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Let $G$ be a group of order $99$ and $X$ a set with $17$ elements. Show that an action of $G$ on $X$ must have at least $6$ fixed points.

I have used the orbit-stabilizer theorem to show that the only possible orbit sizes are $1,3,9$, or $11.$ I have tried applying Burnside's Lemma to get information on the number of orbits but I haven't found anything useful.

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    $\begingroup$ You could have one orbit of length 11 and two of length 3, which would give you no fixed points. This could occur for example when $G = C_{11} \times C_3 \times C_3$. Have you missed out any hypotheses? $\endgroup$
    – Derek Holt
    Nov 8 '20 at 17:51
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    $\begingroup$ @DerekHolt So for this G, there can be no fixed points? I thought the question might be missing a hypothesis or I was misunderstanding something fundamental as I have been stuck all weekend. Thanks for the help. $\endgroup$ Nov 8 '20 at 17:58
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    $\begingroup$ Yes that's right. You have not even assumed that the action is faithful, so you could even have something like 5 orbits of length 3 and two fixed points. $\endgroup$
    – Derek Holt
    Nov 8 '20 at 17:59
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    $\begingroup$ To make Derek counterexample more clear: Consider $G=C_{11} \times C_3 \times C_3$ acting on $A= \{ x_{1},..., x_{11}, y_1,y_2,y_3, z_1,z_2,z_3\}$, where $C_{11}$ permutes the $x'$s ($C_3,C_3$ act trivially here) and $C_3$ permute the $y's$ and $z's$. This action has no fixed point. $\endgroup$
    – N. S.
    Nov 8 '20 at 17:59
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    $\begingroup$ @N.S Ok I understand now. Thanks for the help. Do I close the question? $\endgroup$ Nov 8 '20 at 18:02
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OK, I will turn my comment into an answer.

Suppose that $G = C_{11} \times C_3 \times C_3 = \langle x,y,z \rangle$, where $x,y,z$ have orders $11$, $3$, $3$, respectively.

Then we can define an action of $G$ with no fixed points on a set $X = \{1,2,3, \ldots,17\}$, by letting $x,y,z$ act as the cycles $(1,2,3,\ldots,11)$, $(12,13,14)$, and $(15,16,17)$, respectively (where they all fix all points not in that cycle).

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