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.
Best Answer
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).