midterm1_soln.pdf

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MATH 356 ABSTRACT ALGEBRA

Midterm I (due 5pm February 21st, 2003)

Instructions: This is a closed bo ok, closed notes exam. Use of calculators is not p ermitted. You

have 3 hours. Show all your work for a full credit.

Print name :

Upon finishing please sign the pledge below:

On my honor I have neither given nor received any aid on this exam, and observed the time

limit specified above.

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Problem Max Points Your Score Problem Max Points Your Score

1 10 7 30

2 10 8 20

3 10 9 10

4 10 10 20

5 20 11 25

6 20 12 15

Total 200

1


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(1) By using the multiplication table (cf. Table 1), show that there is only one (up

to isomorphism) group of order 3.

Table 1.

e a be

a

b

Solution. Clearly ab = e since if its a or b then it follows that b = e or a = erespectively. Likewise, b a = e. We need to determine what a a and b b are.a

a cant be a since a

= e. If a

a = e then a

a = a

b and it follows that

a = b. Therefore a a = b. Likewise, b b = a.

(2) Show that R, is not isomorphic to the circle group U = {z C | |z| = 1 }.Solution. The two groups are not isomorphic in the same way C, R, ,which we have done in class. In short, the equation x = 1 is solvable in theformer but not in the latter.

To be more concrete, suppose that there is an isomorphism

: U R.

Since is a homomorphism, (1) = 1. Consider (1) : since ((1) (1)) =(1) (1) = 1, so (1) is a square root of 1. It is not 1, since (1) = 1and is one to one. Therefore (1) = 1.

Now consider (i). We have

1 = (1) = (i i) = (i) (i).So (i) is a square root of -1 in R!

(3) Construct an isomorphism between R/2 and the circle group U.

Solution. Let :R U be defined by (x) = e

2ix/2

. is onto, since forany element eio U, we have

2o2

= eio . Now, lets compute the kernel

of . (x) = e2ix/2 = 1 if and only if x/

2 Z, that is x 2Z = 2.

Therefore R/2 U.

(4) (a) Suppose that every proper subgroup of G is cyclic. Is G cyclic? If not,

give a counterexample.


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Solution. Z2 Z2 is a counterexample.

(b) Suppose that every proper subgroup of G is abelian. Is G abelian? If not,

give a counterexample.

Solution. S3 is a counterexample. Proper subgroups of S3 have orders 1,

2, or 3. Hence they have to be abelian.

(5) Find all subgroups of D3, and draw a subgroup diagram.

Solution. D3 = S3 has one subgroup of order 3 consisting of rotations 1 =2 = {0, 1, 2}. Since any rotation and a reflection generate the whole D3,there is no other subgroup of order 3. To find subgroups of order 2, we note that1, 2, 3 are the only elements of order 2. Therefore order 2 subgroups are

1 = {0, 1}, 2 = {0, 2} and 3 = {0, 3}. The subgroup diagramof S3 is as follows.

S3

oooooooooooooo

EEEE

EEEE

RRRR

RRRR

RRRR

RRRR

RR

1

OOOO

OOOO

OOOO

O1 2

yyyyyyyy

3

lllllllllllllllll

{0}

(6) Recall that an automorphism of a group G is a 1-1 onto homomorphism from

G to itself.

(a) How many automorphisms does Z/15Z have?

Solution. Generators ofZ15 are 1, 2, 4, 7, 8, 11, 13, 14. Hence it has 8

automorphisms.

(b) In general, how many automorphisms does Z/pqZ have for p and q distinctprime numbers?

Solution. Number of automorphisms is equal to the number of positive

integers less than pqthat are relatively prime to pq. We may assume p > q.

Multiples {1, p, 2p, . . . , (q1)p} ofp and multiples {q, 2q , . . . , (p1)q} ofq(less than pq) are the only nonzero elements that are not relatively prime

to pq. These two sets do not intersect : if mp = nq, then p divides n


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but n p 1. Thus the number of elements that are not generators is1 + (p

1 ) + (q

1). Hence the number of generators is pq

1

(p

1)

(q 1) = pqp (q 1) = (p 1)(q 1).

(7) Consider the following elements in S6.

=

1 2 3 4 5 6

4 2 6 5 1 3

, =

1 2 3 4 5 6

2 4 5 1 6 3

(a) Compute 2 and write it as a product of disjoint cycles, and then write

it as a product of transpositions.

Solution.

2 =

1 2 3 4 5 6

2 1 5 3 4 6

Thus 2 = (1 2)(3 5 4) = (1 2)(3 4)(3 5).

(b) List the orbits of 2.

Solution. {1, 2}, {3, 5, 4}, {6}.

(c) Compute the order |2|.

Solution. (3 5 4) is of order 3 and (1 2) is of order 2 and they are disjoint.Hence product of the 2 has order lcm(2, 3) = 6.

(8) Classify the following quotient groups according to the fundamental theorem

of finitely generated abelian groups.

(a) (Z/3Z Z/9Z)/(0, 2).Solution. (Z/3ZZ/9Z)/(0, 2) Z/3Z (Z/9Z) /2 Z/3Z{0} =Z/3Z.

(b) (Z/3Z Z/9Z)/(2, 3).Solution. (2, 3) is of order lcm(3, 3) = 3. Thus the quotient groupin question is of order 39

3= 9. So it is either Z9 or Z3 Z3. Since

(2, 3) = {(2, 3), (1, 6), (0, 0)}, it is clear that 9 (0, 1) is the first multipleof (0, 1) that lands in (2, 3). Hence (0, 1) + (2, 3) is an element of order9 in the quotient group. Hence it must be isomorphic to Z9.


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(9) How many abelian groups (up to isomorphism) are there of order 120?

Solution. 120 = 23

3 5.Z8 Z3 Z5Z4 Z2 Z3 Z5Z2 Z2 Z2 Z3 Z5

(10) Let be the homomorphism from Z Z Z to S20 determined by(1, 0, 0) = (1 4)(5 9 11 20)

(0, 1, 0) = (2 3 6)(7 13 18 19)(0, 0, 1) = (8 10)(12 14 15)

(a) Compute (11, 26, 13).

Solution. (1, 0, 0) is of order lcm(2, 4) = 4, (0, 1, 0) is of order lcm(3, 4) =12 and (0, 0, 1), of order lcm(2, 3) = 6. First we note that (0, 1, 0) =

(0, 1, 0)1 = (6 3 2)(19 18 13 7).(11, 26, 13) = (1, 0, 0)11(0, 1, 0)26(0, 0, 1)13

= (1, 0, 0)3(0, 1, 0)2(0, 0, 1)

= (1 4)(20 11 9 5) (6 2 3)(19 13)(18 7) (8 10)(12 14 15).

(b) Compute Ker ().

Solution. Suppose that (a,b,c) = (1, 0, 0)a(0, 1, 0)b(0, 0, 1)c = e.

Since (1, 0, 0), (0, 1, 0), (0, 0, 1) are all disjoint, it must be that (1, 0, 0)a =

e,

(0, 1, 0)b = e, and (0, 0, 1)c = e. Therefore (a,b,c) = e if and only

if 4 | a, 12 | b and 6 | c. Ker() = 4Z 12Z 6Z.

(11) (a) Show that (Z Z) /(1, 2) is isomorphic to Z.

Solution. The defining equation for the subgroup (1, 2) is b = 2a. Let : Z Z Z be defined by (a, b) = b 2a. This is a homomorphismwith kernel the subgroup defined by b = 2a, that is, (1, 2). is onto,hence (Z Z)/Ker() Z.

(b) Show that (Z Z) /(2, 4) is not isomorphic to Z.


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Solution. It has a torsion element (1, 2) (a torsion element is an element

of finite order), while Z doesnt have any. The quotient group in question

is isomorphic to Z Z2.

(c) Classify (Z Z Z) /(2, 2, 4) according to the fundamental theorem offinitely generated abelian groups.

Solution. The defining equation for the subgroup (2, 2, 4) isa 0 (mod 2)b = a

c = 2a

Let : ZZZ Z2ZZ be the homomorphism defined by (a,b,c) =(a, b a, c 2a). It is onto, and its kernel is determined by the threeequations above. Hence we have

(Z Z Z) /(2, 2, 4) = (Z Z Z) /Ker() Z2 Z Z.

(12) Let : Z+ Z+ be defined by (n) = s where s is the number of positiveintegers less than or equal to n that are relatively prime to n. It is called the

Euler phi-fuction.

(a) Compute (5), (6), and (8).

Solution. (5) = 4, (6) = 2, (8) = 4.

(b) Note that (d) is equal to the number of generators for Z/dZ. Using this,

prove that

n =d|n

(d).

Solution. Consider Zn and the fact that

n = |Zn| = d|n

(number of elements of order d)

But the elements of order d are exactly the generators ofd where d = nd .Hence we have

n =d|n

(d) =d|n

(d).


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(c) Using (b), show that if xm = e has at most m solutions in a finite group

G for each positive integer m, then G is cyclic.

Solution. Let n = |G|. Again we have

n = |G| =d|n

(number of elements of order d)

by Lagranges theorem. Let g and h be elements of order d. Then elements

in g and those in h satisfy xd = e. Since xd = e has at most d solutionsby assumption and |g| = |h| = d, it follows that g = h. Henceif a is an element of order d, then it is an element of the subgroup g.Therefore, the number of elements of order d does not exceed the number

of generators of g, which is (d). If there does not exist an element oforder n, then

n = |G| =

d|n,d=n(number of elements of order d)

d|n,d=n(d)

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