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We claim that

Any group action of a group of order $7$ on a set $S$ of order $2014$ has at least 3 fixed points.

Let $O_{x_i}$ denote the Orbit of $x_i\in S$ and $\text{Stab}_{x_i}$ the stabilizer of $x_i\in S$.

Then by the class equation \begin{align} |S|=&\sum_{i}|O_{x_i}|\\ =&\sum_{i}\frac{|G|}{|\text{Stab}_{x_i}|}\\ =&\sum_{i}\frac{7}{|\text{Stab}_{x_i}|}. \end{align} Since $7$ is prime, $|Stab_{x_i}|=1$ or $7$. Hence, $$|S|=\sum_{i}\frac{7}{|\text{Stab}_{x_i}|}=\sum_i 7+\sum_i 1.$$

Therefore, we can conclude that $|S|$ is equal to some sum of $1's$ and $7's$.

However, from here I'm not sure how to show that there exists at least $3$ fixed points of $X$. Does anyone have suggestions on how to get to the conclusion?

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Hint:

Let $N_1$ be the number of $1$s. The last equation implies that $$N_1\equiv \lvert S\rvert\mod 7.$$

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  • $\begingroup$ From that congruence we have that $N_1\equiv 2014\mod 7\iff N_1\equiv 5\mod 7$. Does this imply that $N_1=5$? Furthermore, what do the number of $1s$ tell us about the fixed points of the group action? $\endgroup$ – user225477 Dec 4 '16 at 14:02
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    $\begingroup$ As $N_1\ge 0$, it implies $N_1\ge 5$. $N_1$ is the number of fixed points. I don't see why it is asserted there are at least $3$ points, as they might have said $5$. $\endgroup$ – Bernard Dec 4 '16 at 14:16

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