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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.

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By Sylow's theorem the number of subgroups of order $11^2$ in $G$ is $1$. Call that group $S$ and observe that a $3$-Sylow-subgroup $P=\langle x\rangle$ acts on $S$. If $P$ acts trivially there is obviously an element of order $33$ (just take $x^3*y$ where $y$ is of order $11$).

So let us assume that $x$ does not act trivially. Now either $S\cong C_{11}^2$ or $S\cong C_{121}$.

In the later case the automorphism group of $S$ has order $110$ which is not divided by $3$ and hence does not allow a non-trivial action of $P$ on $S$.

In the former case the automorphism group is isomorphic to $\mathrm{GL}_2(11)$ of order $120*110$ which is divisible by $3$ exactly once. Hence $x$ acts as an element of order $3$ on $S$ and so $x^3$ acts trivially. Again take $x^3*y$ for any $y$ of order $11$ to obtain an element of order $33$.

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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.

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  • $\begingroup$ $\;1098\neq 1089\;$ $\endgroup$ – Timbuc Jul 3 '14 at 10:09
  • $\begingroup$ @D.N. I don't understand how $\operatorname{Aut}(P)\simeq\mathbb{Z}_{10}$. The order of $P$ is $11^{2} = 121$, so the automorphism group should, I think, have order either $120$ or $13200 = \mid\operatorname{GL}(2,11)\mid$ (in fact, both occur), but neither is divisible by $9$, so I guess the argument goes through anyway. $\endgroup$ – James Jul 3 '14 at 10:44
  • $\begingroup$ @james Yes you are correct. My mistake. Thank you for correcting that mistake. $\endgroup$ – D. N. Jul 3 '14 at 10:54

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