Make sure to include the three rules about being Associative, identity and inver

Question

Make sure to include the three rules about being Associative, identity and inverse.

Please be as detailed as possible.

Problem 1. In the following list, determine which sets and operations form groups, and which do not. If you think it is a group, prove it. If you think it is not, explain which property fails and give a specific example showing why it fails. (If more than one property fails, it is enough to show why one of them fails.) (a) The set of all even integers. Operation: addition. (b) (Q, ?), where Q is the set of all rational numbers. (c) (G, *), where G = {a} and a * a = a. (d) The set of all strictly increasing functions f : [0,1] [0, 1], such that f(0) = 0 and f (1) = 1. Operation: composition of functions.

Explanation / Answer

(A) IT IS A GROUP

Associative: This is the same proof as for integers. Basically, for a,b,c in even integers, (a+b)+c = a+b+c = a+(b+c).

Identity: Since 0 = 2*0, then 0 is even. Now for any even number a, we get 0+a =a = a+0. So 0 is the identity.

Inverse: If a is an even number then a = 2n for some n. Now -2*n = 2*(-n), so -2n is even too and 2n+(-2n) =0 = (-2n)+(2n). Si -2n is the inverse of a.

(B) ITS NOT A GROUP

Notice that substraction is not associative as: (1-2)-3 is (-1)-3 which is -4. However, 1-(2-3) = 1- (-1) = 2.

(C) ITS IS A GROUP (Called the trivial group)

Associative: (a*a)*a = (a)*a = a, and a*(a*a) = a*(a) = a. So the operation is associative.

Identity: a is the identity since the only element in G is a and a*a = a = a*a.

Inverse: The inverse of a is itself since a*a=a=a*a.

(D) ITS A GROUP

Associativity follows from associativity of composition of functions: (f compose g) compose h is the function where (f compose g) compose h of x is f(g(h(x))) and f compose (g compose h) is the function where f compose (g compose h) of x is f(g(h(x)). So it is associative.

Identity: f(x) = x is the identity as it is increasing, f(1)=1, f(0)=0 and for any g(x), we get g compose f (x) =g(f(x))=g(x) and f compose g of x is f(g(x)) = g(x).

Inverse: This is the hardest part. Let f be a strictly increasing function with f(0)=0, f(1)=1. Then since f is strictly increasing if f(x) = f(y), then x=y (if x is not y, then x<y or y<x so f(x)<f(y) or f(y)<f(x)). So f is injective. f is surjective since it is continuous and f(0)=0 and f(1)=1. So any value between 0 and 1 are hit by the intermediate value theorem. Hence f is bijective and has an inverse g, this is ,f compose g = x and g compose f = x.

To show g is strictly increasing, notice that if x < y then f(x)<f(y), so g(f(x)) < g(f(y)). But g(f(x)) = f(g(x)) (they are both x). So f(g(x))<f(g(y)). Since f is strictly increasing then g(x) had to be less than g(y). So g is strictly increasing and since f(1)=1 and f(0)=0 then g(1)=1 and g(0)=0.

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