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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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    Time out, Patience. Digest a little.2012-07-08
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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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    Yah, there for example $(123)(456)$ is of order $9$ element. and yah, the general result I know that is abelian group.but how do I know that order $9$ element will be in $H$?2012-07-08
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    @Patience: Really? The least common multiple of $3$ and $3$ is $9$? Do please think about things more than 10 seconds before posting! The order is not the **product** of the lengths of the cycles, it is the **least common multiple**. Did you even bother to take that element and *check* if the order was $9$? What happens if you raise it to the 3rd power?2012-07-08
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    oops I am sorry, please pardon me dear sir.2012-07-08
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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