Free action by cyclic group. Let $G$ be a group acting on a set $X$. If $g\in G$ has no fixed points, prove or disprove the cyclic group $\left \langle g  \right \rangle$ acts freely on $X$. 
edit: Can also assume $g$ has finite order.
 A: You have edited your question to only talk about finite groups. So I will edit my hint, but keep the same flavour:
Hint: The action of a group $G$ on an object/set $X$ corresponds to a homomorphism $G\rightarrow\operatorname{Aut}(X)$. If this action is free then this map is injective (these are not equivalent conditions though). So, take a group $G$ which acts on an object $X$ where $|G|>|\operatorname{Aut}(X)|$...
In the previous version of the hint, I was suggesting taking $G$ to be infinite. This works for precisely the same reason, but is slightly easier to "see".
Counter-examples:
For a concrete counter-example in the infinite case, let $\mathbb{Z}$ act on a square by rotation. Then the element $1\in\mathbb{Z}$ acts without fixed points. What can you say about the action of the element $4\in\langle 1\rangle=\mathbb{Z}$?
This counter-example ports over to the finite case by picking some natural number $n$ such that $\mathbb{Z}_n$ acts non-trivially by rotations on the square, and such that $4\neq1\pmod n$. Why?
A: Let $G$ be $\mathbb Z/4\mathbb Z$, and $X=\{a,b\}$.
Let $G$ act on $X$ by $1\cdot a = b$ and $1\cdot b=a$ (the whole action is determined by that of $1$).
Hence $1$ has no fixpoint, although $2$ acts as the identity, i.e. $\langle 1\rangle$ doesn't act freely.
