Proving a set under an operation is associative. (Abstract Algebra)

  • Thread starter jdinatale
  • Start date
  • #1
155
0

Homework Statement



godel.jpg


I'm trying to prove that this is a group. I already established elsewhere that it is a binary operation, so now I am onto proving associativity. I've tried many examples and so I'm confident it is associative, but now I just have to prove that.


The Attempt at a Solution



Let [itex]x, y, z \in G[/itex]. Then [itex]x*(y*z) = x*(y + z - [y + z]) = x + y + z - [y + z] - [x + y + z - [y + z]][/itex]

Then [itex](x*y)*z = (x + y - [x + y])*z = x + y + z - [x + y] - [x + y + z - [x + y]] [/itex]

Now the problem is coming up with an equality to show this. We would have to show that [y + z] = [x + y]. I guess you could do a ton of cases where you show what happens when [itex]y + z \geq 1[/itex] and [itex]y + z < 1[/itex], same goes for [itex]x + y[/itex], but I'm now sure that even that would work.
 
Last edited:

Answers and Replies

  • #2
22,089
3,297
Did you see quotient groups?? Perhaps you can prove your structure to be isomorphic to

[tex]\mathbb{R}/\mathbb{Z}[/tex]

Since the above is a group, it follows that your G is also a group.
 
  • #3
155
0
Did you see quotient groups?? Perhaps you can prove your structure to be isomorphic to

[tex]\mathbb{R}/\mathbb{Z}[/tex]

Since the above is a group, it follows that your G is also a group.

No, I've never seen a quotient group or the term isomorphic (although I know what isomorphic means). We haven't even defined a subgroup. This is only the third day of abstract algebra class and he hasn't gotten to either of those terms. We are suppose to be able to prove this just using definitions of a group, definitions of associativity, etc.

I would normally be up for the challenge of proving it using an alternate method like the one you suggested, but the assignment is due tomorrow and I have other assignments to do, so I won't have time to read up on the quotient group stuff yet.
 
  • #4
Dick
Science Advisor
Homework Helper
26,260
619
No, I've never seen a quotient group or the term isomorphic (although I know what isomorphic means). We haven't even defined a subgroup. This is only the third day of abstract algebra class and he hasn't gotten to either of those terms. We are suppose to be able to prove this just using definitions of a group, definitions of associativity, etc.

I would normally be up for the challenge of proving it using an alternate method like the one you suggested, but the assignment is due tomorrow and I have other assignments to do, so I won't have time to read up on the quotient group stuff yet.

You can't show [x+y]=[x+z]. It's not true as you can see by making examples. What you can show is things like [x+[y]]=[x]+[y] when x and y are positive.
 
  • #5
155
0
You can't show [x+y]=[x+z]. It's not true as you can see by making examples. What you can show is things like [x+[y]]=[x]+[y] when x and y are positive.

Thank you for responding in this thread and in my last one. I'm not sure how your example helps. I know you can't show that [x + y] = [x + z]....however, we would have to do this to show that these two equations are equal, since [x +y] and [x + z] are the only difference between the two equations.

Let [itex]x, y, z \in G[/itex]. Then [itex]x*(y*z) = x*(y + z - [y + z]) = x + y + z - [y + z] - [x + y + z - [y + z]][/itex]

Then [itex](x*y)*z = (x + y - [x + y])*z = x + y + z - [x + y] - [x + y + z - [x + y]] [/itex]


I've tried a ton of examples, and I am almost certain that this is associative, but I'm not sure how to show the above equations are equal.
 
  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,046
1,636
Look at [x+y+z-[y+z]] as [a-] where a=x+y+z and b=y+z.
 

Related Threads on Proving a set under an operation is associative. (Abstract Algebra)

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
9
Views
2K
Replies
12
Views
2K
Replies
15
Views
4K
Replies
11
Views
3K
Replies
5
Views
4K
Top