What is a free homotopy class of loops determined by a group element? In a book I encountered this proposition ($H$ is the upper half plane of $\mathbb C$ equipped with the metric $\frac{1}{y^2}dzd\bar z$ and $\Gamma$ is a discrete subgroup of $PSL(2,\mathbb R)$ and acts without fixed points on $H$ as isometries)

Proposition. Let $H/\Gamma$ be a compact Riemann surface. Then for each $\gamma\in\Gamma,\gamma\neq{\rm id}$, the free homotopy class of loops determined by $\gamma$ contains precisely one geodesic (w.r.t the hyperbolic metric).
Proof. Since $H/\Gamma$ is compact, all elements of $\Gamma$ are hyperbolic, i.e. conjugate to the form
$$\gamma=\left(\matrix{\lambda&0\\0&\lambda^{-1}}\right)$$
with $\lambda>0$. And there is precisely one geodesic that is invariant under $\gamma$, namely the imaginary axis. Since the closed geodesics on $H/\Gamma$ are precisely the projections of geodesics on $H$ which are invariant under some nonidentity element of $\Gamma$, and this element of course determine the homotopy class of the geodesic.

Questions:
(1) What is meant by "a free homotopy class of loops determined by $\gamma$". I know what a free homotopy class is, but what does it mean to be "determined by some $\gamma\in\Gamma$"?
(2) "Since the closed geodesics on $H/\Gamma$ are precisely the projections of geodesics on $H$ which are invariant under some nonidentity element of $\Gamma$..."
Attempts:
For (2) I have some idea. Let $p:[a,b]\to H/\Gamma$ be a closed geodesic in $H/\Gamma$ and let $\hat p$ be a lift of it to $H$. Since $p$ is closed,
$$p(a)=p(b)\implies\hat p(a)\equiv\hat p(b)\pmod\Gamma\\
\implies\hat p(a)=\gamma\ \hat p(b)\text{ for some nonidentity }\gamma\in\Gamma$$
Now extend $\hat p$ as a full-length geodesic $s$ in $H$, i.e. a semicircle or straight line perpendicular to $\mathbb R$ so that
$$A=\hat p(a)=\gamma(B),\ B=\hat p(b)$$
are two distinct points on this geodesic. If $s$, as a subset of $H$, is not invariant under $\gamma$, then $s$ and $s'=\gamma s$ are two distinct geodesics in $H$ with $\gamma$ being a bijection between them. Let $\pi:H\to H/\Gamma$ be the natural projection. Since $\pi(A)=\pi(B)$ we can find neighborhoods $U_A,U_B$ of $A,B$ respectively such that $\pi$'s restrictions to them are two isometries near $\pi(A)=\pi(B)$. By restricting to a component if necessary we can assume $\pi(U_A)\cap\pi(U_B)$ is connected, then the map
$$(\pi|_{U_A})^{-1}\pi|_{U_B}:U_B\to U_A$$
can be proved to have the form $z\mapsto\gamma z$ (the proof is a bit tedious so I am not gonna post here). Being an isometry, $(\pi|_{U_A})^{-1}\pi|_{U_B}$ maps a geodesic to a geodesic, hence in $U_A$ we have $s=s'=\gamma s$, contradicting $s,s'$ are two distinct geodesics because they intersect at more than one point.
But I have no idea what is meant by (1), hence no idea about why (2) is useful.
Thanks in advance!
 A: Given $\gamma\in\Gamma$, as $\gamma$ is of hyperbolic type, there is only maximal geodesic $\eta$ invariant under $\gamma$. Prove that if $\pi:H\rightarrow H/\Gamma$ is the projection map, then $\pi(\eta)$ must be a closed geodesic(why?). The class of this loop is the free homotopy class determined by $\gamma$.
Though the ideas behind your attempt to prove (2) are correct, they lack a little bit of rigour. Try filling the gaps of this: 
Suppose $\eta:[a,b]\rightarrow H/\Gamma$ is a closed geodesic, and lift it to $\tilde\eta:[a,b]\rightarrow H$. Then pick up some $\gamma\in \Gamma$ such that $\gamma(\tilde\eta(a))=\tilde\eta(b)$. Suppose, to get a contradiction, $\tilde\eta$ and $\gamma(\tilde\eta)$ are distinct geodesics. Then $\tilde\eta\cup\gamma(\tilde\eta)$ is not a smooth submanifold of $H$ as it must have a "corner" at $\tilde\eta(b)$. Thus as $\pi$ is a local diffeomorphism, this implies $\pi(\tilde\eta\cup\gamma(\tilde\eta))$ cannot be a smooth submanifold of $H/\Gamma$, however this is the geodesic $\eta$, a submanifold of $H/\Gamma$. 
