Proving Infinite Groups: Algebra - Groups

In summary, to prove that the sets {2^k} and {(1+2m)/(1+2n)} form infinite groups with respect to ordinary multiplication, you need to show that they are closed under the operation, have an identity element, and have inverses for every element. For the first set, you can prove closure by showing that 2^(k1)*2^(k2) can be written as 2^(n) for some integer n. The identity element is 2^0=1 and the inverse for any element 2^k is 2^-k. For the second set, you can expand the expression (1+2m)(1+2n) and show that it can be
  • #1
Mattofix
138
0

Homework Statement



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

Homework Equations




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 I am pretty stuck.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
k1 + k2 = k3 which is an integer? therefore closure proven?

2^0=1

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

?
 
  • #5
That's pretty much it. Now try the second one.
 
  • #6
expand...? urghh...i don't know
 
  • #7
Yes. Expand. Like this. (1+2m)(1+2n)=1+2m+2n+4mn=1+2(m+n)^2.
 
  • #8
cant say i understand

why not (1+2m)(1+2m)/(1+2n)(1+2n)?
 
  • #9
Mattofix said:
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?
 

1. What is an infinite group in algebra?

An infinite group in algebra is a set of elements that follow specific rules or operations, such as addition or multiplication. Unlike finite groups, which have a limited number of elements, infinite groups have an infinite number of elements.

2. How do you prove that a group is infinite?

To prove that a group is infinite, you need to show that there is an infinite number of elements in the group. This can be done by demonstrating that the group follows a specific algebraic structure, such as the additive or multiplicative properties of a group, which have an infinite number of solutions.

3. What is the significance of proving a group is infinite?

Proving that a group is infinite is important in mathematics as it allows for the exploration of a larger set of numbers and operations. It also has applications in various fields, such as cryptography and number theory.

4. Can all groups be proven to be infinite?

No, not all groups can be proven to be infinite. Some groups, known as finite groups, have a limited number of elements and cannot be proven to be infinite. However, there are many infinite groups in mathematics, and new ones are constantly being discovered.

5. What are some common techniques for proving a group is infinite?

There are several techniques for proving a group is infinite, including demonstrating that the group follows specific algebraic properties, using induction to show an infinite number of solutions, or constructing an explicit infinite subgroup within the group.

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