3
$\begingroup$

Given $G$ is a group, and $ab=ba$ for $a,b\in G$, we know $o(a)=12,o(b)=18$, in order to calculate $o(ab),o(ab^2),o(a^2b^3)$.

Do we need to assume group is a cyclic group in order to use formula $o(g^k)\mathbb{gcd}(n,k)=n$ for $g$ is a generator and $n=|G|$ ?

Or is there some other way to calculate ? Because in the method above, we don't use value of $o(a),o(b)$ at all.

  • 0
    You would need to know the order of $g$ as an element of $G.$2012-10-01
  • 1
    This is impossible without more information. Knowing $o(a)$ and $o(b)$ is not enough to work out $o(ab)$.2012-10-01
  • 0
    @ChrisEagle could we say the most far position we can touch is a formula of $o(ab)=o(g^{k_1+k_2})=\frac{n}{gcd(n, k_1+k_2)}$ if assume group is cyclic, but what if general group. is there even a formula ?2012-10-01
  • 0
    If $a$ is of order $k$, so is $a^{-1}$, and o(a^{-1})=k$.2012-10-01
  • 2
    The best you can say is that $o(ab) \mid \mathrm{lcm}(o(a),o(b))$. If $\gcd(o(a),o(b))=1$ then you have equality.2012-10-01
  • 0
    More precisely, it holds for two elements in a group if they commute with each other.2012-10-01
  • 1
    With respect to the main question, you have to make use of the fact that the two commute. Cyclicity of the group is not a given so cannot be a solving device. But I agree with Chris Eagle, it seems impossible without more information as to the nature of the group.2012-10-01

1 Answers 1