This one’s real simple, but for some reason I just can’t figure it out. Should be a quick answer, so help me out, please?
Let G be an abelian group, and let a and b be two elements of finite order. Show that |a| divides |ab|.
Eh? Is |a| the order of a? As in ‘The least positive interger n such that a[sup]n[/sup]=1’?
If so, your statement looks false to me: consider the cyclic group G=<1,…,x[sup]4[/sup]>. Let a=b=x. Then |a|=|b|=4 but |ab|=2.
If by |a| you mean the order of a, then it’s false. Counter-example: Z_4 under addition, a = b = 1. |a| = |b| = 4, but |ab| = 2, and 4 does not divide 2.
On preview, Shade beat me to the exact same counter-example, but I’m gonna post this anyway. 
Fair enough. Guess that explains why I can’t prove it.
What is the standard notation for the order of a? Every book I’ve seen uses |a|.
That seems pretty darned standard. The only other notation I’ve ever even seen is o(a), and I only see that rarely.
I didn’t mean to imply it wasn’t, I just wanted to make sure I wasn’t making a stupid mistake. Firstly, I’ve forgotten a lot of notation recently, and secondly, the fact that your statement appeared false made me think maybe you defined order some other way I hadn’t learnt.
Yeah, it’s a bugger trying to prove false things, isn’t it? Fortunatley it didn’t take me too long to think to check.
WOOT! I beat someone to the exact same counterexample, I rock! 
Of course, this raises an immediate question: what is the most general setting where |a| divides |ab| for a, b in an abelian group G?
Based on the counterexample, I’d guess that it might work if |a||b| < |G|, but I’m not running at 100% today, so I’m not feeling good about that.
No, the only way you can be certain that |a| divides |ab| is that |a| and |b| are relatively prime. Easy to find counter-examples to any statement of the sort |a||b| < |G| by looking at Z/nZ for appropriate n.
The most general statement I know is that |a||b|/gcd(|a|,|b|) divides |ab| which divides lcm(|a|,|b|) and both bounds can be attained. Of course, when gcd(|a|,|b|) = 1, then lcm(|a|,|b|) = |a||b| and the two bounds are equal. (gcd = greatest common divisor, lcm = least common multiple).
The conjecture
|a| divides |ab| fails whenever b = a[sup]-1[/sup] and a is not the identity element.