Let X be a set on which the cyclic group with 3 elements acts. If $|X| = 5$, show that it must have at least 2 fixed points. I do not understand how the cyclic group acts on a set. As I know it's a group generated by a single element, but I do not understand how its group action is defined on a set. If you could help me with the action I can try to approach the problem on my own.
 A: Denote by $G= \{ 1, g, g^2\}$ the group with three elements, which is generated by $g$.
Suppose $x \in X$ is not a fixed point. This means that
$gx \neq x$. Moreover
$$g(g^2 x) = g^3 x = 1x = x$$
This means that $g^2x \neq x$.
In a similar way we can prove that $g^2x \neq gx$. In other words, we have an orbit with three distinct elements.
$$Gx = \{ x, gx, g^2x\}$$ has three elements which are not-fixed.
Since orbits are disjoint each other, this means that the set of non-fixed points has cardinality divisible by $3$ (it's a disjoint union of orbits with three elements).
Now, the set of non fixed points can have 0 elements or 3 elements.
Taking the complement, since $|X|=5$, the set of fixed points can have 5 elements or 2 elements.
In both cases, there are at least two fixed points.
A: Let $G$ be a group and $A$ be a non-empty set. Then, a group action is a function $G \times A \to A$ such that

*

*$e \cdot a = a, \forall a \in A$.

*$g_1 \cdot(g_2 \cdot a) = (g_1g_2)\cdot a, \forall g_1,g_2 \in G \text{ and } a \in A$.

This means, any function $G\times A \to A$ satisfying these two conditions is a group action of $G$ on a set $A$. So, the question does not ask for a spesific action, but asks for all possible actions.
Also, we say a point $a \in A$ is fixed by $G$ if $g \cdot a = a, \forall g \in G$.
HINT: Orbit Stabilizer Theorem implies that for a finite group $G$ acting on a set $A$, we have $|G_a|\cdot|O_a| = |G|$ for any element $a \in A$, where $O_a$ is the orbit of $a$ and $G_a$ is the stabilizer of $a$. And a point $a$ is fixed if $|G_a| = |G|$ (or equivalently $|O_a| = 1$). Considering orbits are partitions of set $A$, can you prove your statement?
A: As $3$ is prime, from the orbit-stabilizer theorem follows that the orbits must have size either $1$ or $3$. Therefore, the only allowed orbit equations are $5=1+1+1+1+1$ and $5=1+1+3$. Therefore, whatever the actual action is, there are at least two elements $x\in X$ such that $g\cdot x=x$ for every $g\in G$.
