If $f$ is isometry and triple composition of $f$ is identity, then $f$ has a fixed point.

Question

Let $f$ be isometry. If $f \circ f \circ f$ is identity, prove that $f$ has a fixed point.

I know that $f$, since it is an isometry, is a composition of at most three reflections, but I have no idea how to use it here.

There is another theorem I came accross that says:

If $f$ does not leave any point fixed, then $f$ is either a translation, or the composite of a translation and a rotation, or the composite of a translation, a rotation, and a reflection through a line.

I am not sure how to prove either of those, and whether the latter would maybe be the answer to the former.

Answer 1

I suppose that you are talking about isometries of $\mathbb{R}^2$ with respect to the usual metric. Am I right?

If so, either $f$ preserves the orientation or it reverses it. If it reverses it, so does $f\circ f\circ f$, but this cannot be, since $f\circ f\circ f$ is the identity. Therefore, $f$ preserves the orientation. In other words, it is either a rotation or a translation. If it is a rotation, of course that it has a fixed point. And if it is a non-trivial translation, then $f\circ f\circ f$ is also a non-trivial translation, and therefore not it is the identity.

Answer 2

Let $f(P)=Q$, $f(Q)=R$, so $f(R)=P$, and consider the circumcenter of the triangle $PQR$.