x*^{−}^{ 1} ∈ *K*: we must first show that if *a* ∈ *H*, then *x*^{−}^{ 1}*a*(*x*^{−}^{ 1})^{−}^{ 1} = *x*^{−}^{ 1} *ax* ∈ *H*. Well, *a* = *x*(*x*^{−}^{ 1}*ax*)*x*^{−}^{ 1} and if *x*(*x*^{−}^{ 1} *ax*)*x*^{−}^{ 1} ∈ *H*, then *x*^{−}^{ 1}*ax* ∈ *H* by (ii) above. [Use (ii) with *x*^{−}^{1}*ax* replacing *b*.] Conversely, we must show that if *x*^{−}^{ 1}*ax* ∈ *H*, then *a* ∈ *H*. Well, if *x*^{−}^{ 1}*ax* ∈ *H*, then by (i) above, *x*(*x*^{−}^{ 1}*ax*)*x*^{−}^{ 1} = *a* ∈ *H*.

** E7**We begin by listing all the elements of

_{2}×

_{4}obtained by adding (1, 1) to itself repeatedly: (1, 1); (1, 1) + (1, 1) = (0, 2); (1, 1) + (1, 1) + (1, 1) = (1, 3); (1, 1) + (1, 1) + (1, 1) + (1, 1) = (0, 0). If we continue adding (1, 1) to multiples of itself, we simply obtain the above four pairs over and over again. Thus, (1, 1) is not a generator of

*all*of

_{2}×

_{4}.

This process is repeated for every element of _{2} × _{4}. None is a generator of _{2} × _{4}; hence _{2} × _{4} is not cyclic.

** F1**The table of

*G*is as follows:

Using the defining equations *a*^{2} = *e*, *b*^{3} = *e*, and *ba* = *ab*^{2}, we compute the product of *ab* and *ab*^{2} in this way:

(*ab*)(*ab*^{2}) = *a*(*ba*)*b*^{2} = *a*(*ab*^{2})*b*^{2} = *a*^{2}*b*^{3}*b* = *eeb* = *b*

Complete the problem by exhibiting the computation of all the table entries.

** H3**Recall the definition of the operation + in

__Chapter 3__,

__Exercise F__:

**x**+

**y**has s in those positions where

**x**and

**y**differ, and 0s elsewhere.

**CHAPTER 6**

** A4**From calculus, the function

*f*(

*x*) =

*x*

^{3}– 3

*x*is continuous and unbounded. Its graph is shown below. [

*f*is unbounded because

*f*(

*x*) =

*x*(

*x*

^{2}− 3) is an arbitrarily large positive number for sufficiently large positive values of

*x*, and an arbitrarily large negative number (large in absolute value) for sufficiently large negative values of

*x*.] Because

*f*is continuous and unbounded, the range of

*f*is . Thus,

*f*

*is surjective*. Now determine whether

*f*is injective, and prove your answer.

Graph of *f* (*x*) = *x*^{3} − 3*x*

__A6__*f is injective*: To prove this, note first that if *x* is an integer then *f*(*x*) is an integer, and if *x* is not an integer, then *f*(*x*) is not an integer. Thus, if *f*(*x*) = *f*(*y*), then *x* and *y* are either both integers or both nonintegers. *Case 1*, *both integers*: then *f*(*x*) = 2*x*, *f*(*y*) = 2*y*, and 2*x* = 2*y*; so *x* = *y*. *Case* 2, *both nonintegers*: ⋯ (Complete the problem. Determine whether *f* is surjective.)

** F5**Let

*A*= {

*a*

_{1},

*a*

_{2}, …,

*a*). If

_{n}*f*is any function from

*A*to

*A*, there are

*n*possible values for

*f*(

*a*), namely,

_{x}*a*

_{1},

*a*

_{2}, …,

*a*. Similarly there are

_{n}*n*possible values for

*f*(

*a*

_{2}). Thus there are

*n*

^{2}pairs consisting of a value of

*f*(

*a*

_{1}) together with a value of

*f*(

*a*

_{2}). Similarly there are

*n*

^{3}triples consisting of a value of

*f*(

*a*

_{1}), a value of

*f*(

*a*

_{2}), and a value of

*f*(

*a*

_{3}). We may continue in this fashion and conclude as follows: Since a function

*f*is specified by giving a value for

*f*(

*a*

_{1}), a value for

*f*(

*a*

_{2}), and so on up to a value for

*f*(

*a*), there are

_{n}*n*functions from

^{n}*A*to

*A*. Now, by reasoning in a similar fashion, how many bijective functions are there?

** H1**The following is one example (though not the only possible one) of a machine capable of carrying out the prescribed task:

*A* = {*a*, *b*, *c*, *d*} *S* = {*s*_{0}, *s*_{1}, *s*_{2}, *s*_{3}, *s*_{4}}

The next-state function is described by the following table:

To explain why the machine carries out the prescribed function, note first that the letters *b*, *c*, and *d* never cause the machine to change its state—only *a* does. If the machine begins reading a sequence in state *s*_{0}, it will be in state *s*_{3} after exactly three *a*’s have been read. Any subsequent *a*’s will leave the machine in state *s*_{4}. Thus, if the machine ends in *s*_{3}, the sequence has read exactly three *a*’s. The machine’s state diagram is illustrated below:

__I4__*M*_{1} has only two *distinct* transition functions, which we shall denote by *T _{o}* and

*T*(where

_{e}**o**may be any sequence with on

*odd*number of 1s, and

**e**any sequence with an

*even*number of 1s).

*T*and

_{o}*T*may be described as follows:

_{e}
s_{0}), = s_{1}, |
s_{1}) = s_{0} |

s_{0}), = s_{0}, |
s_{1}) = s_{1} |

Now, *T _{e}* ∘

*T*=

_{o}*T*by part 3. Since

_{oe}**e**is a sequence with an even number of 1s and

**o**is a sequence with an odd number of 1s,

**oe**is a sequence with an odd number of 1s; hence

*T*=

_{oe}*T*. Thus,

_{o}*T*∘

_{e}*T*=

_{o}*T*. Similarly

_{o}*T _{e}* ∘

*T*=

_{e}*T*,

_{e}*T*∘

_{o}*T*=

_{e}*T*, and

_{o}*T*∘

_{o}*T*=

_{o}*T*

_{e}In brief, the table of (*M*_{1}) is as follows:

The table shows that (*M*_{1}) is a two-element group. The identity element is *T _{e}*, and

*T*is its own inverse.

_{o}**CHAPTER 7**

__D2__*f _{n}*

_{ + m}is the function defined by the formula

*f _{n}*

_{ + m}(

*x*) =

*x*+

*n*+

*m*

Use this fact to show that *f _{n}* ∘

*f*=

_{m}*f*

_{n}_{ + m}.

In order to show that *f*_{−}_{ n} is the inverse of *f _{n}*, we must verify that

*f*∘

_{n}*f*

_{−}

_{ n}= ε and

*f*

_{−}

_{ n}∘

*f*=

_{n}*ε*. We verify the first equation as follows:

*f*

_{−}

_{ n}(

*x*) =

*x*+ (−

*n*); hence

[*f _{n}* ∘

*f*

_{−}

_{ n}](

*x*) =

*f*(

_{n}*f*

_{−}

_{ n}(

*x*)) =

*f*(

_{n}*x*−

*n*) =

*x*−

*n*+

*n*=

*x*=

*ε*(

*x*)

Since [*f _{n}* ∘

*f*

_{−}

_{ n}](

*x*) =

*ε*(

*x*) for every

*x*in , it follows that

*f*∘

_{n}*f*

_{−}

_{ n}= ε.

** E3**To prove that

*f*

_{1/a, }

_{−}

_{ b/a}is the inverse of

*f*,

_{a}*, we must verify that*

_{b}*f _{a}*

_{, b}∘

*f*

_{1/a, }

_{−}

_{b}_{/a}= ε and

*f*

_{1/a, }

_{−}

_{ b/a}∘

*f*

_{a}_{, b}= ε

To verify the first equation, we begin with the fact that *f*_{1/a, }_{−}_{b}_{/a}(*x*) = *x*/*a* − *b*/*a*. Now complete the problem.

** H2**If

*f*,

*g*∈

*G*, then

*f*moves a finite number of elements of

*A*, say

*a*

_{1}, …,

*a*, and

_{n}*g*moves a finite number of elements of

*A*, say

*b*

_{1}, …,

*b*. Now if

_{m}*x*is any element of

*A*which is

*not*one of the elements

*a*

_{1}, …,

*a*,

_{n}*b*

_{1}, …,

*b*, then

_{m}[*f* ∘ *g*](*x*) = *f*(*g*(*x*)) = *f*(*x*) = *x*

Thus, *f* ∘ *g* moves at most the finite set of elements {*a*_{1}, …, *a _{n}*,

*b*

_{1}, …,

*b*}.

_{m}**CHAPTER 8**

__A1(__*e*__)__

__A4(__*f*__)__*γ*^{3} = *γ* ∘ *γ* ∘ *γ* = (1 2 3 4 5); *α*^{−}^{ 1} = (4 1 7 3); thus; *γ*^{3}*α*^{−}^{ 1} = *γ*^{3} ∘ *α* ^{−}^{ 1} = 1 (1 2 3 4 5) ∘ (4 1 7 3) = (1 7 4 2 3 5)

__B2__

and so on. Note that *α*^{2}(*a*_{1}) = *a*_{3}, *α*^{2}(*a*_{2}) = *a*_{4}, …, *α*^{2}(*a _{s}*

_{ }

_{−}

_{ 2}) =

*a*. Finally,

_{s}*α*

^{2}(

*a*

_{s}_{ }

_{−}

_{ 1}) =

*a*

_{l}and

*α*

^{2}(

*a*) =

_{s}*a*

_{2}. Thus,

*α*

^{2}(

*a*) =

_{i}*a*

_{?}Complete the problem using addition modulo

*s*, page 27.

** B4**Let

*α*= (

*a*

_{1}

*a*

_{2}⋯

*a*) where

_{s}*s*is odd. Then

*α*

^{2}= (

*a*

_{1}

*a*

_{3}

*a*

_{5}⋯

*a*

_{s}*a*

_{2}

*a*

_{4}⋯

*a*

_{s}_{ }

_{−}

_{ 1}). If

*s*is even, then

*α*

^{2}= ? Complete the problem.

** E2**If

*α*and

*β*are cycles of the same length,

*α*= (

*a*

_{1}⋯

*a*) and

_{s}*β*= (

*b*

_{1}⋯

*b*, let

_{s}*π*be the following permutation:

*π*(

*k*) =

*b*for

_{i}*i*= 1, …,

*s*and

*π*(

*k*) =

*k*for

*k*≠

*a*

_{1}, …,

*a*,

_{s}*b*

_{1}…,

*b*. Finally, let

_{s}*π*map distinct elements of {

*b*

_{1}…,

*b*) − {

_{s}*a*

_{1}, …,

*a*) to distinct elements of {

_{s}*a*

_{1}, …,

*a*} – {

_{s}*b*

_{1}, …,

*b*}. Now complete the problem, supplying details.

_{s}** F1** when

*k*is a positive integer,

*k*<

*s*

For what values of *k* can you have *α ^{k}* = ε?

** H2**Use

__Exercise H1__and the fact that (

*ij*) = (1

*i*)(1

*j*)(1

*i*).

**CHAPTER 9**

** C1**The group tables for

*G*and

*H*are as follows:

*G* and *H* are not isomorphic because in *G* every element is its own inverse (*VV* = *I*, *HH* = *I*, and *DD* = *I*), whereas in *H* there are elements not equal to their inverse; for example, (-*i*)(-*i*) = −1 ≠ 1. Find at least one other difference which shows that *G* *H*.

** C4** The group tables for

*G*and

*H*are as follows:

*G* and *H* are isomorphic. Indeed, let the function *f* : *G* → *H* be defined by

By inspection, *f* transforms the table of *G* into the table of *H*, Thus, *f* is an isomorphism from *G* to *H*.

** El**Show that the function

*f*: →

*E*given by

*f*(

*n*) = 2

*n*is bijective and that

*f*(

*n*+

*m*) =

*f*(

*n*) +

*f*(

*m*).

** F1**Check that (2 4)

^{2}= ε, (1 2 3 4)

^{4}= ε, and (1 2 3 4)(2 4) = (2 4)(1 2 3 4)

^{3}. Now explain why

*G*≅

*G*′.

** H2**Let

*f*:

_{G}*G*

_{1}→

*G*

_{2}and

*f*:

_{H}*H*

_{1}→

*H*

_{2}be isomorphisms, and find an isomorphism

*f*:

*G*

_{1}×

*H*

_{1}→

*G*

_{2}×

*H*

_{2}.

**CHAPTER 10**

** A1(c)**If

*m*< 0 and

*n*< 0, let

*m*= −

*k*and

*n*=

*l*, where

*k*,

*l*> 0. Then

*m*+

*n*= − (

*k*+

*l*). Now,

*a ^{m}* =

*a*

^{−}

^{ k}= (

*a*

^{−}

^{ 1})

^{k}and

*a ^{n}* =

*a*

^{−}

^{ 1}= (

*a*

^{−}

^{ 1})

^{l}Hence

*a ^{m}a^{n}* = (

*a*

^{−}

^{ 1})

*(*

^{k}*a*

^{−}

^{ 1})

*= (*

^{l}*a*

^{−}

^{ 1})

^{k}^{ + l}=

*a*

^{−}

^{ (k + l)}=

*a*

^{m}^{ + n}

** B3**The order of

*f*is 4. Explain why.

** C4**For any positive integer

*k*, if

*a*=

^{k}*e*, then

Conversely, if (*bab*^{−}^{ 1k} = *ba ^{k}b*

^{−}

^{ 1}=

*e*, then

*a*=

^{k}*e*. (Why?) Thus, for any positive integer

*k*,

*a*=

^{k}*e*iff (

*bab*

^{−}

^{ 1})

*=*

^{k}*e*. Now complete the problem.

** D2** Let the order of

*a*be equal to

*n*. Then (

*a*)

^{k}*=*

^{n}*a*= (

^{nk}*a*)

^{n}*=*

^{k}*e*=

^{k}*e*. Now use

__Theorem 5__.

** F2**The order of

*a*

^{8}is 3. Explain why.

** H2**The order is 24. Explain why.

**CHAPTER 11**

** A6**If

*k*is a generator of , this means that consists of all the multiples of

*k*; that is,

*k*, 2

*k*, 3

*k*, etc., as well as 0 and −

*k*, −2

*k*, −3

*k*, etc.:

** B1**Let

*G*be a group of order

*n*, and suppose

*G*is cyclic, say

*G*= ⟨

*a*⟩. Then

*a*, the generator of

*G*, is an element of order

*n*. (This follows from the discussion on the first two pages of this chapter.) Conversely, let

*G*be a group of order

*n*(that is, a group with

*n*elements), and suppose

*G*has an element

*a*of order

*n*. Prove that

*G*is cyclic.

** C4**By

__Exercise B4__, there is an element

*b*of order

*m*in ⟨

*a*⟩, and

*b*∈

*C*. Since

_{m}*C*is a subgroup of ⟨

_{m}*a*⟩, which is cyclic, we know from

__Theorem 2__that

*C*is cyclic. Since every element

_{m}*x*in

*C*satisfies

_{m}*x*=

^{m}*e*, no element in

*C*can have order greater than

_{m}*m*. Now complete the argument.

** C7**First, assume that ord (

*a*) =

^{r}*m*. Then (

*a*)

^{r}*=*

^{m}*a*=

^{rm}*e*. Use

__Theorem 5__of

__Chapter 10__to show that

*r*=

*kl*for some integer

*l*. To show that

*l*and

*m*are relatively prime, assume on the contrary that

*l*and

*m*have a common factor

*q*; that is,

*l* = *jq* *m* = *hq* *h* < *m*

Now raise *a ^{r}* to the power

*h*and derive a contradiction. Conversely, suppose

*r*=

*lk*where

*l*and

*m*are relatively prime. Complete the problem.

** D6**Let (

*a*) be a cyclic group of order

*mn*. If ⟨

*a*⟩ has a subgroup of order

*n*, this subgroup must be cyclic, and generated by one of the elements in ⟨

*a*⟩. Say ⟨

*a*⟩ is a subgroup of order

^{k}*n*. Then

*a*has order

^{k}*n*, so (

*a*)

^{k}*=*

^{n}*a*=

^{kn}*e*. By

__Theorem 5__of

__Chapter 10__,

*kn*=

*mnq*for some integer

*q*(explain why); hence

*k*=

*mq*and therefore

*a*= (

^{k}*a*)

^{m}*∈ ⟨*

^{q}*a*⟩.

^{m}Use this information to show that (*i*) ⟨*a*⟩ has a subgroup of order *n* and (*ii*) ⟨*a*⟩ has *only one* subgroup of order n.

** F3**If there is some

*j*such that

*a*= (

^{m}*a*)

^{j}*=*

^{k}*a*, then

^{jk}*e*=

*a*(

^{m}*a*)

^{jk}^{−}

^{ 1}=

*a*

^{m}^{ }

^{−}

^{ jk}; so by

__Theorem 5__of

__Chapter 10__,

*m*−

*jk*=

*nq*for some integer

*q*. (Explain why.) Thus,

*m*=

*jk*+

*nq*. Now complete the problem, supplying all details.

**CHAPTER 12**

** A2**Let

*x*be any rational number (any element of ), and suppose

*x*∈

*A*. and

_{i}*x*∈

*A*, where

_{j}*i*and

*j*are integers. Then

*i*≤

*x*<

*i*+ 1 and

*j*≤

*x*<

*j*+

*1*. Now if

*i*<

*j*, then

*i*+ 1 ≥

*j*; so

*x*<

*i*+ 1 implies that

*x*<

*j*, which is a contradiction. Thus we cannot have

*i*<

*j*, nor can we have

*j*<

*i*; hence

*i*=

*j*. We have shown that if

*A*. and

_{i}*A*have a common element

_{j}*x*, then

*A*=

_{i}*A*: this is the first condition in the definition of

_{i}*partition*. For the second condition, note that every rational number is an integer or lies between two integers: if

*i*≤

*x*<

*i*+ 1, then

*x*is in

*A*

_{i}__C1__*A _{r}* consists of all the points (

*x*,

*y*) satisfying

*y*= 2

*x*+

*r*; that is,

*A*is the line with slope 2 and y intercept equal to

_{r}*r*. Thus, the classes of the partition are all the lines with slope 2.

For the corresponding equivalence relation, we say that two points (*x*, *y*) and (*u*, *υ*) are “equivalent,” that is,

(*x*, *y*) ~ (*u*, *υ*)

if the two points lie on the same line with slope 2:

Complete the solution, supplying details.

** D5**If

*ab*

^{−}

^{ 1}commutes with every

*x*in

*G*, then we can show that

*ba*

^{−}

^{ 1}commutes with every

*x*in

*G*:

*ba*^{−}^{ 1} *x* = (*x*^{−}^{ 1}*ab*^{−}^{ 1})^{’ 1} = (*ab*^{−}^{ 1}*x*^{−}^{ 1})^{−}^{ 1} (why?)

**CHAPTER 13**

** B1**Note first that the operation in the case of the group is addition. The subgroup (3) consists of all the multiples of 3, that is,

⟨3⟩ = {. . ., −9, −6, −3, 0, 3, 6, 9, …}

The cosets of (3) are (3) + 0 = (3), as well as

⟨3⟩ + 1 = {. . ., −8, −5, −2, 1, 4, 7, 10, …}

⟨3⟩ + 2 = {. . ., −7, −4, −1, 2, 5, 8, 11, …}

Note that ⟨3⟩ + 3 = ⟨3⟩, ⟨3⟩ + 4 = ⟨3⟩ + 1, and so on; hence there are only three cosets of ⟨3⟩, namely,

⟨3⟩ = ⟨3⟩ + 0 ⟨3⟩ + 1 ⟨3⟩ + 2

** C6** Every element

*a*of order

*p*belongs in a subgroup ⟨

*a*⟩. The subgroup ⟨

*a*⟩ has

*p*– 1 elements (why?), and each of these elements has order

*p*(why?). Complete the solution.

** D6**For one part of the problem, use Lagrange’s theorem. For another part, use the result of

__Exercise F4__,

__Chapter 11__.

** E4**To say that

*aH*=

*Ha*is to say that every element in

*aH*is in

*Ha*and conversely. That is, for any

*h*∈

*H*, there is some

*k*∈

*H*such that

*ah*=

*ka*and there is some

*l*∈

*H*such that

*ha*=

*al*. (Explain why this is equivalent to

*aH*=

*Ha*.) Now, an arbitrary element of

*a*

^{−}

^{ 1}

*H*is of the form

*a*

^{−}

^{ 1}

*h*= (

*h*

^{−}

^{ 1}

*a*)

^{−}

^{ 1}. Complete the solution.

__J3__*O*(1) = *O*(2) = {1, 2, 3, 4}; *G*_{1} = {*ε*, *β*}; *G*_{2} = {*ε*, *αβα*}. Complete the problem, supplying details.

**CHAPTER 14**

** A6**We use the following properties of sets: For any three sets

*X*,

*Y*and

*Z*,

(i) (*X* ∪ *Y*) ∩ *Z* = (*X* ∩ *Z*) ∪ (*Y* ∩ *Z*)

(ii) (*X* − *Y*) ∩ *Z* = (*X* ∩ *Z*) − (*Y* ∩ *Z*)

Now here is the proof that *h* is a homomorphism: Let *C* and *D* be any subsets of *A*; then

Now complete, using (i) and (ii).

** C2**Let

*f*be injective. To show that

*K*= {

*e*}, take any arbitrary element

*x*∈

*K*and show that necessarily

*x*=

*e*. Well, since

*x*∈

*K*,

*f*(

*x*) =

*e*=

*f*(

*e*). Now complete, and prove the converse: Assume

*K*= {

*e*} ….

** D6**Consider the following family of subsets of

*G*: {

*H*:

_{i}*i*∈

*I*}, where each

*H*is a normal subgroup of

_{i}*G*. Show that

*H*=

*H*is a normal subgroup of

_{i}*G*. First, show that

*H*is closed under the group operation: well, if

*a*,

*b*∈

*H*, then

*a*∈

*H*and

_{i}*b*∈

*H*for every

_{i}*i*∈

*I*. Since each

*H*is a subgroup of

_{i}*G*,

*ab*∈

*H*for every

_{i}*i*∈

*I*; hence

*ab*∈

*H*. Now complete.

_{i}** E1**If

*H*has index 2, then g is partitioned into exactly two right cosets of

*H*; also

*G*is partitioned into exactly two left cosets of

*H*. One of the cosets in each case is

*H*.

** E6**First, show that

*if x ∈ S and y ∈ S, then xy ∈ S. Well, if x ∈ S, then x ∈ Ha = aH for some a ∈ G. And if y ∈ S, then y ∈ Hb = bH for some b ∈ G. Show that xy*∈

*H*(

*ab*) and that

*H*(

*ab*) = (

*ab*)

*H*and then complete the problem.

** I4**It is easy to show that

*aHa*

^{−}

^{ 1}⊆

*H*. Show it. What does

__Exercise I2__tell you about the number of elements in

*aHa*

^{−}

^{ 1}

** I8**Let

*X*= {

*aHa*

^{−}

^{ 1}:

*a*∈

*G*} be the set of all the conjugates of

*H*, and let

*Y*= {

*aN*:

*a*∈

*G*} be the set of all the cosets of

*N*. Find a function

*f*:

*X*→

*Y*and show that

*f*is bijective.

**CHAPTER 15**

** C4**Every element of

*G*/

*H*is a coset

*Hx*. Assume every element of

*G*/

*H*has a square root: this means that for every

*x*∈

*G*,

*Hx* = (*Hy*)^{2}

for some *y* ∈ *G*. Avail yourself of __Theorem 5__ in this chapter.

** D4**Let

*H*be generated by {

*h*

_{1}, …,

*h*and let

_{n}*G*/

*H*be generated by {

*Ha*

_{1}, …,

*Ha*). Show that

_{m}*G*is generated by

{*a*_{1}, …, *a _{m}*,

*h*

_{1}, …,

*h*}

_{n}that is, every *x* in *G* is a product of the above elements and their inverses.

** E6**Every element of / is a coset + (

*m*/

*n*).

** G6**If

*G*is cyclic, then necessarily

*G*≅

*p*

_{2}. (Why?)

If *G* is not cyclic, then every element *x* ≠ *e* in *G* has order *p*. (Why?) Take any two elements *a* ≠ *e* and *b* ≠ *e* in *G* where *b* is not a power of *a*. Complete the problem.

**CHAPTER 16**

** D1**Let

*f*

*∈*Aut(

*G*); that is, let

*f*be an isomorphism from

*G*onto

*G*. We shall prove that

*f*

^{−}

^{ 1}∈ Aut(

*G*); that is,

*f*

^{−}

^{ 1}is an isomorphism from

*G*onto

*G*. To begin with, it follows from the last paragraph of

__Chapter 6__that

*f*

^{−}

^{ 1}is a bijective function from

*G*onto

*G*. It remains to show that

*f*

^{−}

^{ 1}is a homomorphism. Let

*f*

^{−}(

*c*) =

*a*and

*f*

^{−}

^{ 1}(

*d*) =

*b*, so that

*c*=

*f*(

*a*) and

*d*=

*f*(

*b*). Then

*cd*=

*f*(

*a*)

*f*(

*b*) =

*f*(

*ab*), whence

*f*

^{−}

^{ 1}(

*cd*=

*ab*. Thus,

*f*^{−}^{ 1}(*cd*) = *ab* = *f*^{−}^{ 1}(*c*)*f*^{−}^{ 1}(*d*)

which shows that *f*^{−}^{ 1} is a homomorphism.

** F2**If

*a*,

*b*∈

*HK*, then

*a*=

*h*

_{1}

*k*

_{1}and

*b*=

*h*

_{2}

*k*

_{2}, where

*h*

_{1},

*h*

_{2}∈

*H*and

*k*

_{1},

*k*

_{2}∈

*K*. Then

*ab*=

*h*

_{1}

*k*

_{1}

*h*

_{2}

*k*

_{2}= .

** G3**Note that the range of

*h*is a group of functions. What is its identity element?

** H1**From calculus, cos(

*x*+

*y*) = cos

*x*cos

*y*− sin

*x*sin

*y*, and sin (

*x*+

*y*) = sin

*x*cos

*y*+ cos

*x*sin

*y*.

** L4** The natural homomorphism (

__Theorem 4__,

__Chapter 15__) is a homomorphism

*f*:

*G*→

*G*/⟨

*a*⟩ with kernel ⟨

*a*⟩. Let

*S*be the normal subgroup of

*G*/⟨

*a*⟩ whose order is

*p*

^{m}^{ }

^{−}

^{ 1}. (The existence of

*S*is assured by part 3 of this exercise set.) Referring to

__Exercise J__, show that

*S** is a normal subgroup of

*G*, and that the order of

*S** is

*p*.

^{m}**CHAPTER 17**

** A3**We prove that is associative:

Thus, (*a*, *b*)[(*c*, *d*)(*p*, *q*)] = [(*a*, *b*)(*c*, *d*)](*p*, *q*).

** B2**A nonzero function

*f*is a divisor of zero if there exists some nonzero function

*g*such that

*fg*= 0, where 0 is the zero function (page 46). The equation

*fg*= 0 means that

*f*(

*x*)

*g*(

*x*) = 0(

*x*) for every

*x*∈ . Very precisely, what functions

*f*have this property?

** D1**For the distributive law, refer to the diagram on page 30, and show that

*A*∩ (

*B*+

*C*) = (

*A*∩

*B*) + (

*A*∩

*C*):

*B* + *C* consists of the regions 2, 3, 4, and 7; *A* ∩ (*B* + *C*) consists of the regions 2 and 4. Now complete the problem.

__E3__

** G4**(

*a*,

*b*) is an invertible element of

*A*×

*B*iff there is an ordered pair (

*c*,

*d*) in

*A*×

*B*satisfying (

*a*,

*b*) · (

*c*,

*d*) = (1, 1). Now complete.

** H6**If

*A*is a ring, then, as we have seen,

*A*with addition as the only operation is an abelian group: this group is called the

*additive group*of the ring

*A*. Now, suppose the additive group of

*A*is a cyclic group, and its generator is

*c*. If

*a*and

*b*are any two elements of

*A*, then

*a* = *c* + *c* + ⋯ + *c* (*m* terms)

and

*b* = *c* + *c* + ⋯ + *c* (*n* terms)

for some positive integers *m* and *n*.

** J2**If

*ab*is a divisor of 0, this means that

*ab*≠ 0 and there is some

*x*≠ 0 such that

*abx*= 0. Moreover,

*a*≠ 0 and

*b*≠ 0, for otherwise

*ab*= 0.

** M3**Suppose

*a*= 0 and

^{m}*b*= 0. Show that (

^{n}*a*+

*b*)

^{m}^{ + n}= 0. Explain why, in every term of the binomial expansion of (

*a*+

*b*)

^{m}^{ + n}, either

*a*is raised to a power ≥

*m*, or

*b*is raised to a power ≥

*n*.

**CHAPTER 18**

** A4**From calculus, the sum and product of continuous functions are continuous.

** B3**The proof hinges on the fact that if

*k*and

*a*are any two elements of

*, then*

_{n}*ka* = *a* + *a* + · + *a* (*k* terms)

** C4**If the cancellation law holds in

*A*, it must hold in

*B*. (Why?) Why is it necessary to include the condition that

*B*contains 1?

** C5**Let

*B*be a subring of a field

*F*. If

*b*is any nonzero element of

*B*, then

*b*

^{−}

^{ 1}is in

*F*, though not necessarily in

*B*. (Why?) Complete the argument.

__E5__*f*(*x*, *y*)*f*(*u*, *υ*) = = *f*[*x*, *y*)(*u*, *υ*)] =

Complete the problem.

** H3**If

*a*∈

^{n}*J*and

*b*∈

^{m}*J*, show that (

*a*+

*b*)

^{n}^{ + m}∈

*J*. (See the solution of

__Exercise M3__,

__Chapter 17__.) Complete the solution.

**CHAPTER 19**

** E1**To say that the coset

*J*+

*x*has a square root is to say that for some element

*y*in

*A*,

*J*+

*x*= (

*J*+

*y*)(

*J*+

*y*) =

*J*+

*y*

^{2}.

** E6**A unity element of

*A*/

*J*is a coset

*J*+

*a*such that for any

*x*∈

*A*,

(*J* + *a*)(*J* + *x*) = *J* + *x* and (*J* + *x*)(*J* + *a*) = *J* + *a*

** G1**To say that

*a*∉

*J*is equivalent to saying that

*J*+

*a*≠

*J*; that is,

*J*+

*a*is not the zero element of

*A*/

*J*. Explain and complete.

**CHAPTER 20**

** E5**Restrict your attention to

*A*with addition alone, and see

__Chapter 13__,

__Theorem 4__.

** E6**For

*n*=

*2*you have

Prove the required formula by reasoning similarly and using induction: assume the formula is true for *n* = *k*, and prove for n = k + 1.

**CHAPTER 21**

** B5**Use the product (

*a*− 1)(

*b*− 1).

** C8**In the induction step, you assume the formula is true for n = k and prove it is true for n = k + 1. That is, you assume

*F*_{k + 1} *F*_{k + 2} − *F*_{k}*F*_{k + 3} = (–1)^{k}

and prove

*F*_{k + 2} *F*_{k + 3} − *F*_{k + 1} *F*_{k + 4} = (− 1)^{k + 1}

Recall that by the definition of the Fibonacci sequence, *F*_{n + 2} = *F*_{n + 1} + *F*_{n} for every n > 2. Thus, *F*_{k + 2} = *F*_{k + 1} + *F*_{k} and *F*_{k + 4} = *F*_{k + 3} + *F*_{k + 2}. Substitute these in the second of the equations above.

** E5**An elegant way to prove this inequality is by use of part 4, with

*a*+

*b*in the place of

*a*, and |

*a*| + |

*b*| in the place of

*b*.

** E8**This can be proved easily using part 5.

** F2**You are given that m = nq + r and q = kq

_{1}+ r

_{1}where 0 ≤ r < n and 0 ≤ r

_{1}< k. (Explain.) Thus,

m = n(kq_{1} + r_{1} + r = (nk)q_{1} + (nr_{1} + r)

You must show that nr_{1} + r < nk, (Why?) Begin by noting that k − r_{1} > 0; hence k – r_{1} ≥ 1, so n(k – r_{1}) ≥ n.

** G5**For the induction step, assume k ·

*a*= (k · 1)

*a*, and prove (k + 1) ·

*a*= [(k + 1) · 1]

*a*. From (ii) in this exercise, (k + 1) · 1 = k · 1 + 1.

**CHAPTER 22**

** B1**Assume

*a*> 0 and

*a*|

*b*. To solve the problem, show that

*a*is the greatest common divisor of

*a*and

*b*. First,

*a*is a common divisor of

*a*and

*b*:

*a*|

*a*and

*a*|

*b*. Next, suppose

*t*is any common divisor of

*a*and

*b*:

*t*|

*a*and

*t*|

*b*. Explain why you are now done.

** D3**From the proof of

__Theorem 3__,

*d*is the generator of the ideal consisting of all the linear combinations of

*a*and

*b*.

** E1**Suppose

*a*is odd and

*b*is even. Then

*a*+

*b*is odd and

*a*−

*b*is odd. Moreover, if

*t*is any common divisor of

*a*–

*b*and

*a*+

*b*, then

*t*is odd. (Why?) Note also that if ris a common divisor of

*a*−

*b*and

*a*+

*b*, then

*t*divides the sum and difference of

*a*+

*b*and

*a*−

*b*.

** F3** If

*l*= lcm(

*ab*,

*ac*), then

*l*=

*abx*=

*acy*for some integers

*x*and

*y*. From these equations you can see that

*a*is a factor of

*l*, say

*l*=

*am*. Thus, the equations become

*am*=

*abx*=

*acy*. Cancel

*a*, then show that

*m*= lcm(

*b*,

*c*).

** G8**Look at the proof of

__Theorem 3__.

**CHAPTER 23**

** A4(f)**3

*x*

^{2}− 6

*x*+ 6 = 3(

*x*

^{2}− 2

*x*+ 1) + 3 = 3(

*x*− 1)

^{2}+ 3. Thus, we are to solve the congruence 3(

*x*− 1)

^{2}≡ −3 (mod 15). We begin by solving 3

*y*≡ −3 (mod 15), then we will set

*y*= (

*x*− 1)

^{2}.

We note first that by Condition (6), in a congruence *ax* = *b* (mod *n*), if the three numbers *a*, *b*, and *n* have a common factor *d*, then

(That is, all three numbers *a*, *b*, and *n* may be divided by the common factor *d*.) Applying this observation to our congruence 3*y* ≡ −3 (mod 15), we get

3*y* ≡ −3 (mod 15) is equivalent to *y* ≡ −1 (mod 5)

This is the same as *y* ≡ 4 (mod 5), because in _{5} the negative of 1 is 4.

Thus, our quadratic congruence is equivalent to

(*x* − 1)^{2} ≡ 4 (mod 5)

In _{5}, 2^{2} = 4 and 3^{2} = 4; hence the solutions are *x* − 1 = 2 (mod 5) and *x* − 1 ≡ 3 (mod 5), or finally,

*x* ≡ 3 (mod 5) and *x* ≡ 4 (mod 5)

** A6(d)**We begin by finding all the solutions of 30

*z*+ 24

*y*= 18, then set

*z*=

*x*

^{2}. Now,

30*z* + 24*y* = 18 iff 24*y* = 18 − 30*z* iff 30*z* ≡ 18 (mod 24)

By comments in the previous solution, this is equivalent to 5*z* ≡ 3 (mod4). But in _{4}, = ; hence 5*z* = 3 (mod 4) is the same as *z* ≡ 3 (mod 4).

Now set *z* = *x*^{2}. Then the solution is *x*^{2} ≡ 3 (mod 4). But this last congruence has no solution, because in _{4}, is not the square of any number. Thus, the Diophantine equation has no solutions.

** B3**Here is the idea: By

__Theorem 3__, the first two equations have a simultaneous solution; by

__Theorem 4__, it is of the form

*x*≡

*c*(mod

*t*), where

*t*= lcm(

*m*

_{1},

*m*

_{2}). To solve the latter simultaneously with

*x*≡

*c*

_{3}(mod

*m*

_{3}), you will need to know that

*c*

_{3}≡

*c*[mod gcd(

*t*,

*m*

_{3})]. But gcd(

*t*,

*m*

_{3}) = lcm(

*d*

_{13},

*d*

_{23}). (Explain this carefully, using the result of

__Exercise H4__in

__Chapter 22__.) From

__Theorem 4__(especially the last paragraph of its proof), it is clear that since

*c*

_{3}≡

*c*

_{1}(mod

*d*

_{13}) and

*c*

_{3}≡

*c*

_{2}(mod

*d*

_{23}), therefore

*c*

_{3}≡

*c*[mod lcm(d

_{13},

*d*

_{23})].

Thus, *c*_{3} ≡ *c* [mod gcd(*t*, *m*_{3})], and therefore by __Theorem 3__ there is a simultaneous solution of *x* ≡ *c* (mod *t*) and *x* ≡ *c*_{3} (mod *m*_{3}). This is a simultaneous solution of *x* ≡ *c*_{1} (mod *m*_{1}), *x* ≡ *c*_{2} (mod *m*_{2}), and *x* ≡ *c*_{3} (mod *m*_{3}). Repeat this process *k* times.

An elegant (and mathematically correct) form of this proof is by induction on *k*, where *k* is the number of congruences.

** B5(a)**Solving the pair of Diophantine equations is equivalent to solving the pair of linear congruences 4

*x*≡ 2 (mod 6) and 9

*x*= 3 (mod 12). Now see the example at the beginning of

__Exercise B__.

** D6**Use the definitions, and form the product (

*ab*− 1)(

*ac*− 1)(

*bc*− 1).

** E4**Use the product (

*p*

^{q}^{ }

^{−}

^{ 1}− 1)(

*q*

^{p}^{ }

^{−}

^{ 1}− 1).

** E6**Use

__Exercise D5__.

** E8(a)**Note the following:

(i) 133 = 7 × 19.

(ii) 18 is a multiple of both 7-1 and 19-1.

Now use part 6(*b*).

** F8**Consider (

*n*

^{ϕ}^{(m)}− 1)(

*m*

^{ϕ}^{(n)}− 1).

** H2**In any commutative ring, if

*a*

^{2}=

*b*

^{2}, then (

*a*+

*b*)(

*a*−

*b*) = 0; hence

*a*= ±

*b*. Thus, for any

*a*≠ 0, the two elements

*a*and −

*a*have the same square; and no

*x*≠ ±

*a*can have the same square as ±

*a*.

** H4**() = − 1 because 17 is not a quadratic residue mod 23.

**CHAPTER 24**

** A1**We compute

*a*(

*x*)

*b*(

*x*) in

_{6}[

*x*]:

*a*(*x*)*b*(*x*) = (2*x*^{2} + 3*x* + 1)(*x*^{3} + 5*x*^{2} + *x*) = 2*x*^{5} + 13*x*^{4} + 18*x*^{3} + 8*x*^{2} + *x*

But the coefficients are in _{6}, and in _{6}

13 = 1 18 = 0 and 8 = 2

Thus, *a*(*x*)*b*(*x*) = 2*x*^{5} + *x*^{4} + 2*x*^{2} + *x* in _{6}[*x*].

Note that while the coefficients are in _{6}, the exponents are positive integers and are *not* in _{6}. Thus, for example, in _{6}[*x*] the terms *x*^{8} and *x*^{2} are *not* the same.

** B3**In

_{5}[

*x*], there are 5 × 5 × 5 = 125 different polynomials of degree 2 or less. Explain why. There are 5 × 5 = 25 polynomials of degree 1 or 0; hence there are 125 − 25 = 100 different polynomials of degree 2. Explain, then complete the problem.

** C7**In

_{9}[

*x*],

*x*

^{2}= (

*x*+ 3)(

*x*+ 6) = (2

*x*+ 3)(5

*x*+ 6) = etc. Systematically list all the ways of factoring

*x*

^{2}into factors of degree 1, and complete the problem.

** D6** After proving the first part of the problem, you can prove the second part by induction on the number

*n*of terms in the polynomial. For

*n*= 2, it follows from the first part of the problem. Next, assume the formula for

*k*terms,

and prove for *k* + 1 terms:

[(*a*_{0} + *a*_{1}*x* + ⋯ *a _{k}x^{k}*) +

*a*

_{k}_{ + 1}

*x*

^{k}^{ + 1}]

*= (Complete.)*

^{p}** E5**If

*a*(

*x*) =

*a*

_{0}+

*a*

_{1}

*x*+ ⋯

*a*∈

_{n}x^{n}*J*, then

*a*

_{0}+

*a*

_{1}+ ⋯ +

*a*= 0. If

_{n}*b*(

*x*) is any element in

*A*[

*x*], say

*b*(

*x*) =

*b*

_{0}+

*b*

_{1}

*x*+ ⋯ +

*b*, let

_{m}x^{m}*B*=

*b*

_{0}+

*b*

_{1}+ ⋯ +

*b*. Explain why the sum of all the coefficients in

_{m}*a*(

*x*)

*b*(

*x*) is equal to (

*a*

_{0}+

*a*

_{1}⋯ +

*a*)

_{n}*B*. Supply all remaining details.

** G3**If

*h*is surjective, then every element of

*B*is of the form

*h*(

*a*) for some

*a*in

*A*. Thus, any polynomial with coefficients in

*B*is of the form

*h*(

*a*

_{0}) +

*h*(

*a*

_{1})

*x*+ ⋯ +

*h*(

*a*)

_{n}*x*.

^{n}**CHAPTER 25**

** A4**Assume there are

*a*,

*b*∈

_{5}such that (

*x*+

*a*)(

*x*+

*b*) =

*x*

^{2}+ 2. Note that in

_{5}, only 1 and 4 are squares.

** D4**Let ⟨

*p*(

*x*)⟩ ⊂

*J*where

*J*is an ideal of

*F*[

*x*], and assume there is a polynomial

*a*(

*x*) ∈

*J*where

*a*(

*x*) ∉ ⟨

*p*(

*x*)⟩. Then

*a*(

*x*) and

*p*(

*x*) are relatively prime. (Why?) Complete the solution.

** F2**The gcd of the given polynomials is

*x*+ 1.

**CHAPTER 26**

** A2**To find the roots of

*x*

^{100}−1 in

_{7}[

*x*], reason as follows: From Fermat’s theorem, if

*a*∈

_{7}, then in

_{7},

*a*

^{69}= 1 for every integer

*q*. In particular,

*a*

^{96}= 1; hence

*a*

^{100}=

*a*

^{96}

*a*

^{4}=

*a*

^{4}. Thus any root (in

_{7}) of

*x*

^{100}− 1 is a root of

*x*

^{4}− 1.

** B1**Any rational root of 9

*x*

^{3}+ 18

*x*

^{2}− 4

*x*− 8 is a fraction

*s*/

*t*where

*s* = ± 1, ± 8, ± 2, or ± 4

and

*t* = ± 1, ± 9, or ± 3

Thus, the possible roots are ± 1, ± 2, ± 4, ± 8, ± 1/9, ± 1/3, ± 8/9, ± 8/3, ± 2/9, ± 2/3, ± 4/9, and ± 4/3. Once a single root has been found by substitution into the polynomial, the problem may be simplified. For example, −2 is one root of the above. You may now divide the given polynomial by *x* + 2. The remaining roots are roots of the quotient, which is a quadratic.

** C5** Prove by induction: For

*n*= 1, we need to show that if

*a*(

*x*) is a monic polynomial of degree 1, say

*a*(

*x*) =

*x*–

*c*, and

*a*(

*x*) has one root

*c*

_{1}, then

*a*(

*x*) =

*x*–

*c*

_{1}. Well, if

*c*

_{1}is a root of

*a*(

*x*) =

*x*–

*c*, then

*a*(

*c*

_{1}) =

*c*

_{1}–

*c*= 0, so

*c*=

*c*

_{1}; thus,

*a*(

*x*) =

*x*−

*c*

_{1}.

For the induction step, assume the theorem is true for degree *k* and *k* roots, and prove it is true for degree *k* + 1 and *k* + 1 roots.

** C8**If

*x*

^{2}−

*x*=

*x*(

*x*− 1) = 0, then

*x*= 0 or

*x*− 1 = 0. Does this rule hold in both

_{10}and

_{11}? Why?

** D3**Note that if

*p*is a prime number, and 0 <

*k*<

*p*, then the binomial coefficient is a multiple of

*p*. (See page 202.)

** F1**The fact that

*a*(

*x*) is monic is essential. Explain why the degree of (

*a*(

*x*)) is the same as the degree of

*a*(

*x*).

** H3**Assume there are two polynomials

*p*(

*x*) and

*q*(

*x*), each of degree ≤

*n*, satisfying the given condition. Show that

*p*(

*x*) =

*q*(

*x*).

** I4**If

*a*(

*x*) and

*b*(

*x*) determine the same function, then

*a*(

*x*) −

*b*(

*x*) is the zero function; that is,

*a*(

*c*) –

*b*(

*c*) = 0 for every

*c*∈

*F*.

**CHAPTER 27**

** A1(e)**Let ; then

*a*

^{2}=

*i*− and

*a*

^{4}= −1 −

*i*+ 2 = 1− 2

*i*. Thus,

*a*

^{4}− 1 = −2

*i*, and (

*a*

^{4}− 1)

^{2}= −8. Therefore

*a*is a root of the polynomial (

*x*

^{4}– 1)

^{2}+ 8, that is,

*x*^{8} −2*x*^{4} + 9

** B1(d)**Of the six parts, (

*a*) − (

*f*), of this exercise, only (

*d*) involves a polynomial of degree higher than 4; hence it is the most painstaking. Let

*a*= ; then

*a*

^{2}= 2 + 3

^{1/3}and

*a*

^{2}− 2 = 3

^{1/3}. Thus, (

*a*

^{2}− 2)

^{3}= 3, so

*a*is a root of the polynomial

*p*(*x*) = (*x*^{2} − 2)^{3} − 3 = *x*^{6} − *x*^{6} − 12*x*^{2} − 11

The bulk of the work is to ensure that this polynomial is irreducible. We will use the methods of __Chapter 26__, __Exercises F__ and __E__. By the first of these methods, we transform *p*(*x*) into a polynomial in *Z*_{3}[*x*]:

*x*^{6} − 2 = *x*^{6} + 1

Since none of the three elements 0, 1, 2 in _{3} is a root of the polynomial, the polynomial has no factor of degree 1 in _{3}[*x*]. So the only possible factorings into non constant polynomials are

*x*^{6} + 1 = (*x*^{3} + *ax*^{2} + *bx* + *c*)(*x*^{3} + *dx*^{2} + *ex* + *f*)

or

*x*^{6} + 1 = (*x*^{4} + *ax*^{3} + *bx*^{2} + *cx* + *d*)(*x*^{2} + *ex* + *f*)

From the first equation, since corresponding coefficients are equal, we have:

From (1), *c* = *f* = ±1, and from (5), *a* + *d* = 0. Consequently, *af* + *cd* = *c*(*a* + *d*) = 0, and by (3), *eb* = 0. But from (2) (since *c* = *f*), *b* + *e* = 0, and therefore *b* = *e* = 0. It follows from (4) that *c* + *f* = 0, which is impossible since *c*= *f* = ±1. We have just shown that *x*^{6} + 1 cannot be factored into two polynomials each of degree 3. Complete the solution.

** C4**From part 3, the elements of

_{2}(

*c*) are 0, 1,

*c*, and

*c*+ 1. (Explain.) When adding elements, note that 0 and 1 are added as in

_{2}, since

_{2}(

*c*) is an extension of

_{2}. Moreover,

*c*+

*c*=

*c*(1 + 1) =

*c*0 = 0. When multiplying elements, note the following example: (

*c*+ 1)

^{2}=

*c*

^{2}+

*c*+

*c*+ 1 =

*c*(because

*c*

^{2}+

*c*+ 1 = 0).

** D1**See

__Exercise E3__, this chapter.

** D7**In (), 1 + 1 is a square; so if () ≅ (), then 1 + 1 is a square in (), that is, ∈ (). But all the elements of () are of the form

*a*+

*b*for

*a*,

*b*∈ ∈. (Explain why.) So we would have =

*a*+

*b*. Squaring both sides and solving for (supply details) shows that is a rational number. But it is known that is irrational.

** G2**Let

*Q*denote the field of quotients of {

*a*(

*c*) :

*a*(

*x*) ∈

*F*[

*x*]}. Since

*F*(

*c*) is a field which contains

*F*and

*c*, it follows that

*F*(

*c*) contains every

*a*(

*c*) =

*a*

_{0}+

*a*

_{1}

*c*+ ⋯ +

*a*where

_{n}c^{n}*a*

_{0}, …,

*a*∈

_{n}*F*. Thus,

*Q*⊆

*F*(

*c*). Conversely, by definition

*F*(

*c*) is contained in any field containing

*F*and

*c*; hence

*F*(

*c*) ⊆

*Q*. Complete the solution.

**CHAPTER 28**

** B1**Let

*U*= {(

*a*,

*b*,

*c*): 2

*a*− 3

*b*+

*c*= 0}. To show that

*U*is closed under addition, let

**u**,

**v**∈

*U*. Then

**u**= (

*a*

_{1},

*b*

_{1},

*c*

_{1}) where 2

*a*

_{1}−3

*b*

_{1}+

*c*

_{1}= 0, and

**v**= (

*a*

_{2},

*b*

_{2},

*c*

_{2}) where 2

*a*

_{2}− 3

*b*

_{2}+

*c*

_{2}= 0. Adding,

**u** + **v** = (*a*_{1} + *a*_{2}, *b*_{1} + *b*_{2}, *c*_{1} + *c*_{2})

and

2(*a*_{1} + *a*_{2}) − 3(*b*_{1} + *b*_{2}) + (*c*_{1} + *c*_{2}) = 0

__C5(__*a*__)__*S*_{1} = {(*x*, *y*, *z*): *z* = 2*y* − 3*x*}. A suggested basis for *S*_{1} would consist of the two vectors (1, 0, ?) and (0, 1, ?) where the third component satisfies *z* = 2*y* − 3*x*, that is,

(1, 0, −3) and (0, 1, 2)

Now show that the set of these two vectors is linearly independent and spans *S*_{1}.

** D6**Suppose there are scalars

*k*,

*l*, and

*m*such that

*k*(**a** + **b**) + *l*(**b** + **c**) + *m*(**a** + **c**) = **0**

Use the fact that {**a**, **b**, **c**} is independent to show that *k* = *l* = *m* = 0.

** E5**Let

*k*

_{r}_{ + 1},

*k*

_{r}_{ + 2}, …,

*k*be scalars such that

_{n}*k _{r}*

_{ + 1}

*h*(

**a**

_{r}_{ + 1}) + ⋯ +

*k*

_{n}*h*(

**a**

*) =*

_{n}**b**

Hence

*h*(*k _{r}*

_{ + 1}

**a**

_{r}_{ + 1}+ ⋯ +

*k*

_{n}**a**

*) =*

_{n}**0**

Then *k _{r}*

_{ + 1}

**a**

_{r}_{ + 1}+ ⋯ +

*k*

_{n}**a**

*∈ . Recall that {*

_{n}**a**

_{1}, …,

**a**

*} is a basis for , and complete the proof.*

_{r}** F2**Use the result of

__Exercise E7__.

** G2**Assume that every

*c*∈

*V*can be written, in a unique manner, as

**c**=

**a**+

**b**where

**a**∈

*T*and

**b**∈

*U*. Thus, every vector in

*V*is an element of

*T*+

*U*, and conversely, since

*T*and

*U*are subspaces of

*V*, every vector in

*T*+

*U*is an element of

*V*. Thus,

*V* = *T* + *U*

In order to show that *V* = *T* ⊕ *U*, it remains to show that *T* *U* = {0}; that is, if **c** ∈ *T* *U* then *c* = 0. Complete the solution.

**CHAPTER 29**

** A3**Note first that

*a*

^{2}− = ; hence ∈ (

*a*) and therefore (

*a*) = (,

*a*). (Explain.) Now

*x*

^{3}−

*2*(which is irreducible over by Eisenstein’s criterion) is the minimum polynomial of ; hence [() : ] = 3. Next, in (),

*a*satisfies

*a*

^{2}− 1 − = 0, so

*a*is a root of the polynomial

*x*

^{2}− (1 + ). This quadratic is irreducible over ()[

*x*] (explain why); hence

*x*

^{2}− (1 + ) is the minimum polynomial of

*a*over ()[

*x*]. Thus, [(,

*a*) : (] = 2. Use

__Theorem 2__to complete.

** A4**This is similar to part 3; adjoin first , then

*a*.

** C2**Note that

*x*

^{2}+

*x*+ 1 is irreducible in

_{2}[

*x*].

** D2**Suppose there is a field

*L*properly between

*F*and

*K*, that is,

*F*

*L*

*K*. As vector spaces over the field

*F*,

*L*is a subspace of

*K*. (Why?) Use

__Chapter 28__,

__Exercise D1__.

** F4**The relationship between the fields may be sketched as follows:

[*K*(*b*) : *F*] = [*K*(*b*) : *F*(*b*)] · [*F*(*b*) : *F*] = [*K*(*b*) : *K*] · [*K* : *F*]

** F5**Reasoning as in part 4 (and using the same sketch), show that [

*K*(

*b*) :

*K*] = [

*F*(

*b*) :

*F*], Here,

*b*is a root of

*p*(

*x*).

**CHAPTER 30**

** B3**If (

*a*,

*b*) ∈ × , then

*a*and are rational numbers. By

__Exercise A5__and the definition of , (

*a*, 0) and (

*b*, 0) are constructible from {

*O*,

*I*}. With the aid of a compass (mark off the distance from the origin to

*b*along the

*y*axis), we can construct the point (0,

*b*). From elementary geometry there exists a ruler-and-compass construction of a perpendicular to a given line through a specified point on the line. Construct a perpendicular to the

*x*axis through (

*a*, 0) and a perpendicular to the y axis through (0,

*b*). These lines intersect at (

*a*,

*b*); hence (

*a*,

*b*) is constructible from {

*O*,

*I*}.

** C6**Describe a ruler-and-compass construction of the 30-60-90° triangle.

** D1**From geometry, the angle at each vertex of a regular

*n*-gon is equal to the supplement of 2

*π*/

*n*, that is,

*π*− (2

*π*/n).

** G2**A number is constructible iff it is a coordinate of a constructible point. (Explain, using

__Exercise A__.) If

*P*is a constructible point, this means there are some points, say

*n*points

*P*

_{1},

*P*

_{2}, …,

*P*=

_{n}*P*, such that each

*P*is constructible in one step from × {

_{i}*P*

_{1}, …,

*P*

_{i}_{ }

_{−}

_{ 1}. In the proof of the lemma of this chapter it is shown that the coordinates of

*P*can be expressed in terms of the coordinates of

_{i}*P*, …,

_{x}*P*

_{i}_{ }

_{−}

_{1}using addition, subtraction, multiplication, division, and taking of square roots. Thus, the coordinates of

*P*are obtained by starting with rational numbers and sucessively applying these operations.

Using these ideas, write a proof by induction of the statement of this exercise.

**CHAPTER 31**

** A6**Let be a real cube root of 2. () is not a root field over because it does not contain the complex cube roots of 2.

** B5**First,

*p*(

*x*) =

*x*

^{3}+

*x*

^{2}+

*x*+ 2 is irreducible over

_{3}because, by successively substituting

*x*= 0, 1, 2, one verifies that

*p*(

*x*) has no roots in

_{3}. [Explain why this fact proves that

*p*(

*x*) is irreducible over

_{3}.] If

*u*is a root of

*p*(

*x*) in some extension of

_{3}, then

_{3}(

*u*) is an extension of

_{3}of degree 3. (Why?) Thus,

_{3}(

*u*) consists of all elements

*au*

^{2}+

*bu*+

*c*, where

*a*,

*b*,

*c*∈

_{3}. Hence

_{3}(

*u*) has 27 elements. Dividing

*p*(

*x*) =

*x*

^{3}+

*x*

^{2}+

*x*+ 2 by

*x*−

*u*gives

*p*(*x*) = (*x* − *u*)[*x*^{2} + (*u* + 1)*x* + (*u*^{2} + *u* + 1)]

where *q*(*x*) = *x*^{2} + (*u* + *l*)*x* + (*u*^{2} + *u* + 1) is in *Z*_{3}(*u*)[*x*]. (Why?)

In _{3}(*u*)[*x*], *q*(*x*) is irreducible: this can be shown by direct substitution of each of the 27 elements of _{3}(*u*), successively, into *q*(*x*). So if *w* denotes a root of *q*(*x*) in some extension of _{3}(*u*), then *Z*_{3}(*u*, *w*) includes *u* and both roots of *q*(*x*). (Explain.) Thus, _{3}(*u*, *w*) is the root field of *p*(*x*) over _{3}. It is of degree 6 over _{3}. (Why?)

** C5**We may identify

*F*[

*x*]/⟨

*p*(

*x*)⟩ with

*F*(

*c*), where

*c*is a root of

*p*(

*x*). Then

*F*(

*c*) is an extension of degree 4 over

*F*. (Why?) Now complete the square:

Form the iterated extension *F*(*d*_{1}, *d*_{2}), where *d*_{1} is a root of *x*^{2} − [(*a*^{2}/4) − *b*] and *d*_{2} is a root of *x*^{2} − [(*a*/2) + *d*_{1}]. Explain why (*a*) *F*(*d*_{1}, *d*_{2}) = *F*(*c*) and (*b*) *F*(*d*_{1}, *d*_{2}) contains all the roots of *p*(*x*).

** D3**From page 313, (, ) = ( + ); hence (, , ) = ( + , ). As in the illustration for

__Theorem 2__, taking

*t*= 1 gives

*c*= + + . (Provide details.)

** D5**Use the result of part 3. The degree of (, , ) over is 8 (explain why). Now find a polynomial

*p*(

*x*) of degree 8 having + + as a root: if

*p*(

*x*) is monic and of degree 8,

*p*(

*x*)

*must*be the minimum polynomial of + + over . (Why?) Hence

*p*(

*x*) must be irreducible. Explain why the roots of

*p*(

*x*) are ± ± ± , all of which are in (, , ).

** E4**For

*n*= 6, we have

*x*

^{6}− 1 = (

*x*− 1)(

*x*+ 1)(

*x*

^{2}−

*x*+ 1)(

*x*

^{2}+

*x*+ 1). From the quadratic formula, the roots of the last two factors are in ().

** H2**Note that if

*c*(

*x*) =

*c*

_{0}+

*c*

_{1}

*x*+ ⋯ +

*c*then

_{n}x^{n}and

*h*(*c*(*a*)) = *h*(*c*_{0}) + *h*(*c*_{1})*b* + ⋯ + *h*(*c _{n}*)

*b*

^{n}For *d*(*x*) = *d*_{0} + *d*_{1}*x* + ⋯ + *d _{n}x^{n}*, we have similar formulas. Show that

*h*(

*c*(

*a*)) =

*h*(

*d*(

*a*)) iff

*b*is a root of

*hc*(

*x*) −

*hd*(

*x*).

** I4**The proof is by induction on the degree of

*a*(

*x*). If deg

*a*(

*x*) = 1, then

*K*

_{1}=

*F*

_{1}and

*K*

_{2}=

*F*

_{2}(explain why), and therefore

*K*

_{1}≅

*K*

_{2}. Now let deg

*a*(

*x*) =

*n*, and suppose the result is true for any polynomial of degree

*n*− 1.

Let *p*(*x*) be an irreducible factor of *a*(*x*); let *u* be a root of *p*(*x*) and *υ* a root of *hp*(*x*). If *F*_{1}(*u*) = *K*_{1}, then by parts 1 and 2 we are done. If not, let = *F*_{1}(*u*) and = *F*_{2}(*υ*); *h* can be extended to an isomorphism *h*: → , with *h*(*u*) = *υ*. In [*x*], *a*(*x*) = (*x* − *u*)*a*_{1}(*x*), and in [*x*], *ha*(*x*) = *a*(*x*) = (*x* − *υ*)*ha*_{1}(*x*), where deg *a*_{1}(*x*) = deg *ha*_{1}(*x*) = *n* − 1. Complete the solution.

** J2**Explain why any monomorphism :

*F*(

*c*) → , which is an extension of

*h*, is fully determined by the value of (

*c*), which is a root of

*hp*(

*x*).

** J4**Since

*h*(1) = 1, begin by proving by induction that for any positive integer

*n*,

*h*(

*n*) =

*n*. Then complete the solution.

**CHAPTER 32**

** A2**In the first place, () is of degree 2 over . Next,

*x*

^{2}+ 1 is irreducible over (). (Explain why.) Complete the solution.

** D1**The complex fourth roots of 1 are ±1 and ±

*i*. Thus, the complex fourth roots of 2 are (1), (− 1), (

*i*), and (−

*i*), that is:

Explain why any field containing the fourth roots of 2 contains *α* and *i*, and conversely.

** E1**See

__Chapter 31__,

__Exercise E__.

** E6**Let

*a*α be a real sixth root of 2. Then [(

*α*) : ] = 6. Explain why

*x*

^{2}+ 3 is irreducible over (

*α*) and why [(

*a*,

*i*) : (

*α*)] = 2. If ω = (1/2) + (/2)

*i*(which is a primitive sixth root of 1), then the complex sixth roots of 2 are

*α*,

*αω*,

*αω*

^{2},

*αω*

^{3},

*αω*

^{4}, and

*αω*

^{5}. Any automorphism of (

*α*,

*i*) fixing maps sixth roots of 2 to sixth roots of 2, at the same time mapping

*i*to ±

*i*(and hence mapping

*ω*to

*ω*

^{5}). Provide details and complete the solution.

** H4**Suppose

*a*≠

*h*(

*a*), say

*a*<

*h*(

*a*). Let

*r*be a rational number such that

*a*<

*r*<

*h*(

*a*).

** I4**Every subgroup of an abelian group is abelian; every homomorphic image of an abelian group is abelian.

**CHAPTER 33**

** A2(a)**The polynomial is irreducible over by Eisenstein’s criterion, and it has three real roots and two complex roots. (Explain why.) Argue as in the example of page 340 and show that the Galois group is

*S*

_{5}.

** B3**It must be shown that for each

*i*,

*i*= 0,1, …,

*n*– 1, the quotient group

*J*

_{i}_{ + 1}/

*J*is abelian. It must be shown that

_{i}*J*contains

_{i}*xyx*

^{−}

^{ 1}

*y*

^{−}

^{ 1}for all

*x*and

*y*in

*J*

_{i}_{ + 1}. Provide details and complete.

__C3__*F* has an extension *K* which contains all the roots *d*_{1}, *d*_{2}, …, *d _{p}* of

*x*−

^{p}*a*. In

*K*,

*x*–

^{p}*a*factors into linear factors:

*x ^{p}* −

*a*= (

*x*−

*d*

_{1})(

*x*−

*d*

_{2}) ⋯ (

*x*−

*d*)

_{p}By uniqueness of factorization, *p*(*x*) is equal to the product of *m* of these factors, while *f*(*x*) is the product of the remaining factors.

** D3**If

*f*:

*G*→

*G*/

*K*is the natural homomorphism, then =

*f*

^{−}

^{ 1}() = {

*x*∈

*G*:

*f*(

*x*) ∈ }.

** D7**Prove this statement by induction on the order of

*G*/

*H*. Let |

*G/H*| =

*n*, and assume the statement is true for all groups

*H*′

*G*′, where |

*G*′/

*H*′| <

*n*. If

*G*has no normal subgroup

*J*such that

*H*⊂

*J*⊂

*G*(

*H*≠

*J*≠

*G*), then

*H*is a

*maximal*normal subgroup of

*G*; so we are done by parts 4 and 5. Otherwise, |

*G*/

*J*| <

*n*and |

*J*/

*H*| <

*n*. Complete the solution.