 Solution

Put $P=|p|$ and $\zeta =e^{2\pi i/p}$. The lens space is the quotientThe cyclic group has order $P$. Its action is free: if $g^j$ fixes a point with $z_1\ne 0$, then $p\mid j$; if $z_1=0$, then $z_2\ne 0$ and $p\mid qj$, which again implies $p\mid j$ by coprimality. A finite free smooth action gives a closed orientable three-manifold, and the quotient projection is a covering space.

Since $S^3$ is simply connected, this is the universal covering and its deck group is the cyclic group of order $P$. HenceFor $P=1$ the quotient is $S^3$ and the group is trivial. With this unoriented quotient convention, replacing $p$ by $-p$ reverses the chosen generator and leaves the quotient unchanged; oriented surgery conventions may record the signs differently.