Algebra - Groups

[Mattofix]

1. The problem statement, all variables and given/known data

Prove that the following sets form infinite groups with respect to ordinary multiplication.

a){2^k} where k E Z
b){(1+2m)/(1+2n)} where m,n E Z

2. Relevant equations


3. The attempt at a solution

I sort of know about

closure
associativity
identity
inverses

...and i know that multiplication is somewhere but what do i multiply? - apart form that im pretty stuck.

[Mystic998]

You multiply the elements of the set. To show it's a group, show that the set is closed under the operation (multiplication), that there's an identity, and that arbitrary elements have an inverse. Technically you have to show the operation is associative too, but that's given by the fact that the operation is "ordinary" multiplication.

[Dick]

To prove closure for the easy one a), you just want to show that 2^(k1)*2^(k2) can be written as 2^(n) for some integer n. Can it? There's an identity if there is some integer n such that 2^n=1. Is there? There are inverses if for every integer k1, there is another integer k2 such that 2^k1*2^k2=1. Is there? You don't have to worry about associativity, it's a well known property of the real numbers.

[Mattofix]

k1 + k2 = k3 which is an integer? therefore closure proven?

2^0=1

inverses are just
2^k1*2^-k1=1

?

[Dick]

That's pretty much it. Now try the second one.

[Mattofix]

expand...? urghh...i dont know

[Dick]

Yes. Expand. Like this. (1+2m)(1+2n)=1+2m+2n+4mn=1+2(m+n)^2.

[Mattofix]

cant say i understand

why not (1+2m)(1+2m)/(1+2n)(1+2n)?

[Dick]

cant say i understand

why not (1+2m)(1+2m)/(1+2n)(1+2n)?

That was only a hint. I just did half the problem. You have to tell me if ((1+2m1)/(1+2n1))*((1+2m2)/(1+2n2)) can be written in the form (1+2m3)/(1+2n3) where m3 and n3 are integers, what's their relation to m1,n1,m2 and n2?