ex 3 solutions

7/31/2019 Ex 3 Solutions

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CM121A, Introduction to Abstract Algebra Solution Sheet 3

Solutions to problems 1-4 from exercise sheet 3:

1. Determine whether each of the following binary operations is commutativeand/or associative:

(a) on R defined by x y = xy + 1;Solution: COMMUTATIVE since x y = xy + 1 and y x = yx + 1,so x y = y x for all x, y R.NOT ASSOCIATIVE since

(x y) z = (xy + 1) z = (xy + 1)z + 1 = xyz + z + 1, butx (y z) = x (yz + 1) = x(yz + 1) + 1 = xyz + x + 1.

These might not be the same; take for example x = y = 0, z = 1, so

(x y) z = 2, but x (y z) = 1.(b) on R defined by x y = x + y + 1;

Solution: COMMUTATIVE since xy = x+y +1 = y +x+1 = y xfor all x, y R.ASSOCIATIVE since (x y)z = (x+y +1)z = (x +y +1)+z +1 =x + y + z +2, and similarly x (y z) = x (y + z + 1) = x + y + z + 2,so (x y) z = x (y z) for all x,y,z R.

(c) (composition) on the set of functions from Z to Z;Solution: NOT COMMUTATIVE; take for example the functionsf : Z Z and g : Z Z defined by f(a) = a + 1, g(a) = 2a. Then(f g)(a) = f(g(a)) = f(2a) = 2a + 1, but (g f)(a) = g(f(a)) =g(a + 1) = 2(a + 1) = 2a + 2.

ASSOCIATIVE; we saw in general (end of Section 3.2) that h (g f) = (h g) f.

(d) + on the set of functions from Z to Z (if f : Z Z and g : Z Zare functions, then f + g is the function from Z to Z defined by(f + g)(a) = f(a) + g(a) for a Z);Solution: COMMUTATIVE since (f+ g)(a) = f(a) + g(a) = g(a) +f(a) = (g + f)(a) for all a Z, so f + g = g + f for all functionsf, g : Z Z.ASSOCIATIVE since

((f + g) + h)(a) = (f + g)(a) + h(a) = f(a) + g(a) + h(a)= f(a) + (g + h)(a) = (f + (g + h))(a)

for all a Z, so (f + g) + h = f + (g + h).

(e) + (vector addition) on R3;Solution: COMMUTATIVE since the formula

xyz

+

x

y

z

=

x + x

y + y

z + z

=

x

y

z

+

xy

z

shows that v + v = v + v for all v, v R3.ASSOCIATIVE since the formula

xyz

+

x

y

z

+

x

y

z

=

x + x + x

y + y + y

z + z + z

=

xy

z

+

x

y

z

+

x

y

z

,

shows that (v + v) + v = v + (v + v) for all v, v, v R3.


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(f) (cross product) onR3 (recall

xy

z

x

y

z

=

yz

zy

zx xz

xy yx

).

Solution: NOT COMMUTATIVE; recall that v v = v v forall v, v R3. Taking any v, v such that v v is non-zero, e.g.,

v =

100

and v = 01

0

gives v v = v v.NOT ASSOCIATIVE; taking for example

v = v =

10

0

, v =

01

0

gives (v v) v = 0 = v (v v).

2. Suppose that is a binary operation on a set S. We say e S is a leftidentity for if e a = a for all a S. Similarly, we say e S is a rightidentity for if a e = a for all a S. (So an identity element for isone which is both a left identity and a right identity.)

(a) Prove that if e is a left identity for and e is a right identity for ,then e = e (and so e is in fact an identity element).Solution: We have e = ee = e, where the first equality holds sincee is a left identity for and the second since e is a right identity for.

(b) Give an example of a set S with a binary operation for which thereis no left identity or right identity.Solution: Let S = N with the operation +.

(c) Give an example of an S and with more than one right identity.Solution: Let S be any set with more than one element (Z for

example) and define a b = a for all a, b S. Then every element ofS is a right identity, but none are left identities.

3. Suppose that is a binary operation on a set S with an identity elemente. If a, b S are such that a b = e, then we say a is a left inverse of band b is a right inverse of a.Let S denote the set of functions from Z to Z with the operation of (composition).

(a) Show that the function f defined by f(x) = x is an identity elementfor .Solution: Ifg S, then (fg)(x) = f(g(x)) = g(x) and (g f)(x) =g(f(x)) = g(x) for all x R. So f g = g f = g, showing that f isan identity element.

(b) Find a function in S that has a left inverse, but no right inverse.Solution: Let g(x) = 2x. A left inverse for g is a function h : Z Zsuch that h(g(x)) = f(x) = x for all x Z. The function

h(x) =

x/2, if x is even,(x 1)/2, if x is odd.

has this property.A right inverse for g would be a function j : Z Z such thatg(j(x)) = x for all x Z. This means that 2j(x) = x for all x Z.But since j(x) Z, this is only possible if 2j(x) is even. Therefore ghas no right inverse.

(c) Is the left inverse in your example unique?Solution: No, we could have defined h(x) however we liked for xodd, and it would still be a left inverse for g.Remark: The answer would be no for similar reasons in whateverexample you come up with. Well see why later.

4. Suppose that a,b,c,d Z and m, n N. Prove the following:


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(a) Ifa b mod n and c d mod n, then a c b d mod n.Solution 1: If a b mod n and c d mod n, then a = b + sn andc = d + tn for some s, t Z. Therefore

a c = b + sn (d + tn) = b + d + (s t)n,

so a c b d mod n.Solution 2: We use the analogous statements for sums and productsproved in lecture (Prop. 3.3.4). If c d mod n, then 1 c 1 d mod n since 1 1 mod n. If also a b mod n, then

a + (1) c b + (1) d mod n,

so a c b d mod n.

(b) Ifa b mod n and m|n, then a b mod m.Solution: If a b mod n then n|(a b), and since m|n it followsthat m|(a b), which means that a b mod n.

(c) Ifa b mod n, then am

bm

mod n.Solution: Suppose that a b mod n. We prove that am bm modn by induction on m.If m = 1, this says a b mod n, which is what we assumed, so thereis nothing to prove.Suppose that m > 1 and assume that am1 bm1 mod n. Letc = am1 and d = bm1. By Prop. 3.3.4, ac bd mod n, so am bm mod n.

ex 3 solutions

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