# Thread: Prove G is abelian

1. ## Prove G is abelian

When I get time I take the time to study a little group theory. I am a flyspeck compared to PH and NCA, but here is one I would like some input to see if I have the idea. I will then post my workings.

"Let G be a group which has the following properties:

1: G has no element of order 2

2: $(xy)^{2}=(yx)^{2}, \forall{(x,y)} \in{G}$

By the way, what are the typings for "for all" and "element of"?.

I tried the ones on my list but they wouldn't work.

2. Originally Posted by galactus
By the way, what are the typings for "for all" and "element of"?.

I tried the ones on my list but they wouldn't work.
For all: $\forall$ = $$\forall$$

Element of: $\in$ = $$\in$$

Is that what you were looking for?

-Dan

3. Originally Posted by galactus
When I get time I take the time to study a little group theory. I am a flyspeck compared to PH and NCA, but here is one I would like some input to see if I have the idea. I will then post my workings.

"Let G be a group which has the following properties:

1: G has no element of order 2

2: $(xy)^{2}-(yx)^{2}, \text{for all x,y element of G}$
I think you mean $(xy)^2 = (yx)^2$.
See this.

4. Yes, of course that is what I mean. A stupid typo...duh.

I tried the 'forall and it didn't work. I also tried 'in' and got the same. Let me try this again.

${\forall}, \;\ {\in}$

Yep now it's OK. I must've have done something wrong.

5. I tried to give you a thanks PH, but your button disappears when I try.

What next?.

Anyway, you had an old one I asked about and had forgotten from almost 2 years ago. Cool.

If I may, here is what I done. I think it is similar to what is on the other posting, but I have to admit I didn't look at it in depth.

It is getting late and I do not feel like typing that much at the moment. I will type mine tomorrow. Okey doke.

6. Here is my workings. I hope it's OK. I will post while I have time this morning. Today is the 4th. Things to do

For $x, \;\ y \;\ {\in} \;\ G$,

$x^{2}y=((xy^{-1})y)^{2}y$

$=(y(xy^{-1}))^{2}y$, by part 2.

$=(yxy^{-1})(yxy^{-1})y$

That's

$x^{2}y=yx^{2}$

Next, we have:

$x^{-1}y^{-1}x=x(x^{-1})^{2}y^{-1}x$

$=xy^{-1}(x^{-1})^{2}x$, by part 1.

that's

$x^{-1}y^{-1}x=xy^{-1}x^{-1}$

Similarly, we get:

$y^{-1}x^{-1}y=yx^{-1}y^{-1}$

Then we get:

$(xyx^{-1}y^{-1})^{2}=xy(x^{-1}y^{-1}x)yx^{-1}y^{-1}$

$=xy(xy^{-1}x^{-1})yx^{-1}y^{-1}$

$=xyx(y^{-1}x^{-1}y)x^{-1}y^{-1}$

$=xyx(yx^{-1}y^{-1})x^{-1}y^{-1}$

$=(xy)^{2}(x^{-1}y^{-1})^{2}$

$=(xy)^{2}(yx)^{-2}$

$=(yx)^{2}(yx)^{-2}$

$=1$

${\therefore},$ since G has no elements of order 2, we get

$xyx^{-1}y^{1}=1$

And $xy=yx$, proving that G is abelian.

WHEW!!. There could easily be a typo in all those x's and y's.

7. That looks good except:
Originally Posted by galactus
Next, we have:

$x^{-1}y^{-1}x=x(x^{-1})^{2}y^{-1}x$

$=xy^{-1}(x^{-1})^{2}x$, by part 1.
It has nothing to do with part 1, it rather has to do with the line you previously proved.