Dear friends,
May i check if the answer should be false?
True or false? If a and b are elements of an abelian group, then order of (ab)= lcm (order of (a), order of (b))
Thanks,
weijing
Dear friends,
May i check if the answer should be false?
True or false? If a and b are elements of an abelian group, then order of (ab)= lcm (order of (a), order of (b))
Thanks,
weijing
ord(ab) and lcm(ord(a), ord(b)) have a divisibility relationship:
$\displaystyle (ab)^{lcm(ord(a), ord(b))} \ \overset{(abelian)}{=} \ (a)^{lcm(ord(a), ord(b))}(b)^{lcm(ord(a), ord(b))}$
$\displaystyle = (a)^{ord(a)k_1}(b)^{ord(b) k_2} = ((a)^{ord(a)})^{k_1} \ ((b)^{ord(b)})^{k_2} = (1)^{k_1}(1)^{k_2} = 1.$
so that ord(ab) always divides lcm(ord(a), ord(b).
But are they always equal? (Hint: think about inverses.)