Properties of subgroups of $S_6$

Question

1. I'm trying to prove that a subgroup of $S_6$ of order $9$ isn't cyclic. I used the fact that if it was cyclic than it had an element of order $9$ but there is no element like this in $S_6$. Is it correct?

2. I'm trying to prove that a subgroup of $S_6$ of order $16$ is not abelian. I know it's a $p$-group but not sure how to prove it.

Answer 1

Writing a permutation $\sigma \in S_n$ as a product of $n$ disjoint cicles
$$\sigma = \tau_1 \cdots \tau_k,$$ it is clear that the order of $\sigma$ is the least common multiple of the orders of the $\tau_i$.

From this, it is straightforward to check that the maximun possible order for a permutation in $S_6$ is $6$, so your first claim follows.

By the way, the numerical function $g(n)$ that associates to $n$ the largest order of an element of $S_n$ is called Landau's function, and its values are tabulated here.

Regarding your second claim, a simple computation shows that the subgroup $H$ of $S_6$ generated by $(1 \, 2 \, 3 \,4)$ and $(1 \, 3)$ is isomorphic to the dihedral group $D_{8}$ of order $8$. Then the subgroup $G$ generated by $H$ and $(5 \, 6)$ has order $16$ and is isomorphic to $D_8 \times \mathbb{Z}_2$, in particular it is not abelian.

Since all the subgroup of order $16$ are conjugate in $S_6$ (they are Sylow subgroups) we are done.