Is the Set [1;+∞[ x [1;+∞[ with the Operation (x;y)°(v;w) a Group?

  • Thread starter Alv95
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In summary, the set [1;+∞[ x [1;+∞[ with the operation (x;y)°(v;w) = (x+v-1; yw) is not a group because there are problems with the existence of inverses, as well as the fact that the neutral element (1;1) may not be included in the set for certain values of y.
  • #1
Alv95
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I have to find if the set [itex] [1;+∞[ x [1;+∞[ [/itex] with the operation [itex] (x;y)°(v;w) = (x+v-1; yw) [/itex] is a group

I have already proven Closure, associativity and Identity but I have some problems with invertibility :)

The neutral element that I have found is (1;1)

I did [itex] (x;y)°(x1;y1)= (1;1) [/itex] and I have found [itex]x1=-x+2[/itex] and [itex]y1=1/y[/itex]
The problem is that [itex]1/y[/itex] is not included in the set if [itex]y>1[/itex]...

Any advice? Is it a group? Thank you :)
 
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  • #2
Alv95 said:

Homework Statement


I have to find if the set [itex] [1;+∞[ x [1;+∞[ [/itex] with the operation [itex] (x;y)°(v;w) = (x+v-1; yw) [/itex] is a group


Homework Equations


I have already proven Closure, associativity and Identity but I have some problems with invertibility :)


The Attempt at a Solution


The neutral element that I have found is (1;1)
I did [tex] (x;y)°(x1;y1)= (1;1) [/tex] and I have found x1=-x+2 and y1=1/y
The problem is that 1/y is not included in the set if y>1...

Any advice? Is it a group? Thank you :)

Use the forward slash "/" to close your tex tags.
It looks to me like you have answered your own question. Inverses don't seem to be there.
 
  • #3
So it's not a group? :) The text of the homework that the teacher gave us seemed to imply that it was :)
 
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  • #4
Alv95 said:
So it's not a group? :)

Indeed: it is not a group!
 
  • #5
Thanks ;)
 
  • #6
Just so you get it. The problem is that 1/y is not necessarily in our set. But that is not the only problem. There is a problem with -x+2 as well. That also doesn't necessarily lie in our set! Indeed, if x=2, then -x+2=0 and this is not in our set!
 
  • #7
Alv95 said:
So it's not a group? :) The text of the homework that the teacher gave us seemed to imply that it was :)

C'mon. Show a little confidence in your work. Do you want to build a fence around it and insure it before you are willing to assert the result?
 
  • #8
Well, I prefer to be sure about what I write especially if it is on a new topic that I have just learned at school ;) By seeking help and advices on the internet I hope to improve and strenghten my knowledge and thus be more confident. I didn't expect it to be a problem.
 
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  • #9
Alv95 said:
Well, I prefer to be sure about what I write especially if it is on a new topic that I have just learned at school ;) By seeking help and advices on the internet I hope to improve and strenghten my knowledge and thus be more confident. I didn't expect it to be a problem.

It isn't a problem; I didn't mean to imply that it was. You nailed it on your first try. I was just trying to encourage you to be brave about what you've done. You do good work, so trust it.
 
  • #10
Thanks, I will :)
 

Related to Is the Set [1;+∞[ x [1;+∞[ with the Operation (x;y)°(v;w) a Group?

What is an invertible element in a group?

An invertible element in a group is an element that has a unique inverse element within the group. This means that when the invertible element is multiplied by its inverse, the result is the identity element (usually denoted as e) of the group.

What is the inverse of an invertible element in a group?

The inverse of an invertible element in a group is the unique element within the group that, when multiplied by the invertible element, results in the identity element of the group. This inverse element is denoted as a^-1, where a is the invertible element.

How do you determine if an element is invertible in a group?

To determine if an element is invertible in a group, you must check if it has a unique inverse element within the group. This can be done by multiplying the element with all other elements in the group and checking if any of the results are equal to the identity element. If there is only one such result, then the element is invertible.

What is the order of an invertible element in a group?

The order of an invertible element in a group is the number of times that the element can be multiplied by itself to get the identity element. In other words, it is the smallest positive integer n such that a^n = e, where a is the invertible element and e is the identity element of the group.

Can a non-invertible element exist in a group?

No, in a group, every element must have a unique inverse element. If an element does not have an inverse, then it is not considered to be a part of the group. Therefore, a non-invertible element cannot exist in a group.

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