ex 4 solutions

7/31/2019 Ex 4 Solutions

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CM121A, Abstract Algebra Solution Sheet 4

Solutions to problems 5-7 from exercise sheet 3, and problems 1-4

from exercise sheet 4

Sheet 3:

5. Find all a Z such that a2 1 mod 15.Solution: We know from Prop. 3.3.4 (or the preceding exercise) that ifa b mod 15, then a2 b2 mod 15. So we just need to find the solutionswith 0 a 14, and then the set of all solutions in Z is given by thosecongruence classes modulo 15. Running through the possibilities, we findthat a = 1, 4, 11 and 14 work, so the solutions are:

a 1, 4, 11 or 14 mod 15.

6. Express each of the following elements ofZ101 in the form [r]101 for someinteger r with 0 r 100:

(a) [36]101;Solution: [65]101

(b) [91]101 + [36]101;Solution: [26]101

(c) [91]101[36]101;Solution: [44]101 (Note that since 91 10 mod 101,

91 36 10 36 360 44 mod 101.)

(d) [9136]101.Solution: [1]101 (Note that 91

2 (10)2 1 mod 101, so part (c)of the preceding problem shows that

9136 = (912)18 (1)18 1 mod 101.)

7. Let n be a positive integer.

(a) Prove that multiplication on Zn is associative.Solution: Suppose that [a]n, [b]n, [c]n Zn. By definition of multi-plication on Zn, we have

([a]n[b]n)[c]n = [ab]n[c]n = [(ab)c]n = [a(bc)]n = [a]n[bc]n = [a]n([b]n[c]n)

(where we used the associative law for multiplication on Z to replace(ab)c by a(bc)).

(b) Prove the distributive law on Zn:

[a]n([b]n + [c]n) = [a]n[b]n + [a]n[c]n for all [a]n, [b]n, [c]n Zn.

Solution: Suppose that [a]n, [b]n, [c]n Zn. Then

[a]n([b]n + [c]n) = [a]n[b + c]n = [a(b + c)]n= [ab + ac]n = [ab]n + [ac]n = [a]n[b]n + [a]n[c]n

(where we applied the distributive law on Z to replace a(b + c) by

ab + ac).

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Sheet 4:

1. Which of the following are groups? If it is not, give a reason.

(a) The set of even integers under addition.

Solution: A group.(b) The set of odd integers under multiplication.

Solution: Not a group (3, for example, has no inverse).

(c) The set of complex numbers under multiplication.Solution: Not a group (0 has no inverse).

(d) The set of rational numbers of the form 2n (for n Z) under multi-plication.Solution: A group.

(e) The set of real numbers with the operation defined by xy = xy+1.Solution: Not a group (not associative, from previous exercise sheet).

(f) The set of real numbers with the operation defined by x y =

x + y 1.Solution: A group (associative from previous exercise sheet; identityis 1; inverse of x is 2 x).

(g) The set of vectors (x1, x2, . . . , xn) in Rn under vector addition.

Solution: A group.

(h) The set of vectors (x1, x2, . . . , xn) in Rn under the dot product.

Solution: Not a group (not even a binary operation).

(i) The set of functions f : R R with the operation + (defined by(f + g)(x) = f(x) + g(x)).Solution: A group.

2. List the 12 symmetries of a regular hexagon (give a description and nota-

tion for each). This set forms a group under composition. Construct itsmultiplication table.Solution: Assume the vertices are labeled A, B, C, D, E and F runningclockwise, with a horizontal edge AB on top.

The identity, denoted e.

Rotations clockwise by:

60, denoted 1;

120, denoted 2;

180, denoted 3;

240, denoted 4;

300, denoted 5.

Reflections in the following axes:

through midpoints of AB and DE (vertical), denoted 1;

through vertices B and E, denoted 2;

through midpoints of BC and EF, denoted 3;

through vertices C and F (horizontal), denoted 4;

through midpoints of CD and F A, denoted 5;

through vertices D and A, denoted 6.

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The multiplication table is then:

e 1 2 3 4 5 1 2 3 4 5 6e e 1 2 3 4 5 1 2 3 4 5 6

1 1 2 3 4 5 e 2 3 4 5 6 1

2 2 3 4 5 e 1 3 4 5 6 1 23 3 4 5 e 1 2 4 5 6 1 2 34 4 5 e 1 2 3 5 6 1 2 3 45 5 e 1 2 3 4 6 1 2 3 4 51 1 6 5 4 3 2 e 5 4 3 2 12 2 1 6 5 4 3 1 e 5 4 3 23 3 2 1 6 5 4 2 1 e 5 4 34 4 3 2 1 6 5 3 2 1 e 5 45 5 4 3 2 1 6 4 3 2 1 e 56 6 5 4 3 2 1 5 4 3 2 1 e

3. Recall that M2(R) denotes the set of 2 2 real matrices, and that if

A = a b

c d

, then det A = ad bc.

(a) Prove that matrix multiplication on M2(R) is associative.

Solution: Suppose that A =

a b

c d

, B =

a b

c d

and C =

a b

c d

are elements of M2(R). Then

(AB)C =

aa + bc ab + bd

ca + dc cb + dd

a b

c d

=

aaa + bca + abc + bdc aab + bcb + abd + bdd

caa + dca + cbc + ddc cab + dcb + cbd + ddd

=

a b

c d

aa + bc ab + bd

ca + dc cb + dd

= A(BC)

Therefore matrix multiplication is associative.

(b) Prove that ifA, B M2(R), then det(AB) = det(A)det(B).

Solution: If A =

a b

c d

, B =

a b

c d

, then

AB = aa + bc ab + bd

ca

+ dc

cb

+ dd ,

so

det(AB) = (aa + bc)(cb + dd) (ab + bd)(ca + dc)= aadd + bccb abdc bdca (other terms cancel)= (ad bc)(ad bc)= (det A)(det B).

4. Recall that if n is a positive integer, then

Z

n= { [a]n Zn | a Z, gcd(a, n) = 1}

is a group under multiplication mod n.

(a) List the elements ofZ

15 and find the inverse of each.Solution: Z

15= {[1], [2], [4], [7], [8], [11], [13], [14]}.

Element [1] [2] [4] [7] [8] [11] [13] [14]Inverse [1] [8] [4] [13] [2] [11] [7] [14]

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(b) Let n = 2009. Find the inverse of [1829]n in Z

n.

Solution: Applying the Euclidean Algorithm to 2009 and 1829 gives:

2009 = 1 1829 + 1801829 = 10 180 + 29

180 = 6 29 + 629 = 4 6 + 5

6 = 1 5 + 1.

So in fact gcd(1829, n) = 1, and [1829]n is in Z

n. To find its inverse,

we unwind the above equations:

1 = 6 5 = 6 (29 4 6) = 5 6 29= 5(180 6 29) 29 = 5 180 31 29= 5 180 31 (1829 10 180) = 315 180 31 1829= 315(2009 1829) 31 1829 = 315 2009 346 1829.

So 346 1829 1 mod n, and the inverse of [1829]n is [346]n =

[1663]n.

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ex 4 solutions

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