Theorem. Let $G$ be a finite group of isometries of the plane. There is a point $p$ in the plane such that every element $g$ in $G$ fixes $p$. (That is to say: $g(p) = p$, $\forall g \in G$.)
Isn't the set $ \{e, t_v, (t_v)^{-1} \}$, where $t_v$ is a translation by a non-zero vector $v$, a finite group of the isometries of the plane? And doesn't $t_v$ not preserve any points? I believe this is a counter-example to the above statement, or else I am misunderstanding the statement of the theorem. Any help clearing this up would be greatly appreciated!
Your set isn't a group: it's not closed under composition. In particular, $t_v\circ t_v=t_{2v}$ is not in your set. If you close it under composition, you would need to add $t_{nv}$ for every $n\in\mathbb{Z}$, which would make it infinite.