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For n ≥ 1, let $S_n$ denote the group of all permutations on n symbols. Which of the following statements is true ?

A. S3 has an element of order 4

B. S4 has an element of order 6

C. S4 has an element of order 5

D. S5 has an element of order 6.

My work: Order of group is $n!$.

A and C are false (by Lagrange theorem). Now how can i proceed next intuitively ?

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  • $\begingroup$ Have a gander at this en.wikipedia.org/wiki/Landau%27s_function $\endgroup$ – Donald Splutterwit Aug 21 '18 at 20:08
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    $\begingroup$ An unenlightening approach is to make the assumption that the problem has just one answer and then to note that if B is true, then so is D since $S_4$ is a subgroup of $S_5$. Thus, only D is true. More seriously, what do you know about permutations? Specifically, are you aware of how to determine the order of a given permutation? $\endgroup$ – Karl Kronenfeld Aug 21 '18 at 20:11
  • $\begingroup$ Ya i know order of elemet in permutation group $\endgroup$ – Cloud JR Aug 21 '18 at 20:14
  • $\begingroup$ I think landau function helps $\endgroup$ – Cloud JR Aug 21 '18 at 20:15
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    $\begingroup$ The order of an element is the least common multiple of the orders of its cycle lengths. When evaluating if B is false, try to choose the cycle lengths of a permutation whose lcm is 6. You will find that this is impossible for several numbers whose sum is 4. $\endgroup$ – Mike Earnest Aug 21 '18 at 20:19
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Question B

An element of $\mathcal S_4$ is either a $2$-cycle (of order $2$), a $3$-cycle (having order $3$), a product of two $2$-cycles (having order $2$) or a $4$-cycle (of order $4$). Hence none of the elements of $\mathcal S_4$ may have order $6$.

Question D

A product of two disjoint cycles of order $2$ and $3$ has indeed order $6$ in $\mathcal S_5$. This is the case for example of $(1 \ 2 \ 3)(4 \ 5)$.

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  • $\begingroup$ Is there any general way to look at this tyoe of question $\endgroup$ – Cloud JR Aug 21 '18 at 20:20
  • $\begingroup$ Your answer is illuminating $\endgroup$ – Cloud JR Aug 21 '18 at 20:22
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    $\begingroup$ The permutation given by disjoint (disjoint is important) cycles $\sigma_i$ is given by the $\text{l.c.d.}\{\sigma_i\}$. This is a good exercise to check. $\endgroup$ – mxnoqwerty Aug 21 '18 at 20:25
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    $\begingroup$ @CloudJR Any permutation can be written as a product of disjoint cycles. So looking at disjoint cycles is often useful! Also because cycles ar simple to study. $\endgroup$ – user532133 Aug 21 '18 at 20:27
  • $\begingroup$ L.c.d? Liquid crystal display? lol...do you mean l.c.m.? $\endgroup$ – Cloud JR Aug 21 '18 at 20:27

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