Question

Question on groups and symmetries

Let G = {(x, y, z): x, y, z are elements on R } and define (x1, y1, z1)*(x2, y2, z2) = (x1+x2, y1+x1z2+y2, z1+z2)

1) Show that (G,*) is a group and that it is non-abelian

2) Show that (G,*) does not have any elements of finite order (other than the identity)

Answer 1

@asnaseer could you look at this question please

Answer 2

Let (x1,y1,z1) and (x2,y2,z2) be elements of the group
1.) a group G is said to be abelian if a*b=b*a, where * is the binary operation of the group and a and b are elements of G. now let's check. Let (x1,y1,z1) and (x2,y2,z2). So by the definition of the operation, (x1,y1,z1)*(x2,y2,z2)=(x1+x2, y1+x1z2+y2, z1+z2)
and (x2,y2,z2)*(x1,y1,z1)=(x2+x1,y2+x2z1+y1,z2+z1). Since
x2z1 is not equal to x1z2
y2+x2z1+y1 is not equal to y1+x1z2+y2
(x2+x1,y2+x2z1+y1,z2+z1) is not equal to (x1+x2, y1+x1z2+y2, z1+z2)
(x2, y2, z2)*(x1, y1, z1) is not equal to (x1, y1, z1)*(x2, y2, z2)
then the group is not abelian.
2.) an element a of a group G is said to be of finite order if there exists a positive integer m( or more specifically the smallest among thos positive integers) such that a^m=e where e is the identity of the group and a^m=a*a*a*a....*a(m times)=e. Now the identity of this group is (0,0,0). since (x1,y1,z1)*(0,0,0)=(x1+0,y1+(x1)(0)+0,z1+0)=(x1,y1,z1). Now we know that (x1,y1,z1)^m= (x1,y1,z1)* (x1,y1,z1)* (x1,y1,z1)* (x1,y1,z1)....* (x1,y1,z1) <m times=<x1+x1+x1+...+x1 (m times),y1+y1+y1.....+y1(m times)+x1z1+x1z1+x1z1+....+x1z1(m times),z1+z1+z1....+z1(m times)>=<mx1,my1+mx1z1,mz1> now since the x1 and y1 and z1 are not equal to zero(this is stated in the question) and that m cannot be zero(because m can only be a positive integer), then no matter what the value of m is and for all <x1,y1,z1> element of the group, it can never be equal to the identity element(<mx1,my1+mx1z1,mz1> can never be <0,0,0>=e for any positive integer m). then it has no finite order.

Answer 3

hey sorry but we are showing that (g,*) is a group and in order to show that dont we show it has a inverse as well as the identity?

Answer 4

@anonymoustwo44

Answer 5

ah so your required to prove its a group also?

Answer 6

yeahh

Answer 7

you showed it is non abelian which makes perfect sense but it also asks to show if it is a group as well

Answer 8

ah ok ok. Which of the four conditions have you already proven so far?

Answer 9

i only know how to show the associativity aspect of it by doing (ab)c=c(bc) with a=(x1,y1,z1) b=(x2,y2,z2) and c=(x3,y3,z3) and i dont know the other 2

Answer 10

and isnt it only 3 conditions ? associativity, inverse, and identity?

Answer 11

yes that could be :)) because the extra one could be proven with the associative part which is a*b should be in the set.

Answer 12

ok so for the identity part. the identity will be <0,0,0> since <x1,y1,z1>*<0,0,0>=<x1+0,y1+(x1)(0)+0,z1+0>=<x1,y1,z1>. so it has identity <0,0,0> . Now suppose <x,y,z> is any element in the set. Now we are to show that there is an inverse <x',y',z'>(lets name the inverse this), where <x,y,z>*<x',y',z'>=<x+x',y+xz'+y',z+z'><0,0,0> which implies:
x+x'=0
y+xz'+y'=0
z+z'=0
and so inverse <x',y',z'> for any <x,y,z> in the group is <x',y',z'>=<-x,xz-y,-z> and we know that -x, xz-y, and -z(which is a product of two rational numbers minus a ratioanl number that gives also a rational number) are rational numbers also, so it is still in the set.
Recheck: <x,y,z>*<-x,xz-y,-z>=<x-x,y-xz+xz-y,z-z>=<0,0,0>=e. hence each <x,y,z> element of the set has inverse. Now since you have proven the associative part, we have now shown it satisfies the 3, hence it is a group

Answer 13

Thanks a lot that was REALLY helpful and detailed