Algebra theorem

1. Jul 2, 2013

javedansari

If x and y be elements in a group G such that xy ε Z(G)

prove that xy = yx..

2. Jul 2, 2013

Tenshou

So you are trying to prove commutativity of a group? Well... Let
$xy \in G$ since it must be a obey closure of a group. now, since the product of two elements are in a group let there exist an element $e$ such that $xy= e$, since we are assuming the existing of an element $e$ we should show that this object is "unique" so we set $yc=e$ where $c \in G$ we can now show that, $xy=yxe$ this is clearly true since $\forall x \in G[xe=x]$ is true, clearly $xy=yx(yc)$ using a little bit associativity and hypothesis we get $xy=y(xy)c \implies xy=yc = e$ thus establishing that $\forall x,y \in G [xy=e \implies yx=e]$ since those elements are the same,do the math .

Here is a better one
suppose that the a member $x \in G$ such that $xx=e$ is true, then $xe=ex=e$ furthermore we can show that $x(xx)=(xx)x=e \implies (xx)x =x(xx) =e$ by substitution and associativity it becomes $ex=xe=e$ much easier.

Last edited: Jul 2, 2013
3. Jul 2, 2013

lavinia

The center of the group is those elements that commute with everything.

In particular, xy must commute with x^-1

Not that x and y do not need to be in the center of the group. Can you construct an example?

4. Jul 2, 2013

micromass

I really fail to see how this has anything to do with the problem in the OP. I would appreciate it if you wouldn't give answers on subjects you are not qualified in.

5. Jul 2, 2013

Tenshou

So, if xy doesn't commute with x inverse what would happen to Z(G) would this product not be in the center?

What do you mean? He said prove it, and that was my attempt at it. Did I prove it wrong?

6. Jul 2, 2013

lavinia

Try to figure this out. It isn't difficult.

7. Jul 2, 2013

Tenshou

Sorry I am asking the wrong question. I mean since x and y don't need to be in the center of the group, then why does the product need to be in the center? I mean the group could be centerless, right?

8. Jul 2, 2013

HallsofIvy

No, in any group at least the group identity commutes with all elements. So the "center of the group" consists of at least the identity. In this case, xy is in the center because it is given as the hypothesis of the problem. Sorry, but I can't make heads or tails out of your post #2. You start by arguing that xy must be something, and call it "e". But then state $\forall x \in G [xe= x]$, assuming that e is the identity. That makes no sense at all!

9. Jul 2, 2013

Tenshou

Well, maybe you are right I should have proved $\forall x \in G [xe= x]$ before trying to just assume it. >.< okay, I should have made it a little more obvious. And by centerless I mean that it has only the identity in the center. Can I try again >.< ... in the morning when I can think a little clearer?... but since it is a trivial proof I will show this, so taking a step back from the center of the group and prove that there really does exist this element "e" in G such that the axioms of a group hold.

It is true that if there does exist this element "e" and it god given since the axioms of a Group hold, mainly associativity, the existence of inverse and the identity element "e" so the triple of $(G, \times, e)$is a group $\times : G \to G$, it is known that we can take an subset in G mainly Z(G) and call this the center, this center will be seen to be trivial subset of G. when I wake up. Is this good enough Ivy :3 I must redeem myself!

Oh yeah! the last part is just assuming that x is an inverse of its self. I don't know... maybe it didn't have any relevancy to xy = yx, but it did iff x=y which is... ?

10. Jul 2, 2013

verty

Tenshou: This was Javedansari's question. You have sort of railroaded the thread. I do that too sometimes but usually only when enough hints have been given or the poster has not replied for a couple of days.

And it's not like your posts are getting any clearer or more justified. My recommendation is to try work things out by yourself and if you get stuck, ask questions that are specific to what you wish to know without giving too much away.

But actually, a much stronger recommendation from me is to study your school math instead. Focusing on this higher level stuff is a mistake when we know that there are things more important to know that you must learn.

Now, can we wait for Javedansari to respond? Thanks.

11. Jul 2, 2013

Tenshou

He just made the thread today :/ I wanted to solve this it didn't seem to difficult I mean I have done it before, just with out having this proper(or not) subgroup in there. And I didn't understand what I was doing wrong, and when this question popped up I thought I would try it out and see if I can get some critiques on what I did wrong. And maybe what I could possibly do better.

12. Jul 2, 2013

micromass

This thread was made by javedansari, so it is meant to help him out and to help him solve the problem. This thread is not meant for you to try it out and get help. If you want to try things out and get critiques, fair enough, but then you'll need to make your own thread.

13. Jul 2, 2013

lavinia

the product may not be in the center. But in some cases it might.

A good idea is to construct examples of groups out of low dimensional matrices that are orthogonal transformations and whose entries are zero, one, or minus one.

Take a look ,for instance, at the group generated by these 2x2 matrices.

1 0
0 -1

and

0 -1
1 0

14. Jul 2, 2013

Understood