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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.

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    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

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Using commutativity, the following inelegant method seems to work for the first one (as an example of how to proceed):

For $ab$ we know that $(ab)^{36}=a^{36}b^{36}=1$ so $o(ab)|36$

Suppose $o(ab)=2$ then $(ab)^2=1 \implies (ab)^{12}=1\implies b^{12}=1$ and similarly for if $o(ab)=r$ for some $r|12$ (contradicts $o(b)=18$).

Suppose $o(ab)=9$ then $(ab)^9=1 \implies (ab)^{18}=1\implies a^{18}=a^6=1$ and similarly for if $o(ab)=r$ for some $r|18$ (contradicts $o(a)=12$).