Fundamental group of a compact surface Let $\mathbb{X}$ be a compact surface (possibly with nonempty boundary). If $\pi_1(\mathbb{X})$ has an element of order $2$, can we prove that $\mathbb{X}$ is homeomorphic to the projective plane?
Here is what I thought.

*

*It is clear that $\mathbb{X}$ is closed, otherwise $\pi_1(\mathbb{X})$ would be a free group.


*$\mathbb{X}$ must be nonorientable. If not, the first homology $H_1(\mathbb{X})=\mathbb Z^{2g}$ which contains no torsion and hence $\pi_1(\mathbb{X})$ contains no torsion.


*If $\mathbb{X}$ has genus $g$ (i.e. $g$ Mobius bands), then
$$
\pi_1(\mathbb{X})=\langle a_1,\cdots, a_g \mid \prod_{i=1}^g a_i^2=1\rangle.
$$
However, it is unclear to me how to prove that $g=1$ if $\pi_1(\mathbb{X})$ contains an order 2 element.
 A: One way to prove this is using the uniformization theorem: every closed surface has a Riemannian metric of constant curvature.  It follows that the universal cover is then either $S^2$ (if $g=0$), the Euclidean plane $\mathbb{R}^2$ (if $g=1$), or the hyperbolic plane $\mathbb{H}^2$ (if $g>1$), with the deck transformations acting by isometries.
Now if the fundamental group has an element of order $2$, the universal cover has a deck transformation of order $2$.  But any isometry of $\mathbb{R}^2$ or $\mathbb{H}^2$ of order $2$ has a fixed point, namely the midpoint of the two points forming any orbit.  (This fails for $S^2$ if the two points are antipodal since then they don't have a well-defined "midpoint", and indeed the antipodal map is an isometry of $S^2$ with no fixed points.)  Since a deck transformation cannot have any fixed points, there cannot be a deck transformation of order $2$ if $g>0$.
(More generally, you can make a similar argument using the barycenter of an orbit to show that the fundamental group cannot have any torsion if $g>0$, though it takes some work to define the barycenter and prove it is fixed by isometries in the hyperbolic case.  Alternatively, you can show that the convex hull of an orbit is homeomorphic to a disk and thus there must be a fixed point in it by the Brouwer fixed point theorem.)
