This particular question was asked in my abstract algebra quiz and I was unable to solve it and I am looking for help here.
Let $H$ be a subgroup of $S_8 $ such that order of $H$ is $9$, then $H$ is :
A) cyclic
B) not abelian
C) abelian but not cyclic
D) if $H$ is abelian then it must be cyclic
$H$ cant be cyclic as then it will have an element of order $9 $. Also, by cauchy theorem, there exists an element of order $3$. Disjoint cycles always commute. But how to use that condition here?
But I am unable to think about other options, as i am not really good while analysizing permutation groups. Can you please tell how to approach these kind of problems in permutation groups?
Best Answer
Suppose $H$ is cyclic. Then there exists an element of order $9$ in $S_8$. But $S_8$ does not have any element of order $9$ because a permutation of order $9$ should contain a $9$-cycle. So (A) and (D) are not correct.
Note that for any prime $p$, group of order $p^2$ is always abelian. (Refer here). So (B) is not correct and the only correct answer is (C).