I am studying permutation groups. I understand that a symmetric group $S_n$ is the set of all permutations on $n$ symbols. So, for $S_4$ there will be 24 elements. i can write this like different arrangements of $1,2,3,4$ so $(1 2 3 4),(1 2 4 3)$ .... there will be $24$ such arrangements. Now i have two issues :-

1) Is there a decent way of writing these cycles ? with increasing $n$ it becomes very complex to compute each element

2) for $S_3$ the book says that the elements will be $(123),(12),(13),(23),(132),I$. But i see it like $(123),(132),(321), .... $ so, is there a standard way of creating the arrangement given by the book i.e directly, instead of writing first in the second way and then expressing as product of disjoint cycles ?

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    $\begingroup$ $(132) = (213)$ - cyclically permuting a cycle doesn't change the cycle. So just start with $(1 .. )$ and go from there. Dummit&Foote has a nice little algorithm. $\endgroup$ – Prahlad Vaidyanathan Sep 27 '13 at 14:15
  • $\begingroup$ sorry, please ignore this mistake in the example i gave. I understand this point. BTW is it a normal practice to just memorize these two symmetric sets ? $S_3$ and $S_4$ (to some extent) $\endgroup$ – Aman Mittal Sep 27 '13 at 14:18
  • $\begingroup$ No - don't memorize! Just find a way to build elements of $S_n$ (Hint: Find different partitions of $n$) $\endgroup$ – Prahlad Vaidyanathan Sep 27 '13 at 14:20
  • $\begingroup$ Thanks !! And about the second issue ? $\endgroup$ – Aman Mittal Sep 27 '13 at 14:21
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    $\begingroup$ I think there might be some confusion arising here between a permutation expressed as a cycle (or product of cycles), where $(123)$ is the permutation sending 1 to 2, 2 to 3, and 3 to 1, and a permutation expressed as the image of $1,2,\dots,n$, where $(123)$ is the identity permutation. $\endgroup$ – aPaulT Sep 27 '13 at 14:57

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