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Suppose that the order of $g$ is $6.$ Determine the orders of $g^2,g^3,g^4$ and $g^5$.

For the first two I have that $(g^2)^3=g^6 = e$ and $(g^3)^2 = g^6 = e$ so the order of $g^2$ is $3$ and the order of $g^3$ is $2$.

For the two latter ones I tried to follow similar reasoning trying to get $g^6$ to appear somehow so that I can get $e$ to the power of something which is just $e$. What I have is that since $\operatorname{lcm}(4,6)=12$ $$(g^4)^{12}=g^{48}=(g^6)^{8} = e$$ and likewise since $\operatorname{lcm}(5,6)=30$ $$(g^5)^{30}=g^{150}=(g^6)^{25}=e$$ is there some kind of deeper connection here with the least common multiple and the order or am I just confusing something.

Shaun
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Zwertp
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  • @BillDubuque There. I deleted my answer. Now there is an entry on the site marked as duplicate, but with two answers worst than mine. Will this entry be deleted according to the new policies? – jjagmath Nov 08 '21 at 17:58
  • @jjagmath If you have something novel to add then the correct thing to do is to post an answer in one of the dupes (there are many on this topic, e.g. see the "Linked" questions there). As for poor answers, you can either post comments that help to improve them, or downvote or flag them (e.g. as low quality or not an answer). – Bill Dubuque Nov 08 '21 at 18:01

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