What Is the Order of \(a*b\) Given the Orders of \(a\) and \(b\)?

[Ancient_Nomad]

Hello,

I was thinking about how to predict the order of an element a*b if the orders of a and b are known where a and b are elements of some group.

One textbook I have gives the result (without proof) that if a and b commute and their orders are relatively prime, then order(a*b) = order(a)*order(b). But I have been unable to prove this result. Can someone help me out with this and explain if there is any result if a and b are not relatively prime.

Also, to me it seems logical that nothing specific can be said about the order of a*b if a and b do not commute. Can someone please tell me if I am right or correct me if not.TIA

 

[CompuChip]

Note that
(a * b)^n = a * b * a * b * a * ... * b * a * b
with n copies of a * b. Now if a and b commute, you can pull all the a's through all the b's and write it as
(a * b)^n = a^n * b^n.

From there you should be able to prove that order(a * b) = lcm(order(a), order(b)).

If they don't commute there is not much that can be said... unless you have special relations that do allow you to "commute" elements (for example, in dihedral groups, $r s = s r^{-1}$).

 

[Citan Uzuki]

From there you should be able to prove that order(a * b) = lcm(order(a), order(b))

This is not true. Let $$a=b=(\mathrm{1 2}) \in S_2$$. Then a and b commute, o(a) = o(b) = 2, so lcm(o(a), o(b)) = 2, but o(ab) = 1. Now, it is true that the order of ab divides lcm(o(a), o(b)), but it need not be equal to it.

One textbook I have gives the result (without proof) that if a and b commute and their orders are relatively prime, then order(a*b) = order(a)*order(b). But I have been unable to prove this result.

It is easy to show that if a and b commute, the order of ab divides o(a)*o(b). To show that o(a)*o(b) divides o(ab), we let n=o(ab) and note that $a^nb^n=(ab)^n=1$, so $a^n=b^{-n}$. Now, it can be shown that $o(a^n) \mid o(a)$ (hint: apply Lagrange's theorem to the subgroups generated by a^n and a), and likewise that $o(b^{-n}) \mid o(b)$. But $a^n=b^{-n}$, so $o(a^n) = o(b^{-n}) \mid o(b)$. Thus, o(a^n) divides both the order of a and the order of b. But o(a) and o(b) are coprime, so this can only happen if o(a^n) = 1, which implies that a^n must be the identity. So o(a) | n, and also since $b^{-n} = a^n = 1$, o(b) | -n | n. Finally, note that if two coprime numbers both divide n, their product divides n, thus o(a)*o(b) | n. Since we already know n | o(a)*o(b), it follows that o(ab) = n = o(a)*o(b).