TITLE: Existence of an element in a group of certain order if an element of other order exists
QUESTION [6 upvotes]: Show that if a group $G$ of order $1089=3^2\cdot 11^2$ contains an element of order $9$ then it also contains an element of order $33$.
I tried to see what would Sylow theorems tell for this problem but i could not conclude anything about the order$(33)$ of an element in the group.
How to approach this problem? Any hint would be appreciated. Thank you.

REPLY [3 votes]: A more general result would be that any group of order 1089 contain an element of order 33.
Let $P$ be the Sylow 11-subgroup of $G$. Then, $P$ is normal in $G$. Suppose $P = C_G(P)$. Then, by N/C lemma (Rotman, An introduction to Group theory, Theorem 7.1 (i)), $G/P$ is isomorphic to a subgroup of the group of $Aut(P)$, which is of order 110 or 13200. Since $|G/P|=9$, we have 9 divides 110 or 13200, which is a contradiction. So $P \neq C_G(P)$. Let $y \in C_G(P) \setminus P$ and $x \in P$ such that order of $x$ is 11, such an element $x$ always exist. Since $P$ is normal in $G$, 3 divides order of $y$ and so there exist an integer $m$ such that order of $x^m$ is 3  and $x^my$ is an element of order 33.