# Thread: Algebra: abelian finite group

1. ## Algebra: abelian finite group

12 i) Let A be an abelian finite group, and let a,b be elements in A.
Show that the order of ab is a divisor of the lcm( (ord(a), ord(b) ).

Okay, I tried:
It follows by definition that all elements in A have finite order.
lcm( (ord(a), ord(b) ) = x, so x divides ord(a) and x divides ord(b).

I have to show that the order of ab is a divisor of x.
My first thought:
I have to show that "the order of ab" is a divisor of "x".
Since "x" divides both ord(a) and ord(b) it follows:
ord(a) / x / order of ab = ord(a) / (x*order of ab)
ord(b) / x / order of ab = ord(b) / (x*order of ab)
Well, I guess I'm going the wrong way because it doesn't make sense.
Can someone help me get further?
Tnx. & sorry if I got the topic ("pre-algebra and algebra" wrong!)

2. Let's try some examples:

if $\displaystyle a^5=1$ and $\displaystyle b^7=1$, then clearly $\displaystyle (ab)^{35}=1$.

if $\displaystyle a^4=1$ and $\displaystyle b^6=1$, then clearly $\displaystyle (ab)^{12}=1$.

You're on the right track, it's just that ord(a) | lcm(ord(a),ord(b)), not the other way around.

Also, I think it should be ord(ab) = lcm(ord(a),ord(b)). Do you have an example where ord(ab) < lcm(ord(a),ord(b))?

3. Originally Posted by qmech
You're on the right track, it's just that ord(a) | lcm(ord(a),ord(b)), not the other way around.

Also, I think it should be ord(ab) = lcm(ord(a),ord(b)). Do you have an example where ord(ab) < lcm(ord(a),ord(b))?
Hm, I don't really get what you're trying to show me here. Like; why do you think ord(ab) = lcm(ord(a),ord(b)) ? What's that got to do with it?

Ill try.
ord(ab) = lcm(ord(a),ord(b))
I have to show that ord(ab) | lcm((ord(a), ord(b))
And yeah, if ord(ab) is lcm((ord(a), ord(b)), it's easypeasy, because x divides x, so that's the proof that ord(ab) | lcm( (ord(a), ord(b)).
But then, how to show that ord(ab) = lcm(ord(a), ord(b)) ?

4. I think I can use Lagrange's theorem, but how? I have no idea.