# Subgroup of order $9$ of $S_6$

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. – Gerry Myerson Jul 8 '12 at 23:42
Why are all your questions in 4 true or false? – William Jul 8 '12 at 23:44

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 $9$?
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 $H$? – La Belle Noiseuse Jul 8 '12 at 23:50
@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? – Arturo Magidin Jul 8 '12 at 23:51
Dear sir, There is no way to get an 9 order element in $S_6$.so Only 1 is correct – La Belle Noiseuse Jul 9 '12 at 7:29