practicesolns2.pdf

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Solutions to Practice Exam 2Modern Algebra

Friday, October 14, 2011

Remember that in general you cannotassume that multiplication of group elements is

commutative. You may not use a calculator on this exam. Show all your work.

1. Consider the set S= {10, 14, 2009, 9, 17, 2, 23, 8, 0}. Find the equivalence classes ofSunder the equivalence relation of congruence modulo 3. [10 points]

Solution. The equivalence classes are

{10, 8},{14, 2009, 17, 2, 23}, and{9, 0}.

2. Let a,b,x,yZ

and suppose a xand b y. Prove that ab xy. [10 points]

Proof. Since a x, we know x = au for some u Z. Likewise since b y, we know

y= bv for some v Z. Thus xy= (au)(bv) = (ab)(uv), so ab xy.

3. Let p be prime, let a and b be integers, and suppose a2 b2 (mod p). Prove thata b (mod p) or a b (mod p).

(Hint:Recall the lemma that ifp xy, then p xor p y.) [10 points]

Proof.

a2

b2

(mod p) a2

b2

0 (mod p) p a2 b2

p (a b)(a+b)

p a b or p a+b

a b 0 (mod p) or a+b 0 (mod p) a b (mod p) or a b (mod p).

4. Find the greatest common divisor of 6331 and 741. Be sure to show your work.[10 points]

Solution. Well apply the Euclidean algorithm:

6331 741 = 8 R403741 403 = 1 R338403 338 = 1 R 65338 65 = 5 R 13gcd

65 13 = 5 R 0

So (6331, 741) = 13.


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Practice Exam 2 Solutions Page 2 of 4 Modern Algebra, Fall 2011

5. Give an explicit counterexample to disprove the following statement:Let a,b,c,d Z. If (a, b) = 1 and (c, d) = 1, then (ac, bd) = 1. [10 points]

Solution. There are infinitely many counterexamples. One would be a= d= 2 andb= c = 3, so that (a, b) = (2, 3) = 1 and (c, d) = (3, 2) = 1, but

(ac,bd) = (2 3, 3 2) = (6, 6) = 6 = 1.

6. On the set {x R :x = 0}, define a relation by saying

a biff a

b Q,

where as usual Q denotes the set of rational numbers. For example, 3 7 since37

= 37 Q. Prove that is an equivalence relation. [10 points]

Proof. We need to prove the is reflexive, symmetric, and transitive.

Reflexive:Let 0 =

aR

. Then

a

a Q

a

a

.

Symmetric: Let 0 = a, b R such that a b. Then a

b Q; since Q is closed under

taking multiplicative inverses, b

a Q also, so b a.

Transitive: Let 0= a, b, c R such that a b c. Then a

b Q and

b

c Q. Since

Q is closed under multiplication,

a

c =

a

b

b

c Q

so a c.Since is reflexive, symmetric, and transitive, we conclude that is an equivalencerelation.

7. Consider the set of positive integers N. Ifa, b N, then we can factor them uniquelyas

a= 2em and b= 2fn,

where e 0, f 0, and m and n are odd. Define an equivalence relation on N bysaying a bife= f in that factorization.

Let [x] denote the equivalence class ofx under , and letSbe the set of all equivalence

classes. [10 points]

(a) If we try to define an operation onSby saying [x][y] = [xy], is this operationwell-defined? If so, prove it; if not, give a specific counterexample with actualnumbers.

Proof. Yes, the operation is well-defined. Suppose [x] = [x] and [y] = [y];then x = 2ea and x = 2eb and y = 2fc and y = 2fd for some odd numbersa,b,c,d. Then [x] [y] = [xy] = [2ea2fd] = [2e+f(ad)], where ad is odd; likewise


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[x] [y] = [xy] = [2eb2fd] = [2e+f(bd)], where bd is odd. Then by definition,[2e+f(ad)] = [2e+f(bd)], so [x] [y] = [x] [y]. We have proved that if

[x] [y] = [xy]

[x

] [y

] = [x

y

]

then [xy] = [xy], so is well-defined.

(b) If we try to define an operation on S by saying [x] [y] = [x+ y], is thisoperation well-defined? If so, prove it; if not, give a specificcounterexamplewith actual numbers.

Solution. No, it is not well-defined. For example, [3] [3] and [5] [13], but[3] [5] = [8] = [16] = [3] [13]. Schematically,

[3] [5] = [8]

[3] [13] = [16]

8. LetA and B be groups, let a A, and let b B. Consider the element (a, b) A B.

Prove that o((a, b)) o(a)o(b). [10 points]

Proof.

(a, b)o(a)o(b) =

ao(a)o(b), bo(a)o(b)

=

(ao(a))o(b), (bo(b))o(a)

=

(eA)o(b), (eB)

o(a)

= (eA, eB).

Thus o((a, b)) o(a)o(b).

9. In the group S5 Z3, find the order of the element

(125)(34), [1]

. [10 points]

Solution. We calculate

(125)(34), [1]

1=

(125)(34), [1]

(125)(34), [1]2

=

(152), [2]

(125)(34), [1]

3=

(34), [0]

(125)(34), [1]4

=

(125), [1]

(125)(34), [1]

5=

(152)(34), [2]

(125)(34), [1]6

=

(1), [0]

Since 6 is the smallest number n such that

(125)(34), [1]n

is the identity, we conclude

that o

(125)(34), [1]

= 6.

10. Here is the Cayley table for D8: [10 points]


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R0 R90 R180 R270 H V D1 D2R0 R0 R90 R180 R270 H V D1 D2R90 R90 R180 R270 R0 D2 D1 H V

R180 R180 R270 R0 R90 V H D2 D1

R270 R270 R0 R90 R180 D1 D2 V HH H D1 V D2 R0 R180 R90 R270V V D2 H D1 R180 R0 R270 R90

D1 D1 V D2 H R270 R90 R0 R180D2 D2 H D1 V R90 R270 R180 R0

Find the subgroup R270, Hand list all of its elements. Be sure to show your work.

Solution. Let X= R270, H.

(a) Since R270 X, we know R0, R90, R180, R270 X.

(b) Since H, R90 X, we know D1 = H R90 X.

(c) Since H, R180 X, we know V =H R180 X.

(d) Since H, R270 X, we know D2= H R270 X.

We conclude that X= {R0, R90, R180, R270, H , D1, V , D2} =D8.

XC. Suppose A and B are subgroups of a group G, and suppose their union A B is alsoa subgroup ofG. Prove that A B or B A. [10 bonus points]

Proof. Suppose not; then A B and B A. Thus there exists some a B Aandsome b B A. Now a, b A B, which by hypothesis is a subgroup ofG, so theirproduct ab A B. Thus ab A or ab B. But ifab A, then ab= a A b=

a1

a

A, a contradiction; likewise ifab B, then ab = b

B a = b

b1

B, acontradiction.

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