Product of elements with finite order in a group need not have finite order...

Show by example that the product of elements with finite order in a group need not have finite order. What if the group is abelian?

I think it's something like g^k = g^s

Then (g^k)(g^-k) = (g^s)(g^-k) = g^s-k = e

Therefore, there is a smallest integer m = s-k s.t g^m = e.??

You need to stick with (infinite) non-abelian groups for such an example.
(Otherwise, since ab = ba, we find that |ab| is still finite, bounded above by lcm(|a|, |b|).
As David pointed out below, we can't have equality.)

A quick example is G = . (Note I say nothing about xy = yx, etc.)
Then, xy has infinite order, because I gave no way to simplify products of the form xy or yx.

For instance, (xy)^2 = (xy)(xy) = xy, and that's all we can do to the element to simplify it.
(If (xy)^2 = e, then this forces xy = yx, which is a contradiction of how x and y are defined.)

Similarly, for any positive integer n, (xy)^n is a distinct element in G as well not equal to e.

I hope this helps!

kb it is NOT true that if G is abelian |ab| = lcm(|a|,|b|).

for example, let G = C4 = = {e,x,x^2,x^3}

then |x| = |x^3| = 3, and lcm(3,3) = 3, but:

|(x)(x^3)| = |e| = 1.

the result you have given hold in general only when the orders are co-prime. at best, we can usually say the order is bounded by the lcm.

everything else you said is true.