abstract algebra problem set solutions

7/26/2019 Abstract Algebra Problem set solutions

1/4

Spencer Leonardis

1-14-2015

Discussion ?

Math 111A

1.1. Let S and Tbe sets and let f : S Tbe a function. Prove the following:

(a) IfUis a set and g : TUis a function such that g f is injective then also f is injective.

Proof. Let s1,s2 S and assume that s1 =s2. Since g f is injective we have g(f(s1))= g(f(s2))

so we must have f(s1) =f(s2). Thus f is injective.

(b) IfR is a set and h :RS is a function such that fhis surjective then also f is surjective.

Proof. Assume that fhis surjective. Then for every t Tthere exists an r R, for which

f(h(r))= t. Let s S. Then for each r R we have h(r)=s. This shows there is ans Ssuch that

f(h(r))=f(s)= t. Thus f is surjective.

(c) Show that ifg,h : TS are functions satisfyinggf= ids and fh= id t then f is bijective

and g= h=f1.

(d) Show that for subsetsV1andV2 one has (i) f1(V1V2)=f

1(V1)f1(V2) and

(ii) f1(V1V2)=f1(V1)f

1(V2).

Proof (i). First we show that f1(V1V2)f1(V1)f

1(V2). Assume thata f1(V1V2). Then

f(a) V1V2implying that f(a) V1 or f(a)V2. If f(a) V1, thena f1(V1) and if f(a)V2,

thena f1(V2). Thusa f1(V1)f

1(V2). Now leta f1(V1)f

1(V2). This means that

a f1(V1) ora f1(V2). Ifa f

1(V1), then f(a) V1 and ifa f1(V2), then f(a)V2. This

means that f(a)V1V2, soa f1(V1V2). We have shown that f

1(V1)f1(V2)f

1(V1V2)

and since it was previously demonstrated that f1(V1V2)f1(V1)f

1(V2), we may conclude

that f1(V1V2)= f1(V1)f

1(V2).

Proof (ii). First we show f1(V1V2)f1(V1)f

1(V2). Leta f1(V1V2). Then f(a) V1V2

implying that f(a) V1 and f(a) V2. This means that a f1(V1) anda f

1(V2). Thus

a f1(V1) f1(V2). Now suppose that a f

1(V1)f1(V2). Thena f

1(V1) and a f1(V2).

This means that f(a) V1 and f(a) V2, implying that f(a) V1V2. Thusa f1(V1V2). We

have shown that f1(V1)f1(V2)f

1(V1V2) and since it was previously demonstrated that

f1(V1V2)f1(V1) f

1(V2), it follows that f1(V1V2)=f

1(V1)f1(V2).

1


2/4

(e) Show that for subsetsU1 and U2 one has (i) f(U1U2)=f(U1)f(U2) and

(ii) f(U1U2)f(U1)f(U2). Give an example where the last inclusion is not an equality.

Proof (i). Suppose that t f(U1U2)= {f(a) : a U1U2}. Then t = f(s) for some s U1U2.

Thuss U1or s U2. Ifs U1, then t =f(s) {f(a) : a U1}=f(U1). Ifs U2, then

t=f(s) {f(a) : a U2}=f(U2). Thus f(s)= t f(U1)f(U2). This shows that f(U1U2)

f(U1)f(U2). Now assume that t f(U1)f(U2). This means that t f(U1) ort f(U2). If

t f(U1), thent =f(s) for some s U1. Ift f(U2), then t =f(s) for some s U2. Thus t =f(s)

for some s U1U2 and t =f(s) {f(a) : a U1U2}=f(U1U2). This shows that

f(U1)f(U2)f(U1U2) and since f(U1U2)f(U1)f(U2), it follows that

f(U1U2)=f(U1)f(U2).

Proof (ii). Lett f(U1U2)= {f(a) : a U1U2}. This means t =f(s) for some s U1U2.

Sinces U1, we have t = f(s) {f(a) : a U1}=f(U1). Since s U2, we have

t=f(s) {f(a) : a U2}=f(U2). This demonstrates that t f(U1)f(U2). Therefore

f(U1U2)f(U1)f(U2).

1.2. Assume that n N. Letd0,d1, . . . ,dr {0,...,9}be its decimals read from left to right; that is,

n= d0+d110+32102+ +dr10

r.

(a) Show that n d0+ +dr mod 9.

Proof. Ifn d0+ +dr mod 9, then we have

n (d0+ +dr)= d0+10d1+102d2+ +10

rdr (d0+ +dr)= 9d for some d Z. we may

rearrange the preceding equality as (d0d0)+ (10d1d1)+ (102d2d2)+ + (10

rdrdr)

= () 0+9d1+99d2+ + (10r1)dr = 9d. Since the coefficients ofd0, . . . ,dr on the right hand

side of () are all equal to 10r1 for some r N0, we must show that 9 | (10r1).

We will use induction on r N0. Settingr = 0, we obtain 9 | 0 which is true since 0 = 9dfor d = 0.

Assume 9 | 10r1. Then there exists d Z for which 10r1= 9d. Notice that 10r+11

= 10r101= 10(9d+1)1= 10 9d+9= 9(10d+1). This demonstrates that 10r+11= 9(10d+1)

= 9cfor some c Z. We have shown that if 9 | 10r1 then 9 | 10r+11, completing the induction.

Since 9 | 10r1, the coefficients ofd0, . . . ,dr in () are all divisible by 9.

Thus 9 | 0d0+9d1+99d2+ + (10r1)dr = n (d0+ +dr). Since 9 | n (d0+ +dr), it follows

thatn (d0+ +dr) mod 9.

2


3/4

(b) Show that n is divisible by 9 if and only ifd0+ +dr is divisible by 9.

Proof. Suppose 9 | d0+d110+32102+ +dr10

r. Then there exists d Z such that

() d0+10d1+102d2+ +10

rdr = 9d. Since 9 | (10r1) by part (a), we may rearrange () as

d0+ (10d19d1)+ (102d299d2)+ +10

rdr (10r1)dr = d0+ +dr

= 9dd111d2

10r1

9 dr

= 9a

This demonstrates thatd0+ +dr = 9afor some a Z, so 9 | d0+ +dr.

Now assume that 9 | d0+ +dr. Then there exists d Z such that () d0+ +dr = 9d. We may

manipulate () to obtain

d0+ (9d1+d1)+ (99d2+d2)+ + (10r1)dr +dr= d0+10d1+10

2d2+ +10rdr

= n= 9d+9d1+99d2+ + (10r1)dr

= 9

d+d1+11d2+ +

10r1

9dr

= 9a.

This shows that n = 9afor some a Z, so 9 | n.

(c) Show that n is divisible by 3 if and only ifd0+ +dr is divisible by 3.

Proof. Suppose that 3 | n. Then there exists d Z such that ()d0+10d1+102d2+ +10

rdr = 3d.

We may rearrange () asd0+ +dr = 3d9d199d2 (10r1)dr. Since the coefficients of

d0, . . . ,dr are divisible by 9 by part (a), each coefficient is also divisible by 3.

Thus we have d0+ +dr = 3

d3d133d2

10r1

3dr

= 3a. This showsd0+ +dr = 3a

for some a Z, so 3 | d0+ +dr. Now assume that 3 | d0+ +dr. Then there existsd Z such

that ()d0+ +dr = 3d. We may express () as

n= 3d+9d1+99d2+ + (10r1)dr = 3

d+3d1+33d2+ +

10r1

3dr

= 3a. Thusn = 3afor

some a Z, so 3 | n.

(d) Show that n d0d1+d2+ mod 11.

Proof. Ifi N0, then the coefficient of each d2i {d0, . . . ,dr}is divisible by v 102i1.

(e)

(f)

2.1.

(a) Is commutative?

No, is not commutative since there exists b,e S such thatb e= c = b=eb.

(b) Is associative?

Observe from the table thatb d = db=e. Since b e =e b, it follows that

b (db) = (bd)b. Thus is not associative.

3


4/4

2.2. Suppose that is an associative and commutative binary operation on a set S . Show that

the subset T:= {a S: aa= a}ofS is closed under .

Proof. Suppose that a,b S. Since is a binary operation inS we have a b S. Now let a,b T.

Then (ab) (ab)= (ab) (ba)= a ((bb)a)= a (ba)= a (ab)= (aa)b= ab.

We have shown that for a,b Tone has (ab) (ab)= ab. Thusa b T.

2.3 : Let f :QQ be defined by f(x)= 3x1.

(a) Show that f is bijective and compute f1.

Suppose that x1,x2 Q and 3x11= 3x21. Then 3x1 = 3x2 which reduces to x1 =x2. This shows

f is injective. Let b Q. Then b =c

d for some c Z andd N. Seta :=

b+1

3. Then

f(a)=f

b+1

3

= 3

b+1

3

1= b+11= b, showing that f is surjective. Since f is both injective

and surjective, it is also bijective. To find the inverse of fwe interchange variables to obtain

x= 3y1. Solving for yyields y=f1(x)=1

3x+

1

3.

(b) Find a binary operation on Q such that f : (Q,+) (Q,) is an isomorphism.

Since by part (a) fis bijective, we must define a binary operation on Q such that

f : (Q,+) (Q,) satisfies f(a+b)=f(a)f(b).

Define :QQQ by (a,b) a+ b c. The identity element of is c =f(0)=1. This means

thata b= a+ b+1 and f(a+ b)= 3(a+ b)1

= 3a+3b2+1

= (3a1)+ (3b1)+1

= (3a1) (3b1)=f(a)f(b).

Thus there exists an isomorphism f :QQ and : (a,b) a+ b+1 is the binary operation on Q

such that (Q,+)= (Q,).

(c) Find a binary operation on Q such that f : (Q,) (Q,+) is an isomorphism.

Since by part (a) fis bijective, we must define a binary operation on Q such that

f : (Q,) (Q,+) satisfies f(ab)=f(a)+f(b).

Define :QQQ by (a,b) a+ b c. The identity element of is c =f1(0)=1

3. This means

thata b= a+ b1

3and f(ab)= 3

a+ b

1

3

1= 3a+3b2

= (3a1)+ (3b1)=f(a)+f(b).

Thus there exists an isomorphism f :QQ and : (a,b) a+ b1

3 is the binary operation on Q

such that (Q,)= (Q,+).

4. Assume that (S,) and (T,) are isomorphic binary structures.

(a) Show that (S,) is commutative if and only if (T,) is commutative.

Proof.

(b) Show that (S,) is associative if and only if (T,) is associative.

4

abstract algebra problem set solutions

Documents