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Consider the permutation group $S_6$ and let $H\subseteq S_6$ be a subgroup of $9$ elements

  1. It is abelian but not cyclic

  2. It is cyclic

  3. It is not abelian

  4. If H is abelian then it is cyclic.

Wel I know a general result that group of order $p^2$ is abelian where $p$ is a prime number, hence $H$ is abelian.but I dont know whether $H$ is cyclic or not, is it?thank you for the help.

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    Why are all your questions in 4 true or false?2012-07-08

1 Answers 1

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Hint. If you write an element of $S_6$ as a product of disjoint cycles, the order of the element is the least common multiple of the length of the cycles. Is there a way to get an element of order $9$?

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    Dear sir, There is no way to get an$9$order element in $S_6$.so Only 1 is correct2012-07-09