# Given a group is abelian, o(x) and o(y) are finite, show o(x*y) is finite.

• Feb 19th 2011, 11:48 AM
Relmiw
Given a group is abelian, o(x) and o(y) are finite, show o(x*y) is finite.
Does anyone have any ideas on this one that could help get me started?
Here is about as far as I've come so far.

we know:
there are integers m, n such that x^m = e and y^n = e
x*y = y*x

we want to show:
there is an integer j where (x*y)^j = e

the question also wants me to show that that o(xy) divides o(x)*o(y). I'm thinking that the fact that i'm going to be able to show:
there is an integer a where j * a = m * n
is going to be helpful with the first question, but i haven't figured out how yet
• Feb 19th 2011, 12:00 PM
Ackbeet
Try j = lcm(o(x),o(y)).
• Feb 19th 2011, 01:05 PM
Relmiw
Quote:

Originally Posted by Ackbeet
Try j = lcm(o(x),o(y)).

The section in my book does make mention of gcd, relatively prime and the euclidean algorithm, where for integers m and n, m = qn + r.
We did use lcm in class once too.
I've tried to work with the least common multiple of the order of x and the order of y, but i can't see how that relates to x^m * y^m.

Thanks for your help so far
• Feb 19th 2011, 01:14 PM
Ackbeet
Well, because

$\displaystyle o(x)|j:=\text{lcm}(o(x),o(y))$ and $\displaystyle o(y)|j$, you have

$\displaystyle j=m\,o(x),$ and $\displaystyle j=n\,o(y),$ for some integers $\displaystyle m,n.$

Then, because the group is abelian, can't you say that

$\displaystyle (xy)^{j}=x^{j}y^{j}=(x^{o(x)})^{m}(y^{o(y)})^{n}?$

What happens next?
• Feb 19th 2011, 02:12 PM
Relmiw
If the group wasn't abelian, would (xy)^j then become y^j * x^j?

Everything else you've said makes sense
$\displaystyle (x^(o(x)))^m * (y^(o(y)))^n ==> e^m * e^n ==> e$

So (xy)^j ==> e and the order of x*y is lcm(o(x), o(y)), and thus finite.
Thank you.
• Feb 19th 2011, 02:41 PM
Ackbeet
Quote:

Originally Posted by Relmiw
If the group wasn't abelian, would (xy)^j then become y^j * x^j?

No, it wouldn't look like that. I suppose you'd have to write out this:

$\displaystyle (xy)^{j}=\underbrace{(xy)(xy)\dots(xy)}_{j\;\text{ times}}.$

Unless $\displaystyle x$ and $\displaystyle y$ happen to commute, you can't slip x's past y's unless you know the multiplication table for the group.

Quote:

Everything else you've said makes sense
$\displaystyle (x^{o(x)})^m * (y^{o(y)})^n ==> e^m * e^n ==> e$

So (xy)^j ==> e and the order of x*y is lcm(o(x), o(y)), and thus finite.
Thank you.
You're welcome! Have a good one.

Incidentally, on this forum, if you see a really cool bit of LaTeX code that you want to copy, double-click it, and the source code will pop up. Great time-saver!
• Feb 19th 2011, 03:27 PM
Relmiw
Ah, ok. I see what you are saying about (xy) * (xy) * (xy), ...

I was just working at showing o(xy) divides o(x)*o(y) when i started thinking about just letting j = o(x) * o(y).
I don't think there is a need to use the least common multiple. The question doesn't ask for the order of (xy), only that it is finite.
It would certainly make the second part of the question easier if it were true that j = o(x) * o(y)

I wasn't able to get the LaTeX code to copy like you mentioned. Is that the mathematical text? I double click but nothing happens.
• Feb 19th 2011, 03:29 PM
Ackbeet
Either way works. The least common multiple is actually the order of xy, whereas the product o(x)*o(y) might not be (depending on whether o(x) and o(y) are relatively prime or not). Showing that o(xy) is finite is a bit easier than showing that o(xy) = lcm(o(x),o(y)).

Cheers.