An embedding of $\Bbb C^2/\Bbb Z_{p^2}$ into $\Bbb C^3/\Bbb Z_p$ onto the subvariety $\{xy=z^p\}$ Let $p,q$ be coprime integers with $1\leq q<p$. Let $\Gamma_{p,q}$ be the action of the cyclic group $\Bbb Z_{p^2}$ on $\Bbb C^2$ generated by $(z_1,z_2)\mapsto (\zeta z_1, \zeta^{pq-1} z_2)$ where $\zeta=e^{2\pi i / p^2}$. Simiarly let $G_{p,q}$ be the action of $\Bbb Z_p$ on $\Bbb C^3$ generated by $(x,y,z)\mapsto (ax,a^{-1}y,a^qz)=(\zeta^px,\zeta^{-p}y,\zeta^{pq}z)$ where $a=e^{2\pi i /p}=\zeta^p$. According to Example 2.5 (p.5) of this paper (https://arxiv.org/pdf/1606.08656.pdf), there is an embedding of $\Bbb C^2/\Gamma_{p,q}$ into $\Bbb C^3/G_{p,q}$ onto the subvariety $\{xy=z^p\}$, but I can't see why. Can we explicitly construct such an embedding?
 A: First of all, note that the variety $V=\{xy=z^p\} \subset \mathbb C^3$ is invariant under $G_{p,q}$. Draw the following diagram:
$$\require{AMScd}
\begin{CD}
\mathbb C^2 @>{g}>> V\\
@V{\sigma}VV @VV{\pi}V \\
\mathbb C^2/\Gamma_{p,q} @>{f}>> V/G_{p,q}
\end{CD}$$
Because the actions of the groups are free outside the coordinate axis, $\pi , \sigma$ are ramified coverings  of $p$ and $p^2$ sheets, respectively. With that in mind my first thought was to find a $p$-valued map $g$ (I realized later that this makes no sense at all) from $\mathbb C^2$ to $V$ that would respect the group action.
In this situations the best is going for the simplest maps first, so I try $g(z_1, z_2)=(z_1, z_2, (z_1z_2)^{1/p})$ but then that does not behave well with the group actions: the equivalent element $(\zeta z_1,\zeta^{pq-1}z_2)$ is sent to $(\zeta z_1, \zeta^{pq}(\zeta^{-1} z_2), \zeta^q(z_1z_2)^{1/p})$. If you look more closely, if we had $\zeta^p$ instead of $\zeta$ everything would work, so the next try is $g(z_1, z_2) =(z_1^p, z_2^p, z_1z_2)$. This clearly respects the group actions, and also, going back to counting ramifications, it makes sense because it is a $p$-ramified covering of $V$, so if $f$ were an embedding, then $\pi g=f \sigma$ should be a $p^2$-ramified covering.
It is clear that $g$ is surjective and therefore so is $f$. On the other hand, if $g(z_1, z_2)$ is $G_{p,q}$-related to $g(w_1, w_2)$ then there is some $k$ such that
$$
z_1^p \zeta ^kp = w_1^p \quad , \quad z_2^p\zeta^{-kp}=w_2^p \quad , \quad z_1z_2 \zeta ^{kpq} = w_1w_2.
$$
Taking $p$-th roots, we get $l,m$ such that $w_1 = z_1 \zeta^{k+lp}$, $w_2 = z_1 \zeta^{-k+mp}$ and by the third equation (i assume $z_1z_2\neq 0$), $\zeta^{kpq} = \zeta^{lp+mp}$. After some manipulation it is easy to see that $(z_1, z_2)$ and $(w_1, w_2)$ are $\Gamma_{p,q}$-related, so $f$ is bijective.
Now, I am not familiar with which theorems you can use to prove that some bijective holomorphic map between complex varieties (not manifolds, I know that a bijective holomorphic map between complex manifolds is a biholomorphism) has an inverse. Maybe you can construct the inverse locally, with $p$-th roots (note that the maps we are looking for should come from $p$-valued maps from $V$ to $\mathbb C^2$) but I hope this is sufficient to convince you that $g(z_1, z_2) =(z_1^p, z_2^p, z_1z_2)$ is the desired map.
I have tried to explain my thought process because I am by no means an expert in this area and I'm just extrapolating my knowledge from Riemann Surfaces; I hope everything is clear, and if I said anything wrong, feel free to correct me
