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For $n\ge1$, let $S_n$ denote the group of all permutations on $n$ symbols.

Which of the following statements is true?

A. $S_3$ has an element of order 4

B. $S_4$ has an element of order 6

C. $S_4$ has an element of order 5

D. $S_5$ has an element of order 6

How to approach this question? Any hints?

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closed as off-topic by Derek Holt, mechanodroid, Namaste, jvdhooft, kingW3 Oct 12 '17 at 13:28

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  • 4
    $\begingroup$ hint: think of the elements as a multiplication of disjoint cycles. $\endgroup$ – drhab Oct 12 '17 at 8:18
  • $\begingroup$ @drhab there is no element of order $6$ in $S_4$ and order $6$ element in $S_5$ is in form $(a_1a_2)(a_3a_4a_5)$. Thanks $\endgroup$ – Vincent Oct 12 '17 at 9:01
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Assertion A. is false, because $\#S_3=6$ and $4\nmid6$. For a similar reason, C. is false.

For B. and D., the approach mentioned in the comments (think of the elements as a multiplication of disjoint cycles) is the way to go.

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  • $\begingroup$ Thanks! can you suggest online resources from where such question can be practiced? $\endgroup$ – Vincent Oct 12 '17 at 9:03
  • $\begingroup$ @VKSingh I'm not aware of any. $\endgroup$ – José Carlos Santos Oct 12 '17 at 9:04

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