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I'm taking a introductory class of Abstract Algebra, and is having hard time on applying the definition of a cyclic group onto a group of permutations. Because so far I've only learned how to find cyclic subgroups for groups of multiplication and addition, for which I find either the values of the generator to nth power or n multiply to the generator, such that n belongs to Z. So how is such concept of "n copies of the generator"(the phrase from my teacher) applied or adapted to a group of permutations?

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  • $\begingroup$ It is good form to accept an answer which people put effort in. Let me know if any of it is still unclear. $\endgroup$ – kodlu Feb 21 '18 at 23:17
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Composition is the group operation. So $$\sigma=(12)(3)$$ in $S_3$ has order 2, e.g.

In general the order of $\sigma$ is the minimum $n$ such that $$ \underbrace{\sigma(\sigma(\cdots \sigma(}_{n~times}\cdot)\cdots))=e $$ where $e$ is the identity permutation. To make it clearer, write the permutation in list notation. So $$\sigma=(213),\sigma^2=(123)=e$$ and the cyclic subgroup is $\{\sigma,e\}.$

Why dont you compute the cyclic group generated by $(134)(25)$ in $S_5$ as an exercise. Hint: It has order 10.

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  • $\begingroup$ So is it just keeping applying the permutation of the generator to itself? $\endgroup$ – user917099 Feb 20 '18 at 21:15
  • $\begingroup$ see my edit at the bottom of the answer. $\endgroup$ – kodlu Feb 20 '18 at 22:05
  • $\begingroup$ Honestly I haven't learned the list notation. So what you put up there seems to be a cycle to me. But many thanks because your first line fully reminded me the basic fact so I can think further through it. $\endgroup$ – user917099 Feb 21 '18 at 23:38
  • $\begingroup$ OK, the original was cycle notation so let's take $(134)(25)$ this can also be written as $\left(\begin{array}{ccccc} 1&2&3&4&5\\3&5&4&1&2\end{array}\right)$ where the position in the first row goes to the position in the second row. List notation just omits the first row since it's always in the order 12345 and writes the second row. So it becomes $(35412)$. Hope that helps. $\endgroup$ – kodlu Feb 22 '18 at 0:12
  • $\begingroup$ Doesn't the cycle (134)(25) have order 6? i.e. LCM(3,2) $\endgroup$ – SH7890 Nov 1 '18 at 23:32

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