# 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.

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.

• 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? Nov 8 '20 at 17:51
• @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. Nov 8 '20 at 17:58
• 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. Nov 8 '20 at 17:59
• 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. Nov 8 '20 at 17:59
• @N.S Ok I understand now. Thanks for the help. Do I close the question? Nov 8 '20 at 18:02

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).