\begin{document}

\title{The Locus of Curves with $D_n$-Symmetry inside $\mathbb{m}athfrak M_g$ }
\date{}
\author{Binru Li, Sascha Weigl}
\mathbb{m}aketitle

\begin{abstract}
\noindent
The aim of this paper is to determine the irreducible components of  $\mathbb{m}athfrak{M}_{g}(D_n)$, the locus  inside $\mathbb{m}athfrak M_g$ of the curves admitting an effective action by the dihedral group $D_n$. 
This is done by classifying pairs $(H,H')$ of distinct subgroups of the mapping class group $Map_g$, such that both $H$ and $H'$ are isomorphic to $D_n$ and the fixed point locus of $H$ inside the Teichm\"uller space $\mathbb{m}athcal T_g$ is contained in the fixed point locus of $H'$.

\end{abstract}

\section{Introduction}
Given a finite group $H$, denote by $\mathbb{m}athfrak{M}_g(H)$ the locus inside $\mathbb{m}athfrak{M}_g$ (the coarse moduli space of curves of genus $g\geq 2$) of the curves admitting an effective action by the group $H$. A good approach to understanding the irreducible components of $\mathbb{m}athfrak{M}_g(H)$ is to view $\mathbb{m}athfrak{M}_g$ as the quotient of the Teichm\"uller space $\mathbb{m}athcal T_g$ by the natural action of the mapping class group $Map_g$: 
$$\mathbb{P}^1i: \mathbb{m}athcal T_g \to \mathbb{m}athcal T_g/Map_g = \mathbb{m}athfrak M_g .$$ 
 Observe that $$\mathbb{m}athfrak{M}_{g}(H) = \bigcup \limits_{[\rho]} \mathbb{m}athfrak M_{g,~\rho}(H),$$
where $\rho:H\hookrightarrow Map_g$ is an injective homomorphism, $\mathbb{m}athfrak{M}_{g,\ \rho}(H)$ is the image of the fixed locus of $\rho(H)$ under the natural projection $\mathbb{P}^1i$ and $\rho \sim \rho'$ iff they are equivalent by the equivalence relation generated by the automorphisms of $H$ and the conjugations by $Map_g$. We call this equivalence class an {\em unmarked topological type} (cf. \cite{CLP2}, section 2). Since each $\mathbb{m}athfrak M_{g,~\rho}(H)$ is an irreducible (Zariski) closed subset of $\mathbb{m}athfrak{M}_g$ (cf. \cite{CLP2}, Theorem 2.3), in order to determine the irreducible components of $\mathbb{m}athfrak{M}_{g}(H)$, it suffices to determine the maximal loci of the form $\mathbb{m}athfrak M_{g,~\rho}(H)$, i.e. to figure out when one locus contains another. \\
The case where $H$ is a cyclic group was investigated in \cite{Cor} and \cite{Cat1}. In \cite{CLP2} the authors have defined a new homological invariant which allows them to tell when two homomorphism $\rho$ and $\rho'$ are not equivalent; for the case of $H=D_n$, the dihedral group, they also found one representative for each unmarked topological type.\\
In this paper, we focus on the case $H=D_n$, and solve the following problem: for which $\rho$ and $\rho'$, does $\mathbb{m}athfrak M_{g,~\rho'}(D_n)$ contain $\mathbb{m}athfrak M_{g,~\rho}(D_n)$? Hence we determine the loci $\mathbb{m}athfrak{M}_{g,\ \rho}(D_n)$ which are not maximal whence the irreducible decomposition of $\mathbb{m}athfrak{M}_g(D_n)$. The above problem is equivalent to the classification of subgroups $H,H'$ of $Map_g$ $(g\geq 2)$, where $H$ and $H'$ satisfy the following condition: $$(*)~H,H'  \simeq D_n, H\neq H' ~\mathbb{m}box{and}~Fix(H)\subset Fix(H') .$$

\noindent
For any finite subgroup $H\subset Map_g$, set $\delta_H:=$dim$Fix(H)$ and let $G:=G(H):=\bigcap_{C\in Fix(H)}Aut(C)$ ($Fix(H)$ corresponds to the complex structures for which the action of $H$ is holomorphic, whereas $G(H)$ is the common automorphism group of all the curves in $Fix(H)$). 
 If $H=G(H)$ we call $H$ {\em full}.\\
It is easy to see that condition $(*)$ is equivalent to the condition\\
$(**)$ $H$ is isomorphic to $D_n$ and not full, $G(H)$ has a subgroup $H'$ which is isomorphic to $D_n$ and different from $H$.  \\

For any curve $C\in Fix(G)$, we have a Galois cover $p: C\rightarrow C/G=:C'$ which is branched in $r$ ($r$ can be zero) points $P_1,...,P_r$ on $C'$ with branching indices $m_1,...,m_r$. By Theroem \ref{MSSV}, in our case $C'$ is always $\mathbb{P}^1$. The cover map $p$ is determined by a surjective homomorphism $f$ from the orbifold fundamental group $T(m_1,...,m_r):=<\gamma _1,...,\gamma _r |\Pi \gamma_i=1,\gamma_i^{m_i}=1>$ to $G$ (cf. \cite{Cat2}, section 5). The vector $v:=(f(\gamma_1),...,f(\gamma_r))$ is called the \emph{Hurwitz vector} associated to $f$ (See section 5 for more details). Then two Hurwitz vectors $v$ and $v'$ determine the same topological type if and only if they are equivalent for the equivalence relation generated by the action of $Aut(G)$ and by sequences of braid moves. (See Definition \ref{equivalent HV}).\\

Our main result is the following:

\begin{thm}
Let $H,H'$ be subgroups of $Map_g$, satisfying condition $(*)$. Then $G(H) \simeq D_n \times \mathbb{m}athbb{Z}/2$ and $H$ corresponds to $D_n \times \{0\}$. The group $H'$ and the topological action of the group $G(H)$ (i.e. its Hurwitz vector) are as listed in the tables of section 2.
\end{thm}

\noindent
The structure of this paper is as follows:

\noindent
In section 2 we present our results through tables.\\
In section 3 we quote a Theorem from \cite{MSSV} (cf. Theorem \ref{MSSV}), which contains the possible cases (which we call \emph{cover type}) where $H \subsetneq G \subset Map_g$ and $\delta_G=\delta_H$. From this Theorem, using the Riemann-Hurwitz formula, we obtain pairs of dimensions $(\delta_H,\delta_{H'})$, which can occur under condition (**).

 We will also see that $C/G\simeq \mathbb{m}athbb{P}^1$ and $[G:H]=2$ except for one case.\\
In section 4 we will understand group theoretically which cases of $H$ and $G$ can happen under condition (**). This is done by classifying the index 2 subgroups of $G$, where $G$ is a finite group containing two distinct index 2 subgroups which are isomorphic to $D_n$. The cases there are called the \textit{group types}. \\
In section 5 we classify the equivalence classes of Hurwitz vectors of the map $C\rightarrow C/G\simeq \mathbb{m}athbb{P}^1$ for each cover type and group type, by giving one representative vector for each equivalence class.

\section{Results}
We present our results through tables. There will be one table for each normal form of Hurwitz vectors for the covering $C \to C/G$, obtained in section 5. For the reader's convenience we present a short list of notation:\\

\begin{tabular}{cl}

$\nu$ & Hurwitz vector for the covering $C \to C/G$ \\

$v_{G/H'}$ &  Hurwitz vector for the double covering $C/H' \to C/G = \mathbb{P}^1$ \\

$g_{C/H'}$& Genus of $C/H'$ \\

$\delta_{H'}$& Dimension of $Fix(H')$\\

$v_{H'}$& Hurwitz vector for the covering $C \to C/H'$

\end{tabular}

\noindent
We will use the following subgroups of $D_n \times \mathbb{m}athbb{Z}/2$, where  $D_n =<x,y~|~x^n=y^2=1, yxy^{-1} = x^{-1}>$ and $e$ denotes the neutral element of $D_n$.\\

\begin{center}
\begin{tabular}{c|c}
Subgroup&Generators\\
\hline
$K$ & $(x,0)$\\

$H_{1,1}$ & $K,(e,1)$\\

$H_{1,2}$ & $K,(y,1)$\\

$H_{1,3}$& $(x^2,0),(y,0),(e,1)$\\

$H_{1,4}$&$(x^2,0),(y,0),(x,1)$\\

$H_{1,5}$&$(x^2,0),(yx,0),(e,1)$\\

$H_{1,6}$&$(x^2,0),(yx,0),(x,1)$\\

\end{tabular}
\end{center}

\noindent
For compactness, we make the following conventions:\\
 Whenever the groups $H_{1,4},H_{1,6},H_{1,3},H_{1,5}$ occur, we assume that $n = 2m$, in the last 2 cases we additionally assume $m$ to be odd. If $H_{1,1}$ appears we are in the case $n = 2$. We identify the groups $H_{1,3}$ and $H_{1,5}$ with $D_n$ by sending their respective generators in the given order to $x^{m+1},y,x^m$.\\
The cover types are those which appear in Theorem \ref{MSSV}. 

\begin{theorem}\label{main result}
Let $H,H'$ be subgroups of $Mapg_g$, satisfying condition $(*)$. Then $G(H) \simeq D_n \times \mathbb{m}athbb{Z}/2$, $H$ corresponds to $D_n \times \{0\}$. The group $H'$ and the topological action of the group $G(H)$ (i.e. its Hurwitz vector) are as listed in the following tables.
\end{theorem}
We obtain immediately the following corollary:
\begin{corollary}
The locus $\mathbb{m}athfrak{M}_{g,\ \rho}(D_n)$ is maximal iff its topological type $[\rho]$ is different from those which are determined by $C\rightarrow C/H$ in the following tables.
\end{corollary}
\begin{remark}
Given a cover $C\rightarrow C/H$, the data consisting of $g_{C/H}$ and the branching indices are called the signature of the cover. In $\cite{BCGG}$, section 3 the authors computed the signatures for the possible non-maximal loci of the form $\mathbb{m}athfrak{M}_{g,\ \rho}(D_n)$, which is a corollary of our result.
\end{remark}

 
 \begin{center} \begin{large}Cover type I)\\
  ($\delta_H =3$, $g_{C/H}=2$, $C\rightarrow C/H$ is unramified)\end{large} \end{center}
  \[v=(((y,1),(y,1),(yx,1),(yx,1),(e,1),(e,1))\]

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$H'$&$v_{G/H'}$&$g_{C/H'}$&$\delta_{H'}$&$v_{H'}$\\
\hline
$H_{1,2}$&(0,0,0,0,1,1)&0&5&(y,y,y,y,yx,yx,yx,yx)\\
\hline
$H_{1,3}$&(0,0,1,1,0,0)&0&5&$(yx^m,yx^{m-2},yx^m,yx^{m+2},x^m,x^m,x^m,x^m)$\\
\hline
$H_{1,4}$&(1,1,0,0,1,1)&0&5&(yx,yx,yx,yx,y,y,y,y)\\
\hline
$H_{1,5}$&(1,1,0,0,0,0)&0&5&$(yx^m,yx^m,yx^m,yx^m,x^m,x^m,x^m,x^m)$\\
\hline
$H_{1,6}$&(0,0,1,1,1,1)&1&4&(e,yx;y,y,y,y)\\
\hline
\end{tabular}
\end{center}

\[v = ((y,1),(yx^{m},1),(yx,1),(yx,1),(x^{m},1),(e,1)),\ n=2m\]

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$H'$&$v_{G/H'}$&$g_{C/H'}$&$\delta_{H'}$&$v_{H'}$\\
\hline
$H_{1,2}$&$(0,0,0,0,1,1)$&0&5&$(y,y,yx^m,x^my,yx,yx,yx,yx)$\\
\hline
$H_{1,3}$&$(0,1,1,1,1,0)$&1&4&$(x^{m+1},x^{m-1};x^m,x^m,yx^m,yx^m)$\\
\hline
$H_{1,4}$($m$ odd)&$(1,0,0,0,0,1)$&0&5&$(yx^m,x^my,yx,yx^3,yx,xy,x^m,x^m)$\\
$H_{1,4}$($m$ even)&$(1,1,0,0,1,1)$&1&4&$(x^m,x^my;yx,yx,yx,yx)$\\
\hline
$H_{1,5}$&$(1,0,0,0,1,0)$&0&5&$(yx^{\frac{m^2-1}{2}},yx^{\frac{m^2-1}{2}},yx^m,yx^m,yx^m,yx^m,x^m,x^m)$\\
\hline
$H_{1,6}$($m$ odd)&(0,1,1,1,0,1)&1&4&$(x^{m+1},x^{m-1};x^m,x^m,y,y)$\\
$H_{1,6}$($m$ even)&(0,0,1,1,1,1)&1&4&$(e,x^{m-1}y;y,y,x^my,yx^m)$\\
\hline

\end{tabular}
\end{center}

\[v = ((y,1),(yx^{m},1),(yx^2,1),(yx^2,1),(x^{m},1),(e,1)),\ n=2m,\ m ~\mathbb{m}box{odd.}\]

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$H'$&$v_{G/H'}$&$g_{C/H'}$&$\delta_{H'}$&$v_{H'}$\\
\hline
$H_{1,2}$&$(0,0,0,0,1,1)$&0&5&$(y,y,yx^m,yx^m,yx^2,yx^2,yx^2,yx^2)$\\
\hline
$H_{1,3}$&$(0,1,0,0,1,0)$&0&5&$(x^m,x^m,yx^m,yx^{-1},yx^{-3},yx^{-1},yx)$\\
\hline
$H_{1,4}$&$(0,0,0,0,1,1)$&0&5&$(y,y,yx^m,yx^m,yx^2,yx^2,yx^2,yx^2)$\\
\hline
$H_{1,5}$&$(1,0,1,1,1,0)$&1&4&$(x^2,x^{-2};x^m,x^m,x^my,x^my)$\\
\hline
$H_{1,6}$&(0,1,0,0,0,1)&0&5&$(y,y,x^2y,x^6y,x^2y,yx^2,x^m,x^m)$\\
\hline
\end{tabular}
\end{center}

For $n=2$ we have two extra cases:
\[ v = ((y,1),(y,1),(x,1),(x,1),(e,1),(e,1)) \]

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$H'$&$v_{G/H'}$&$g_{C/H'}$&$\delta_{H'}$&$v_{H'}$\\
\hline
$H_{1,1}$&$(1,1,0,0,0,0)$&0&5&$(x,x,x,x,y,y,y,y)$\\
\hline
$H_{1,2}$&$(0,0,1,1,1,1)$&1&4&$(e,x;y,y,y,y)$\\
\hline
$H_{1,3}$&$(0,0,1,1,0,0)$&0&5&$(yx,yx,yx,yx,x,x,x,x)$\\
\hline
$H_{1,4}$&$(1,1,0,0,1,1)$&1&4&$(e,y;x,x,x,x)$\\
\hline
$H_{1,5}$&$(1,1,1,1,0,0)$&1&4&$(e,yx;x,x,x,x)$\\
\hline
$H_{1,6}$&(0,0,0,0,1,1)&0&5&$(y,y,y,y,x,x,x,x)$\\
\hline
\end{tabular}
\end{center}

\[v = ((y,1),(yx,1),(x,1),(x,1),(x,1),(e,1))\]

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$H'$&$v_{G/H'}$&$g_{C/H'}$&$\delta_{H'}$&$v_{H'}$\\
\hline
$H_{1,1}$&$(1,1,0,0,0,0)$&0&5&$(yx,yx,yx,yx,yx,yx,y,y)$\\
\hline
$H_{1,2}$&$(0,0,1,1,1,1)$&1&4&$(e,e;y,y,yx,yx)$\\
\hline
$H_{1,3}$&$(0,1,1,1,1,0)$&1&4&$(y,y;x,x,yx,yx)$\\
\hline
$H_{1,4}$&$(1,0,0,0,0,1)$&0&5&$(yx,yx,x,x,x,x,x,x,x,x)$\\
\hline
$H_{1,5}$&$(1,0,1,1,1,0)$&1&4&$(e,y;x,x,x,x)$\\
\hline
$H_{1,6}$&(0,1,0,0,0,1)&0&5&$(y,y,x,x,x,x,x,x,x,x)$\\
\hline
\end{tabular}
\end{center}

\begin{center} \begin{large}Cover type II) \\
($\delta_H=2$, $g_{C/H}=1$)\end{large} \end{center}
$(1)$ $c_5=2$.
$$v=((y,1),(yx,1),(yx,1),(e,1),(y,0)),\ v_H=(x,x^{-1};y,y).$$
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$H'$&$v_{G/H'}$&$g_{C/H'}$&$\delta_{H'}$&$v_{H'}$\\
\hline
$H_{1,2}$&$(0,0,0,1,1)$&0&3&$(y,y,yx,yx,yx,yx)$\\
\hline
$H_{1,3}$&$(0,1,1,0,0)$&0&3&$(yx^m,yx^{-1},x^m,x^m,y,yx^{m+1})$\\
\hline
$H_{1,4}$&$(1,0,0,0,1)$&0&3&$(yx,yx,yx,yx,y,y)$\\
\hline
$H_{1,5}$&$(1,0,0,0,1)$&0&3&$(yx^m,yx,yx^m,yx^{-1},x^m,x^m)$\\
\hline
$H_{1,6}$&$(0,1,1,1,1)$&1&2&$(e,yx;y,y)$\\ 
\hline
\end{tabular}
\end{center}

$(2)$ $c_5>2$.
$$v=((y,1),(yx^{-1},1),(e,1),(e,1),(x,0)),\ c_5=n,\ v_H=(x^{-1},y;x,x).$$
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$H'$&$v_{G/H'}$&$g_{C/H'}$&$\delta_{H'}$&$v_{H'}$\\
\hline
$H_{1,2}$&$(0,0,1,1,0)$&0&3&$(y,y,yx^{-1},yx^{-3},x,x)$\\
\hline
$H_{1,3}$&$(0,1,0,0,1)$&0&4&$(yx^m,yx^{-1},x^m,x^m,x^m,x^m,x^{m+1})$\\
\hline
$H_{1,4}$&$(1,0,1,1,1)$&1&3&$(y,y;x^2,yx^3,yx)$\\
\hline
$H_{1,5}$&$(1,0,0,0,1)$&0&4&$(yx^{-1},yx^{m-2},x^m,x^m,x^m,x^m,x^{m+1})$\\
\hline
$H_{1,6}$&$(0,1,1,1,1)$&1&3&$(yx^{-1},yx^{-1};x^2,yx^2,y)$\\ 
\hline
\end{tabular}
\end{center}

$$v=((y,1),(yx^{m-1},1),(x^m,1),(e,1),(x,0)),\ n=2m,\ v_H=(x^{m-1},yx^m;x,x).$$
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$H'$&$v_{G/H'}$&$g_{C/H'}$&$\delta_{H'}$&$v_{H'}$\\
\hline
$H_{1,2}$&$(0,0,1,1,0)$&0&3&$(y,y,yx^{m-1},yx^{m-3},x,x)$\\
\hline
$H_{1,3}$&$(0,0,1,0,1)$&0&4&$(yx^m,yx^{m+2},yx,yx,x^m,x^m,x^{-2})$\\
\hline
$H_{1,4}\ (m\ odd)$ &$(1,1,0,1,1)$&1&3&$(x^{m-1},y;x^2,x^m,x^m)$\\
$H_{1,4}\ (m\ even)$ &$(1,0,1,1,1)$&1&3&$(yx^{m},y;x^2,yx^{m+3},yx^{m+1})$\\
\hline
$H_{1,5}$&$(1,1,1,0,1)$&1&3&$(x^{m+1},y;x^{-2},x^m,x^m)$\\
\hline
$H_{1,6}\ (m\ odd)$&$(0,0,0,1,1)$&0&4&$(y,yx^{-2},yx^{m-1},yx^{m-1},x^m,x^m,x^2)$\\ 
$H_{1,6}\ (m\ even)$&$(0,1,1,1,1)$&1&3&$(yx^{-1},yx^{m-1};x^2,yx^2,y)$\\ 
\hline
\end{tabular}
\end{center}

$$v=((y,1),(yx^{m-2},1),(x^m,1),(e,1),(x^2,0)),\ n=2m,\ m\ \mathbb{m}box{odd},\ c_5=m,$$
 $$v_H=(x^{m-2},yx^m;x^2,x^2),$$
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$H'$&$v_{G/H'}$&$g_{C/H'}$&$\delta_{H'}$&$v_{H'}$\\
\hline
$H_{1,2}$&$(0,0,1,1,0)$&0&3&$(y,y,yx^{m-2},yx^{m-6},x^2,x^2)$\\
\hline
$H_{1,3}$&$(0,1,1,0,0)$&0&3&$(yx^m,yx^{m-4},x^m,x^m,x^2,x^2)$\\
\hline
$H_{1,4}$&$(1,0,0,1,0)$&0&3&$(yx^{m-2},yx^{m-6},x^m,x^m,x^2,x^2)$\\
\hline
$H_{1,5}$&$(1,0,1,0,0)$&0&3&$(y,yx^{-4},x^m,x^m,x^2,x^2)$\\
\hline
$H_{1,6}$&$(0,1,0,1,0)$&0&3&$(y,yx^{-4},x^m,x^m,x^2,x^2)$\\ 
\hline
\end{tabular}
\end{center}

For $n=2$ we have one extra case.\\
$$v=((yx,1),(x,1),(e,1),(e,1),(y,0)),\ v_H=(y,yx;y,y).$$
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$H'$&$v_{G/H'}$&$g_{C/H'}$&$\delta_{H'}$&$v_{H'}$\\
\hline
$H_{1,1}$&$(1,0,0,0,1)$&0&3&$(x,x,y,y,y,y)$\\
\hline
$H_{1,2}$&$(0,1,1,1,1)$&1&2&$(x,x;yx,yx)$\\
\hline
$H_{1,3}$&$(1,1,0,0,0)$&0&3&$(x,x,x,x,y,y)$\\
\hline
$H_{1,4}$&$(0,0,1,1,0)$&0&3&$(yx,yx,x,x,y,y)$\\
\hline
$H_{1,5}$&$(0,1,0,0,1)$&0&3&$(yx,yx,x,x,x,x)$\\
\hline
$H_{1,6}$&$(1,0,1,1,1)$&1&2&$(yx,yx;x,x)$\\ 
\hline
\end{tabular}
\end{center}

\begin{center} \begin{large}Cover type III-a)\\
 ($\delta_H =1$, $g_{C/H}=1$)\end{large} \end{center}
$$v=((y,1),(yx^{-1},1),(e,1),(x,1)),\ 2d_4=n=2m,\ v_H=(x^{-1},y;x^2).$$
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$H'$&$v_{G/H'}$&$g_{C/H'}$&$\delta_{H'}$&$v_{H'}$\\
\hline
$H_{1,2}$&$(0,0,1,1)$&0&2&$(y,yx^2,yx^{-1},yx^{-1},x^2)$\\
\hline
$H_{1,3}$&$(0,1,0,1)$&0&2&$(yx^m,yx^{-1},x^m,x^m,x^{m+1})$\\
\hline
$H_{1,4}$&$(1,0,1,0)$&0&1&$(yx^{-1},yx^{-3},x,x)$\\
\hline
$H_{1,5}$&$(1,0,0,1)$&0&2&$(yx^{-1},yx^{m-2},x^m,x^m,x^{m+1})$\\
\hline
$H_{1,6}$&$(0,1,1,0)$&0&1&$(x,x,y,yx^{-2})$\\ 
\hline
\end{tabular}
\end{center}

\begin{center} \begin{large}Cover type III-b) \\
($\delta_H=1$, $g_{C/H}=0$)\end{large} \end{center}
$$v=((yx,1),(e,1),(y,0),(x,0)),\ c_4=n=2m,\ v_H=(y,yx^{-2},x,x).$$
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$H'$&$v_{G/H'}$&$g_{C/H'}$&$\delta_{H'}$&$v_{H'}$\\
\hline
$H_{1,2}$&$(0,1,1,0)$&0&1&$(yx,yx^{-1},x,x)$\\
\hline
$H_{1,3}$&$(1,0,0,1)$&0&2&$(x^m,x^m,y,yx^{m-1},x^{m+1})$\\
\hline
$H_{1,4}$&$(0,1,0,1)$&0&2&$(yx,yx^{-1},y,y,x^2)$\\
\hline
$H_{1,5}$&$(0,0,1,1)$&0&2&$(yx^m,yx^{-1},x^m,x^m,x^{m+1})$\\
\hline
$H_{1,6}$&$(1,1,1,1)$&1&1&$(yx,x;x^2)$\\ 
\hline
\end{tabular}
\end{center}

\section{A rough classification}
In this section we determine the possible pairs of dimensions $(\delta_H, \delta_{H'})$, for distinct subgroups $H$ and $H'$ of $Map_g$ which satisfy condition (**). \\
Given $C\in Fix(H)$, assume that $C\rightarrow C/H$ is a cover branched on $r$ points. We have that $\delta_H=3g_{G/H}-3+r$ (cf. \cite{CLP2}, Theorem 2.3).\\
The case $\delta_H=\delta_{H'}$ was done in Corollary 7.2 of [CLP2]. We only consider the case $\delta_H<\delta_{H'}$.\\
We recall Lemma 4.1 of [MSSV]:

\begin{theorem}\label{MSSV} (MSSV)\\
Let $H\subsetneqq G$ be two (finite) subgroups of $Map_g$, $\delta_H=\delta_G=:\delta$. Then one of the following holds:\\
$I)$ $\delta_{H}=3$, $[G$:$H]=2$, $C\rightarrow C/G$ is a covering of $\mathbb{m}athbb{P}^1$ branched on 6 points $P_1,\dots,P_6$, and with branching indices all equal to 2. Moreover the subgroup H corresponds to the unique genus two double cover of $\mathbb{m}athbb{P}^1$ branched on the 6 points.\\
$II)$ $\delta_H=2$, $[G$:$H]=2$, and $C\rightarrow C/G$ is a covering of $\mathbb{m}athbb{P}^1$ branched on five points, $P_1,\dots,P_5$, with branching indices $2,2,2,2,c_5$. Moreover the subgroup H corresponds to a double cover of $\mathbb{m}athbb{P}^1$ branched on the 4 points $P_1,\dots, P_4$ with branching index 2.\\
$III)$ $\delta_{H}=1$, there are 3 possibilities:\\
$III-a)$ H has index 2 in G, and $C\rightarrow C/G$ is a covering of $\mathbb{m}athbb{P}^1$ branched on 4 points, $P_1,\dots,P_4$, with branching indices $2,2,2,2d_4,$ where $d_4>1$. Moreover the subgroup H corresponds to the unique genus one double cover of $\mathbb{m}athbb{P}^1$ branched on the 4 points $P_1,\dots, P_4$.\\
$III-b)$ H has index 2 in G, and $C\rightarrow C/G$ is a covering of $\mathbb{m}athbb{P}^1$ branched on 4 points, $P_1,\dots,  P_4$, with branching indices $2,2,c_3,c_4$, where $c_3\leq c_4$ and $c_4>2$. Moreover the subgroup $H$ corresponds to a genus zero double cover of  $\mathbb{m}athbb{P}^1$ branched on two points with branching index 2.\\
$III-c)$ $H$ is normal in $G$, $G/H\cong (\mathbb{m}athbb{Z}/2)^2$, moreover $C\rightarrow C/G$ is a covering of $\mathbb{m}athbb{P}^1$ branched on 4 points $P_1,\dots, P_4$, with branching indices $2,2,2,c_4$, where $c_4>2$. Moreover the subgroup $H$ corresponds to the unique genus zero cover of $\mathbb{m}athbb{P}^1$ with group $(\mathbb{m}athbb{Z}/2)^2$ branched on the 3 points $P_1,P_2,P_3$ with branching index 2.
\end{theorem}
We call the cases in Theorem \ref{MSSV} the \emph{cover type} (of $H$ and $G$).\\
Since we have condition $(**)$, which implies $\delta_G = \delta_H$, we can apply Theorem \ref{MSSV}. Moreover we apply the Riemann-Hurwitz formula to each cover type to find the possible pairs $(\delta_H,\delta_{H'})$.

\begin{corollary}\label{MSSV Cor}
Assume $(**)$ and moreover $\delta_H<\delta_{H'}$. Then the following pairs of dimensions $(\delta_H,\delta_{H'})$ can occur:\\

\noindent
$I)$ $(3,4)$, $(3,5)$.\\
$II)$ $(2,3)$, $(2,4)$.\\
$III-a)$ $(1,2)$.\\
$III-b)$ $(1,2),(1,3)$.\\
$III-c)$ None.
\begin{proof}
$I)$ $\delta_H=3$.\\
 By the Riemann-Hurwitz formula,
$$2g(C)-2=|G|(-2+6 \cdot \frac{1}{2})=|H'|(2(g_{C/H'}-1)+k/2)$$
 where $k$ is the number of  branching points of $C\rightarrow C/H'$.\\
It is easy to see that $(g_{C/H'},k)=(2,0),(1,4)$ or $(0,8)$, corresponding to $\delta_{H'}=3,4,5$. Since we require $\delta_{H}<\delta_{H'}$, the possible pairs are (3,4) and (3,5).\\
$II)$ $\delta_H=2$.\\
 In this case $C/H'\rightarrow \mathbb{m}athbb{P}^1$ is a double covering branched on at most 5 points. Using Riemann-Hurwitz, there are two cases:\\
(i)  $g_{C/H'}=0$ and $C/H' \rightarrow \mathbb{m}athbb{P}^1$ is branched on 2 of the 5 points with branching indices 2,2. \\
If $c_5=2$ or $P_5$ is not a branching point, we have $\delta_{H'}=3$;\\
Otherwise $c_5$ is even and bigger than 2 and $P_5$ is a branching point, we get $\delta_{H'}=4$. \\
(ii) $g_{C/H'}=1$ and $C/H' \rightarrow \mathbb{m}athbb{P}^1$ is branched on 4 of the 5 points with branching indices 2,2,2,2.\\
The only possible case in which $\delta_{H'}>2$ is that $c_5$ is even and bigger than 2 and $P_5$ is one of the branching points. In this case $\delta_{H'}=3$.\\
$III)$ $\delta_H=1$.\\
$III-a)$
Similar to case $II)$, one gets $g_{C/H'}=0$, and $C/H'\rightarrow \mathbb{m}athbb{P}^1$ is a double cover with one of the branching points $P_4$ and $\delta_{H'}=2$.\\
$III-b)$
$i)$ If $c_3=2$, the only possibility is $c_4$ even, $g_{C/H'}=0$ and $C/H'\rightarrow \mathbb{m}athbb{P}^1$ is a double cover with one of the branching points $P_4$, here $\delta_{H'}=2$.\\
$ii)\  c_3>2$, there are three possibilities:\\
$\alpha)$ $c_3$ or $c_4$ is even, one and only one point of $P_3$,$P_4$ is a branching point. This case is similar to $III-b)-i)$, $\delta_{H'}=2$.\\
$\beta)$ Both $c_3$ and $c_4$ are even, $g_{C/H'}=0$, and $C/H'\rightarrow \mathbb{m}athbb{P}^1$ is a double cover branching on $P_3$,$P_4$. We have $\delta_{H'}=3$.\\
$\gamma)$ Both $c_3$ and $c_4$ are even, $g_{C/H'}=1$, and $C/H'\rightarrow \mathbb{m}athbb{P}^1$ is a double cover branching on 4 points $P_1,\dots, P_4$. We have $\delta_{H'}=2$.\\
$III-c)$ We will give the proof in section \ref{admissible f's}, Lemma \ref{group type 3}.
\end{proof}
\end{corollary}

Remark: Cor. \ref{MSSV Cor} is valid for any $H,H'$ with  the same index in $G$ except for the case $III-c)$.
\section{Index 2 subgroups of G}
From Theorem \ref{MSSV} we  know that $[G$:$H]=2$ except for $III-c)$. Such a pair is given by an exact sequence
$$1 \to H \to G \to \mathbb{m}athbb{Z}/2 \to 1.$$
This type of extensions, where $H=D_n$ and $G$ has another subgroup $H'$ isomorphic to $D_n$, has been classified in \cite{CLP2}, Proposition 7.4. There are 3 cases, which we call \emph{group types}:\\

\noindent
\underline{$Group\ type\ 1)$} $G\cong D_n\times \mathbb{m}athbb{Z}/2$, $H$ corresponds to the subgroup $D_n\times \{0\}$.\\
\underline{$Group\ type\ 2)$} $n=2d$, $G\cong D_{2n}=<z,y|z^{2n}=y^2=1,yzy=z^{-1}>$, $H=<x:=z^2,y>$.\\
\underline{$Group\ type\ 3)$} $n=4h$, where $h$ is odd, and G is the semidirect product of $H\cong D_n$ with $<\beta_2>\cong \mathbb{m}athbb{Z}/2$, such that conjugation by $\beta_2$ acts as follows:
$$y\mathbb{m}apsto yx^2,x\mathbb{m}apsto x^{2h-1}.$$
For each group type, we will determine the index 2 subgroups of $G$ and find out which of them are isomorphic to $D_n$.\\

\noindent
\underline{$Group\ type\ 1)$} Recall the standard presentation $D_n=<x,y|x^n=y^2=1,yxy^{-1}=x^{-1}>$ and let $C_n:=\mathbb{m}athbb{Z}/n$.\\
We have to understand the index 2 subgroups $K$ of $D_n$, such that $K\triangleleft G$, where $K$ corresponds to $H \cap H'$. \\
$a)$ $K= C_n\times 0$ (This is the only case when n is odd).\\
Since $G/K\cong (\mathbb{m}athbb{Z}/2)^2$, there are two more index 2 subgroups $H_{1,1}:=<K,(e,1)>$,\\
$H_{1,2}:=<K,(y,1)>\cong D_n$.\\
$b)$ If $n=2d$, there are two more cases, $K=<(x^2,0),(y,0)>$ or $K=<(x^2,0),(yx,0)>$ 
(both isomorphic to $D_d$). \\
Here we have 4 more index 2 subgroups, $H_{1,3}:=<(x^2,0),(y,0),(e,1)>$, $H_{1,4}:=<(x^2,0),(y,0),(x,1)>$, $H_{1,5}:=<(x^2,0),(yx,0),(e,1)>$,  $H_{1,6}:=<(x^2,0),(yx,0),(x,1)>$. On checks easily that $H_{1,4}$ and $H_{1,6}$ are isomorphic to $D_n$ and that $H_{1,3}$ and $H_{1,5}$ are isomorphic to $D_n$ if and only if $d$ is odd.\\
\underline{$Group\ type\ 2)$} Using similar arguments as for group type 1), we obtain 2 more index 2 subgroups: $H_{2,1}=C_{2n}$, $H_{2,2}=<z^2,yz> \cong D_n$.\\
\underline{$Group\ type\ 3)$}
There are 6 more index 2 subgroups: $H_{3,1}=<C_n,(e,\beta_2)>$, $H_{3,2}=<C_n,(y,\beta_2)>$, $H_{3,3}=<(x^2,0),(y,0),(e,\beta_2)>$, $H_{3,4}=<(x^2,0),(y,0),(x,\beta_2)>$, $H_{3,5}=<(x^2,0),(yx,0),(e,\beta_2)>$, $H_{3,6}=<(x^2,0),(yx,0),(x,\beta_2)>$,
 and only $H_{3,3}$ is isomorphic to $D_n$ ( since $H_{3,3}=<(y,\beta_2),(e,\beta_2)>$).
 
 
\section{Hurwitz vectors for $C \to C/G$}\label{admissible f's}

We start by recalling some general theory of Galois covers of Riemann surfaces (cf. \cite{Cat2}, section 5).\\
Let $H$ be a finite group (not necessarily isomorphic to $D_n$) which acts effectively on a curve $C$ of genus $g\geq 2$, we obtain a Galois cover $p: C\rightarrow C/H:=C'$ branched on $r$ points with branching indices $m_1,...,m_r$. Denote by $g'$ the genus of $C'$, the {\em orbifold fundamental group} of the cover is a group with the following presentation:
$$T(g';m_1,...,m_r):=<\alpha_1,\beta_1,...,\alpha_{g'},\beta_{g'};\gamma_1,...,\gamma_r\ |\ \Pi[\alpha_j,\beta_j]\cdot \Pi\gamma_i=1, \gamma_i^{m_i}=1>$$
The cover $C\rightarrow C/H$ is (topologically) determined by a surjective morphsim 
$$f: T(g';m_1,...,m_2)\rightarrow G,$$ such that $f(\gamma_j)$ has order $m_j$ inside $G$. We call $v:=[f(\alpha_1),f(\beta_1),...,f(\alpha_{g'}),f(\beta_{g'});f(\gamma_1),...,f(\gamma_r)]$ the {\em Hurwitz vector} associated to $f$.\\
In this section we study the Hurwitz vectors of each cover type $C\rightarrow C/G$ in Theorem $\ref{MSSV}$. Hence we have that $C/G\simeq \mathbb{m}athbb{P}^1$, and we set $T(m_1,...,m_r):=T(0;m_1,...,m_r)$. \\
Given a morphism $f:T(m_1,...,m_r)\rightarrow G$, the Hurwitz vector associated to $f$ is not uniquely determined, since we can choose different presentations for $T(m_1,...,m_r)$. For instance consider $T(m_1,...,m_r)$ with the presentation $<\gamma_1,...\gamma_r|\Pi\gamma_i=1,\gamma_i^{m_i}=1>$, for any $1\leq k<r$, we have a set of generators $\{\delta_i\}$, where $\delta_i:=\alpha_i$ if $i\neq k,k+1$; $\delta_k:=\alpha_{k}\alpha_{k+1}\alpha^{-1}_{k}$ and $\delta_{k+1}:=\alpha_k$, this induces an isomorphism between $T(m_1,...,m_r)$ and $T(l_1,...,l_r)$, where $l_i=m_i$ if $i\neq k,k+1$; $l_k=m_{k+1}$ and $l_{k+1}=m_k$. Different choices of the generators correspond to the following braid group action on the set of Hurwitz vectors.\\
Recall that Artin's {\em braid group on $r$ strands} has the presentation 
$$\mathbb{m}athcal B_r := <\sigma_1,...,\sigma_{r-1} | \forall 1\leq i\leq r-2,\ \sigma_i\sigma_{i+1}\sigma_i = \sigma_{i+1}\sigma_i\sigma_{i+1}; \forall   |j -i| \geq 2,\sigma_i\sigma_j=\sigma_j\sigma_i>. $$  
The group $\mathbb{m}athfrak{B}_r$ acts on the set of Hurwitz vectors of length $r$ as follows:
$$(v_1,...,v_i,v_{i+1},...,v_r) \stackrel{\sigma_i}{\mathbb{m}apsto} (v_1,...,v_iv_{i+1}v_i^{-1},v_{i},...,v_r).$$
On the other hand, for any $h\in Aut(G)$, we can compose $f$ with $h$, this induces a $Aut(G)$-action on the set of Hurwitz vectors: 
given $v=(v_1,...,v_r)$ a Hurwitz vector, define $h(v) := (h(v_1),...,h(v_r))$.\\
Since these actions (by $\mathbb{m}athfrak{B}_r$ and by $Aut(G)$) commute, they induce an action of the group $\mathbb{m}athcal B_r \times Aut(G)$ on the set of Hurwitz vectors of length $r$.
\begin{definition}\label{equivalent HV}
Given two $G$-Hurwitz vectors $v,v'$ of length $r$, we say that $v$ and $v'$ are equivalent if they are in the same $ \mathbb{m}athcal{B}_r\times Aut(G)$-orbit.
\end{definition}
 \begin{remark}
 Two Hurwitz vectors $v$ and $v'$ determine the same unmarked topological type iff they are equivalent (cf. \cite{CLP2}, section 2).
\end{remark}
\begin{definition}
Let $C\rightarrow C/G\cong \mathbb{m}athbb{P}^1$ be a Galois cover of a given group type and cover type. We call a homomorphism $f:T(m_1,\dots, m_r)\rightarrow G$ \underline{admissible} if it satisfies the following two conditions:\\
$(1)$ $f$ is surjective, $T(m_1,...,m_r)$ is isomorphic to the orbifold fundamental group of $C\rightarrow C/G$ and $f(\gamma_i)$ has order $m_i$ in $G$.\\
$(2)$ $f_{H}:=\mathbb{P}^1i_{H}\circ f: T(m_1,\dots, m_r)\rightarrow G/H$ corresponds to the cover $C/H\rightarrow \mathbb{m}athbb{P}^1$, where $\mathbb{P}^1i_H:G\rightarrow G/H$ is the quotient homomorphism.
\end{definition}
\begin{definition}
Let $f:T(m_1,...m_r)\rightarrow G$ and $f':T(l_1,...l_r)\rightarrow G$ be admissible for a given cover type and group type. We say $f$ is equivalent to $f'$ if their corresponding Hurwitz vectors are in the same $\mathbb{m}athfrak{B}_r\times Aut(G)_H$-orbit, where $Aut(G)_H$ denotes the subgroup of $Aut(G)$ which leaves $H$ invariant.
\end{definition}
\begin{remark}
An admissible $f$ determines both the covers $C\rightarrow C/G$ and $C\rightarrow C/H$, hence we require the equivalence relation to be generated by $\mathbb{m}athfrak{B}_r$ and $Aut(G)_H$. It can happen that two admissible homomorphisms have equivalent Hurwitz vectors, but are not equivalent (cf. Remark \ref{Rk}). 
\end{remark}

\begin{example}\label{Ex}
Cover type $III-b)$ and group type $1)$ (cf. Corollary \ref{MSSV Cor})
\end{example}
\noindent
$i)$ $c_3=2$, assume $n$ even and $c_4=n$.\\ 
Consider $f:T(2,2,2,c_4)\rightarrow D_n\times \mathbb{m}athbb{Z}/2$: $\gamma_1\mathbb{m}apsto (yx,1)$, $\gamma_2\mathbb{m}apsto (e,1)$, $\gamma_3\mathbb{m}apsto (y,0)$, $\gamma_4\mathbb{m}apsto (x,0)$.\\
$\delta_{H_{1,2}}=\delta_{H_{1,6}}=1$, $\delta_{H_{1,4}}=2$.\\
$ii)$ $c_3>2$, assume we have an admissible $f$, it is easy to see that $f(\gamma_3)=(x^{i_3},0)$, $f(\gamma_4)=(x^{i_4},0)$. $f(\gamma_1),f(\gamma_2)\in\{(yx^k,1),k\in \mathbb{m}athbb{Z}; (x^{n/2},1)(if\ n\ is\ even) \}$. Since $\Pi f(\gamma_i)=1$, there are only two possibilities:\\
$(a)$ $f(\gamma_1),f(\gamma_2)=(x^{n/2},1)$, which implies $\mathbb{m}box{Im}(f)\subset <(x,0),(0,1)>$, a contradiction.\\
$(b)$ $f(\gamma_1)=(yx^{i_1},1)$, $f(\gamma_2)=(yx^{i_2},1)$, which implies $\mathbb{m}box{Im}(f)\subset <(x,0),(y,1)>$, again a contradiction.\\

Now we classify all admissible $f$'s  for the covering $C \to C/G$,  in the following way: For each cover type and group type, we construct all possible Hurwitz vectors according to their branching behavior, as given in Theorem \ref{MSSV}. \\

 

\begin{lemma}
Group type $2)$ has no admissible $f$ for any cover type.
\begin{proof}
 \underline{Cover type I)} \\
Assume we have an admissible $f:T(2,2,2,2,2,2)\rightarrow D_{2n}$, then $f_{H}(\gamma_i)=1$, $i=1,\dots,6$, which implies that
 $f(\gamma_i)\in \{yz^{2k+1}, z^{2l+1},k,l\in \mathbb{m}athbb{Z} \}$. Moreover $f(\gamma_i)$ has order two, thus $f(\gamma_i)\in\{yz^{2k+1},k\in \mathbb{m}athbb{Z} \}$. We find that $\mathbb{m}box{Im}(f)\subset H_{2,2}$, a contradiction.\\
 \underline{ Cover type II) }\\
If there exists an admissible $f:T(2,2,2,2,c_5)\rightarrow D_{2n}$, we get $f(\gamma_i)\in \{yz^{2k+1},k\in \mathbb{m}athbb{Z} \}, i=1,2,3,4$ and $f(\gamma_5)\in \{z^{2l},l\in \mathbb{m}athbb{Z} \} $(since $\Pi f(\gamma_i)=1$), which implies that $\mathbb{m}box{Im}(f)\subset H_{2,2}$, a contradiction.\\
 \underline{Cover type III-a)} \\
Given an admissible $f:T(2,2,2,2d_4)\rightarrow D_{2n}$, we get $f(\gamma_i)\in\{yz^{2k+1},k\in \mathbb{m}athbb{Z} \}, i=1,2,3$, and $f(\gamma_4)\in \{z^{2l+1},l\in \mathbb{m}athbb{Z} \}$. However, $\Pi f(\gamma_i)\neq 1$, a contradiction.\\
 \underline{Cover type III-b)}\\
$i)$ $c_3=2$. We have $f(\gamma_i)=yz^{2k_i+1},i=1,2, f(\gamma_3)=yz^{2k_3}$ or $z^n$,  $f(\gamma_4)=z^{2k_4}$. If $f(\gamma_3)=yz^{2k_3}$ we find $\Pi f(\gamma_i)\neq 1$; otherwise $f(\gamma_3)=z^n$, which implies $\mathbb{m}box{Im}(f)\subset <yz,z^2>$. In both cases we have no admissible $f$.\\
$ii)$ $c_3>2$. We have $(f(\gamma_1),f(\gamma_2),f(\gamma_3),f(\gamma_4))=(yz^{2k_1+1},yz^{2k_2+1},z^{2k_3},z^{2k_4})$. We see $\mathbb{m}box{Im}(f)\subset <yz,z^2>$, a contradiction.
\end{proof}
\end{lemma}

\begin{lemma}\label{group type 3}
Group type $3)$ has no admissible $f$ for any cover type.
\begin{proof}
  First we determine the order 2 elements of type $(a,\beta_2)$ in $G$. One computes easily that $(x^j,\beta_2)^2=(x^{2jh},0)$ and $(yx^k,\beta_2)^2=(x^{2kh-2k+2},0)\neq (e,0)$. Therefore we conclude that $(a,\beta_2)$ is of order two $\Leftrightarrow$ $a=x^j$ and $j$ is even. \\
 \underline{Cover type I)}\\ 
Now assume we have an admissible $f$, which implies that $f(\gamma_i)=(x^{2j_i},\beta_2)$. However these elements are contained in the proper subgroup $<(x^{2},0),(e,\beta_2)>$, we see $f$ can not be surjective, a contradiction.\\
 \underline{Cover type II)}\\
If there exists an admissible $f$, we must have $f(\gamma_i)=(x^{2j_i},\beta_2),i=1,2,3,4$, and since $\Pi f(\gamma_i)=1$ it follows that $\mathbb{m}box{Im}(f)\subset <(x^2,0),(e,\beta_2)>$, a contradiction. \\
 \underline{Cover type III-a)}\\
Assume we have an admissible $f$, we see that $f(\gamma_i)=(x^{2j_i},\beta_2),i=1,2,3$. Since $\Pi f(\gamma_i)=1$ it follows that $\mathbb{m}box{Im}(f)\subset <(x^2,0),(e,\beta_2)>$, again a contradiction. \\
 \underline{Cover type III-b)}\\ 
$i)$ $c_3=2$.   We must have $f(\gamma_1)=(x^{2j_1},\beta_2)$, $f(\gamma_2)=(x^{2j_2},\beta_2)$, $f(\gamma_3)=(x^{2h},0)$ or $(yx^k,0)$, $f(\gamma_4)=(x^l,0),l\neq 2h$. If $f(\gamma_3)=(x^{2h},0)$, then $\mathbb{m}box{Im}(f)\subset <(x,0),(0,\beta_2)>$; if $f(\gamma_3)=(yx^k,0)$ we see $\Pi f(\gamma_i)\neq 1$. In both cases we can not get an admissible $f$.\\
$ii)$ $c_3>2$. Given an admissible $f$, we have $f(\gamma_1)=(x^{2j_1},\beta_2)$, $f(\gamma_2)=(x^{2j_2},\beta_2)$, $f(\gamma_3)=(x^{k_3},0)$ and
$f(\gamma_4)=(x^{k_4},0)\ (k_3,k_4\neq 2h)$. One sees immediately that $\mathbb{m}box{Im}(f)\subset <(x,0),(0,\beta_2)>$, a contradiction.
\end{proof}
\end{lemma} 

\begin{lemma}
Cover type $III-c)$ has no admissible f.
\begin{proof}
Assume that we have an admissible $f:T(2,2,2,c_4)\rightarrow G$. \\
Let $(b_1,b_2,b_3,b_4):=(f(\gamma_1),f(\gamma_2),f(\gamma_3),f(\gamma_4))$. We have \\
$(1)$ $b_1^2=b_2^2=b_3^2=1$. Since $b_4\in H$ and $order(b_4)=c_4>2$, we see that $b_4$ must lie in the cyclic group,  say $b_4=x^k$, we also find $n>2$.\\
$(2)$ The fact that $H$ is normal in $G$ implies that $b_ixb_i=x^{k_i},i=1,2,3$, therefore $x^kb_i=b_ix^{kk_i},i=1,2,3$. \\
$(3)$ $b_1b_2b_3b_4=1\Rightarrow$ $b_1b_2=x^{-k}b_3$, moreover $(b_1b_2)^2=x^{-k}b_3x^{-k}b_3=x^{-k-kk_3}$. \\
Any element in $\mathbb{m}box{Im}(f)$ has the form $\Pi_{i=1}^l \beta_i $, where $\beta_i\in \{x^k,x^{-k}, b_1,b_2,b_3\}$. Since $b_1b_2b_3b_4=1$, without loss of generality we can assume $\beta_i\in \{x^k,x^{-k}, b_1,b_2\}$, which means that every element in $\mathbb{m}box{Im}(f)$ is a word in these four elements. \\
Using (2), we can "move" the $x^{\mathbb{P}^1m k}$ terms to the end. Taking $(1)$ into account, we see that the elements are of the forms $(b_1b_2)^sx^t$, $b_2(b_1b_2)^sx^t$ or
 $(b_1b_2)^sb_1x^t$, now use $(3)$, one sees immediately that elements in $\mathbb{m}box{Im}(f)$ have the form $x^j$, $b_1x^j$,  $b_2x^j$ or $b_3x^j$. It turns out that $H\nsubset \mathbb{m}box{Im}(f)$, a contradiction.
\end{proof}
\end{lemma}

From the preceeding, we know that the only group type to consider is Group type I). We denote by $(e,0)$ the neutral element of $D_n \times \mathbb{m}athbb{Z}/2$, where $\mathbb{m}athbb{Z}/2$ is additively generated by $1$.\\

For the action of the braid group on the set of Hurwitz vectors we make use of Lemma 2.1 in \cite{CLP1}.

\begin{lemma}\label{Lemma 2.1}

Every Hurwitz vector of length $r$  with elements in $D_n$ of the form $$v=(v_1,...,yx^a,yx^b,yx^c,...,v_r)$$ is equivalent to $v'=(v_1,...,yx^{a'},yx^{a'},yx^{c'},...,v_r)$ or $v''=(v_1,...,yx^{a'},yx^{b'},yx^{b'},...,v_r)$ via braid moves that only affect the triple $(yx^a,yx^b,yx^c)$. 

\end{lemma}

\begin{lemma} Classification of cover type I)\

In this case the only admissible Hurwitz vector for \textbf{$n$ odd} is

\[ v = ((y,1),(y,1),(yx,1),(yx,1),(e,1),(e,1)).\]
 
For \textbf{$n$ even} (n=2m) there are the following possibilities: 

\[v = ((y,1),(y,1),(yx,1),(yx,1),(e,1),(e,1)),\]

\[v = ((y,1),(yx^{m},1),(yx,1),(yx,1),(x^{m},1),(e,1)),\]

\[v = ((y,1),(yx^{m},1),(yx^2,1),(yx^2,1),(x^{m},1),(e,1)), m ~\mathbb{m}box{odd.}\]

For $n=2$ there are the following:

\[ v = ((y,1),(y,1),(x,1),(x,1),(e,1),(e,1)),\]
\[v = ((y,1),(yx,1),(x,1),(x,1),(x,1),(e,1))\]

\begin{proof}

Since the cover $C/H \to \mathbb{P}^1$ branches in 6 points (cf. [MSSV]) we need a Hurwitz vector with second component equal to $1$. So we have 

\[v = ((y^{k_1}x^{l_1},1),(y^{k_2}x^{l_2},1),(y^{k_3}x^{l_3},1),(y^{k_4}x^{l_4},1),(y^{k_5}x^{l_5},1),(y^{k_6}x^{l_6},1))\] 

The first observation is that the condition $<v> = G$ implies that there must exist $j$, s.t. $k_j = 1$.
Therefore up to automorphism we can assume 

\[v = ((y,1),(y^{k_2}x^{l_2},1),(y^{k_3}x^{l_3},1),(y^{k_4}x^{l_4},1),(y^{k_5}x^{l_5},1),(y^{k_6}x^{l_6},1))\] 

We consider the two cases $n$ odd and $n$ even separately.

\begin{itemize}

\item[i)] \underline{n odd}: Not all $k_j$ can be equal to $1$. Otherwise we cannot generate the element $(y,0)$. Now the only element of order two of the form $(x^l,1)$ in $G$ is $(e,1)$. So because of the product one condition $v$ either looks like

\[v = ((y,1),(yx^{l_2},1),(yx^{l_3},1),(yx^{l_4},1),(e,1),(e,1))\] 

or

\[v = ((y,1),(y,1),(e,1),(e,1),(e,1),(e,1)),\] 

the latter being excluded, since $G\neq <v>$.\\

The product one condition gives $l_2 + l_4 \equiv l_3 \mathbb{m}od n$. The condition $<v> = G$ implies $\gcd(l_2,l_3,l_4,n) = \gcd(l_2,l_4,n) = 1$. Since the second factor $\mathbb{m}athbb{Z}/2$ of $G$ is abelian, we can apply Lemma \ref{Lemma 2.1} to achieve that $l_3 = l_4$. Now $v$ looks like 

\[ v = ((y,1),(yx^{l_2},1),(yx^{l_4},1),(yx^{l_4},1),(e,1),(e,1)) \]

and again by product one we obtain $l_2 \equiv 0 \mathbb{m}od n$ and therefore $1 = \gcd(l_2,l_4,n) = \gcd(l_4,n)$. \\

So we can apply the automorphism $(x^{l_4},0)  \mathbb{m}apsto (x,0) ,(y,0) \mathbb{m}apsto (y,0)$ to $v$ and we can take 

\[ v = ((y,1),(y,1),(yx,1),(yx,1),(e,1),(e,1)) \]

as a Hurwitz vector for the covering $C \to \mathbb{P}^1$.

\item[ii)]\underline{n even}: Recall the general form:

\[v = ((y,1),(y^{k_2}x^{l_2},1),(y^{k_3}x^{l_3},1),(y^{k_4}x^{l_4},1),(y^{k_5}x^{l_5},1),(y^{k_6}x^{l_6},1))\]

Again, first we distinguish the possible Hurwitz vectors by the (even and positive) number of $k_{j}$ that are equal to $1$. We call the element $y^kx^l$ a reflection if $k\equiv 1(\mathbb{m}od2).$\\
In the current case there exists $m = n/2$, which gives the extra order $2$ element $(x^m,1) \in G$. As in the odd case, 6 reflections cannot occur. For the case of 2 reflections, assume, up to ordering,

\[v = ((y,1),(yx^{l_2},1),(x^{l_3},1),(x^{l_4},1),(x^{l_5},1),(x^{l_6},1)).\]

As before, $(l_3,l_4,l_5,l_6) = (0,0,0,0)$ is impossible. In the cases $(l_3,l_4,l_5,l_6) = (m,m,0,0)$ and $(l_3,l_4,l_5,l_6) = (m,m,m,m)$ we get $l_2 = 0$. In the first case we can only have $<v> = G$ if $n=2$. Also in the second case we must have $n=2$ but the elements $(y,1)$ and $(x,1)$ cannot generate $G$ since the element $(e,1)$ is missing. In the cases $(l_3,l_4,l_5,l_6) = (m,m,m,0)$ and $(l_3,l_4,l_5,l_6) = (m,0,0,0)$ we get $l_2 = m$ which also implies that $n=2$. So if $n>2$ these cases don't occur. The corresponding Hurwitz vectors are:

\[v = ((y,1),(y,1),(x,1),(x,1),(e,1),(e,1)),\]
\[v = ((y,1),(yx,1),(x,1),(x,1),(x,1),(e,1))\]
and
\[v = ((y,1),(yx,1),(x,1),(e,1),(e,1),(e,1)),\]

the third one being equivalent to the second one by an automorphism of $G$ that fixes $D_n$.

Assume, for the case of 4 reflections, up to ordering

\[v = ((y,1),(yx^{l_2},1),(yx^{l_3},1),(yx^{l_4},1),(x^{l_5},1),(x^{l_6},1)).\]

Here we have the 3 cases: $l_5 = l_6 = m$, $l_5 = l_6 = 0$ and $l_5 = m$, $l_6 = 0$.\\

In the first 2 cases from the product-one condition we get $l_2 + l_4 \equiv l_3 \mathbb{m}od n$.
To generate $G$ we must have $\gcd(l_2,l_3,l_4,n) = \gcd(l_2,l_4,n) = 1$.

Using Lemma \ref{equivalent HV} again, we arrive at 

\[v = ((y,1),(yx^{l_2},1),(yx^{l_4},1),(yx^{l_4},1),(x^{m},1),(x^{m},1))\]

resp.

\[v = ((y,1),(yx^{l_2},1),(yx^{l_4},1),(yx^{l_4},1),(e,1),(e,1))\]

and so we get $l_2 \equiv 0 \mathbb{m}od n$. Now we have $\gcd(l_2,l_4,n) = \gcd(l_4,n) = 1$ and we can apply the automorphism $x^{l_4}  \mathbb{m}apsto x, y \mathbb{m}apsto y$ to $v$ to arrive at

\[v = ((y,1),(y,1),(yx,1),(yx,1),(x^{m},1),(x^{m},1))\]

resp.

\[v = ((y,1),(y,1),(yx,1),(yx,1),(e,1),(e,1)).\]

Using the morphism $(e,1) \mathbb{m}apsto (x^m,1), (y,0) \mathbb{m}apsto (yx^{-m},0)$ we see that these two are equivalent.

It remains to consider the case $l_5 = m$ and $l_6 = 0$, i.e.

 \[v = ((y,1),(yx^{l_2},1),(yx^{l_3},1),(yx^{l_4},1),(x^{m},1),(e,1)).\]
 
 We apply Lemma 2.1, [CLP1] again and it follows $l_2 = m$. So we get
 
\[v = ((y,1),(yx^{m},1),(yx^{l},1),(yx^{l},1),(x^{m},1),(e,1))\]

where $\gcd(l,m) = 1$.\\

We have two sub cases, i.e. $\gcd(l,n) = 1$ and $\gcd(l,n) = 2$. In the first case we can use the automorphism $x^l \mathbb{m}apsto x, y \mathbb{m}apsto y$ to obtain 
\[v = ((y,1),(yx^{m},1),(yx,1),(yx,1),(x^{m},1),(e,1)).\]

In the second case (where $m$ must be odd) we can achieve

\[v = ((y,1),(yx^{m},1),(yx^2,1),(yx^2,1),(x^{m},1),(e,1)).\]

\end{itemize} 

\end{proof}

\end{lemma}

\begin{lemma}
Classification of cover type II)\\
Up to equivalence, the admissible $f$ is given by the Hurwitz vector:\\
$(1)$ $c_5=2$, 
$$v=((y,1),(yx,1),(yx,1),(e,1),(y,0)),$$
$(2)$ $c_5>2$,
$$v=((y,1),(yx^{-1},1),(e,1),(e,1),(x,0)),c_5=n,$$
$$v=((y,1),(yx^{m-1},1),(x^m,1),(e,1),(x,0)),n=2m,c_5=n,$$
$$v=((y,1),(yx^{m-2},1),(x^m,1),(e,1),(x^2,0)),n=2m,m\ is\ odd,c_5=m,$$
\begin{proof}
 Assume we have an admissible $f:T(2,2,2,2,c_5)\rightarrow D_n\times \mathbb{m}athbb{Z}/2$.\\
we must have:
$$v:=(f(\gamma_1),f(\gamma_2),f(\gamma_3),f(\gamma_4),f(\gamma_5))=((a_1,1),(a_2,1),(a_3,1),(a_4,1),(a_5,0))$$
There are two cases:\\
$(1)$ $\underline{c_5=2}$.\\
As in the previous argument, we do the classification in terms of the number of reflections in $\{a_i\}$, which can be either 2 or 4.\\
$(i)$ There are 2 reflections.\\
$(a)$ $a_5$ is a reflection, W.L.O.G we can assume $a_1$ is another reflection, and $a_1=yx^l,a_5=y$. $a_2,a_3,a_4\in \{e,x^{n/2}$(if n is even)$\}$.\\
There are 4 cases (up to an order change): $\alpha)\ (a_2,a_3,a_4)=(e,e,e)$, $\beta)\ (a_2,a_3,a_4)=(x^{n/2},e,e)$, $\gamma)\ (a_2,a_3,a_4)=(x^{n/2},x^{n/2},e)$, $\delta)\ (a_2,a_3,a_4)=(x^{n/2},x^{n/2},x^{n/2})$.\\
Case $\alpha)$,$\delta)$ we get no admissible $f$ since $f$ can not be surjective.\\
For case $\beta),\gamma)$ (where n is even) we get $f$ is admissible $\iff$ $n=2$.\\
$(b)$ $a_5$ is not a reflection, first we conclude that $n$ must be even and $a_5=x^{n/2}$. Using similar arguments as in $a)$, one finds that
$$v=((y,1),(yx^l,1),(a_3,1),(a_4,1),(x^{n/2},0)),\ a_3,a_4\in \{e,x^{n/2}\}.$$
There are three cases, and one checks easily that in each case $f$ is admissible if and only if $n=2$.\\
$(ii)$ There are 4 reflection.\\
$a)$ $a_5$ is a reflection. W.L.O.G we assume
$$v=((yx^{l_1},1),(yx^{l_2},1),(yx^{l_3},1),(a_4,1),(y,0)),\ a_4\in \{e, x^{n/2}\textit{(if n is even)}\}.$$
Again we apply Lemma \ref{equivalent HV} so that we can assume $l_2=l_3$. Since $f$ is admissible, (using similar arguments as in the previous Lemma,) 
we have:\\
Case $\alpha)$ If $a_4=e$, then $l_1\equiv 0\ (n)$, $\gcd(l_2,n)=1$. Under the automorphism $x^{l_2}\mathbb{m}apsto x$, $y\mathbb{m}apsto y$, we get
$$v\sim((y,1),(yx,1),(yx,1),(e,1),(y,0)).$$
Case $\beta)$ $n=2m$ and $a_4=x^m$. One gets $l_1\equiv m\ (2m)$, and $\gcd(l_2-m,2m)=1$. Using the automorphism $x^{l_2-m}\mathbb{m}apsto x$, $y\mathbb{m}apsto y$, then we can achieve\\
$$v\sim((yx^m,1),(yx^{m+1},1),(yx^{m+1},1),(x^m,1),(y,0)).$$
Using the automorphism (of $G$): $(x,0)\mathbb{m}apsto (x,0)$, $(y,0)\mathbb{m}apsto (y,0)$, $(e,1)\mathbb{m}apsto (x^m,1)$, one finds that Case $\beta)$ is equivalent to Case $\alpha)$.\\
$b)$ $a_5$ is not a reflection.\\
In this case $n$ must be even, and $v=((y,1),(yx^{l_2},1),(yx^{l_3},1),(yx^{l_4},1),(x^{n/2},0))$. It is easy to see that $f$ can not be surjective since $(y,0)$ is not contained in the image.\\
Up to now we have got all the admissible $f$'s for the case $n=2$. (Since $n=2$ implies that $c_5=2$). One checks easily that they are equivalent to each other, since in this case $G$ is abelian.\\
$(2)$ $\underline{c_5>2}$.\\
$a_5$ must lie in the cyclic subgroup, say $a_5=x^k\ (k\neq \frac{n}{2}$ if n is even $)$.  \\
$(i)$ There are 2 reflections, W.L.O.G. we assume
$$v=((y,1),(yx^l,1),(a_3,1),(a_4,1),(x^k,0)),\ a_3,a_4\in \{e,x^{n/2}(if\ n\ is\ even)\}$$
There are 3 cases:\\
Case $\alpha)$ $(a_3,a_4)=(e,e)$.\\
We get $l+k\equiv 0\ (\mathbb{m}od n)$ and $\gcd(k,n)=1$. Applying the automorphism $x^k\mathbb{m}apsto x$, $y\mathbb{m}apsto y$ we get
$$v\sim((y,1),(yx^{-1},1),(e,1),(e,1),(x,0)).$$
Moreover we see that $c_5=n$.\\
Case $\beta)$ $n=2m$ and $(a_3,a_4)=(x^m,e)$.\\
We get $l+k\equiv m\ (\mathbb{m}od 2m)$ and $\gcd(k,m)=1$. \\
If $\gcd(k,n)=1$ (which is the unique case if $2|m$), 
$$v\sim ((y,1),(yx^{m-1},1),(x^m,1),(e,1),(x,0))$$
Here we find $c_5=n$.\\
Otherwise $\gcd(k,n)=2$ (which may happen only when $2\nmid m$),
$$v\sim ((y,1),(yx^{m-2},1),(x^m,1),(e,1),(x^2,0))$$
and we have $c_5=m$.\\
Case $\gamma)$ $n=2m$ and $(a_3,a_4)=(x^m,x^m)$.\\
We get $l+k\equiv 0\ (n)$ and $\gcd(k,n)=1$.
$$v\sim ((y,1),(yx^{-1},1),(x^m,1),(x^m,1),(x,0)),c_5=n$$
Using the automorphism $(x,0)\mathbb{m}apsto (x,0)$, $(y,0)\mathbb{m}apsto (yx^{-m},0)$, $(e,1)\mathbb{m}apsto (x^m,1)$, one finds case $\gamma)$ is equivalent to Case $\alpha)$.\\
$(ii)$ There are 4 reflections.\\
One checks easily that $f$ can not be surjective since $(y,0)\notin \mathbb{m}box{Im}(f)$.
\end{proof}
\end{lemma}

\begin{lemma}
Classification of cover type III-a)\\
We have that $n=2m$ and $d_4=m$. Up to equivalence there is a unique admissible $f$ given by the Hurwitz vector:
$$v=((y,1),(yx^{-1},1),(e,1),(x,1)).$$
\begin{proof}
Assume $f:T(2,2,2,2d_4)\rightarrow D_n\times \mathbb{m}athbb{Z}/2$ is admissible.
$$v:=(f(\gamma_1),f(\gamma_2),f(\gamma_3),f(\gamma_4))=((a_1,1),(a_2,1),(a_3,1),(a_4,1)).$$
$d_4>1\Rightarrow a_5=x^k$ $(k\neq n/2\ if\ n\ is \ even)$.\\
There can only be 2 reflections among $a_1,a_2,a_3$. W.L.O.G. we can assume
$$v=((y,1),(yx^l,1),(a_3,1),(x^k,1)),a_3\in \{e,n/2(if\ n\ is\ even)\}$$
Case $a)$ $a_3=e$.\\
We get $l+k\equiv 0\ (\mathbb{m}od n)$ and $\gcd(k,n)=1$,
$$v\sim ((y,1),(yx^{-1},1),(e,1),(x,1))$$
In this case $2d_4=n$, it turns out that $n$ must be even.\\
Case $b)$ $n=2m$ and $a_3=x^{m}$.\\
We get $l+k\equiv m\ (2m)$ and $\gcd(l,n)=1$,
$$v\sim ((y,1),(yx,1),(x^m,1),(x^{m-1},1)).$$
Using the automorphism $(x,0)\mathbb{m}apsto (x^{-1},0)$, $(y,0)\mathbb{m}apsto (yx^{-m},0)$, $(e,1)\mathbb{m}apsto (x^m,1)$, we find that Case $b)$ is equivalent to Case $a)$.
\end{proof}
\end{lemma}

\begin{lemma}
Classification of cover type III-b)\\
We have that $c_3=2$ and $c_4=n$. Up to equivalence there is a unique  admissible $f$ given by the Hurwitz vector:
$$v=((yx,1),(e,1),(y,0),(x,0)).$$
\begin{proof}
From Example \ref{Ex} we see if that a type $III-b)$ cover has group type $1)$, $c_3$ must be 2, combining with the proof of Corollary  \ref{MSSV Cor} one obtains that the case $(\delta_H,\delta_{H'})=(1,3)$ does not occur. \\
Let $f:T(2,2,2,c_4)\rightarrow D_n\times \mathbb{m}athbb{Z}/2$ be admissible. We must have
$$v:=(f(\gamma_1),f(\gamma_2),f(\gamma_3),f(\gamma_4))=((a_1,1),(a_2,1),(a_3,0),(a_4,0))$$
Since $c_4>2$ we get $a_4=x^k$. It is obvious that there are two (and only two) reflections among $a_1,a_2,a_3$.\\
$(1)$ $a_3$ is not a reflection. $n$ must be even (let $n=2m$) and $a_3=x^m$.
W.L.O.G we assume 
$$v=((y,1),(yx^l,1),(x^m,0),(x^k,0)).$$
It is easy to see $(y,0)\notin \mathbb{m}box{Im}(f)$, therefore in this case there is no admissible $f$.\\
$(2)$ $a_3$ is a reflection. W.L.O.G we assume 
$$v=((yx^l,1),(a_2,1),(y,0),(x^k,0)),a_2\in \{e,n/2(if\ n\ is\ even)\}.$$
$(i)$ $a_2=e$, we get $k\equiv l\ (n)$ and $\gcd(k,n)=1$,
$$v\sim (yx,1),(e,1),(y,0),(x,0)), c_4=n.$$
$(ii)$ $n=2m$ and $a_2=x^m$, we get $k\equiv l+m\ (2m)$, $\gcd(k,n)=1$,
$$v\sim (yx^{m+1},1),(x^m,1),(y,0),(x,0)), c_4=n.$$
Using the automorphism $(x,0)\mathbb{m}apsto (x,0)$, $(y,0)\mathbb{m}apsto (y,0)$, $(e,1)\mathbb{m}apsto (x^m,1)$, we see that Case $(ii)$ is equivalent to Case $(i)$.
\end{proof}
\end{lemma}

\begin{remark} \label{Rk}
If we drop the restriction on $f_H$, it is easy to check that the Hurwitz vectors in $III-a)$ and $III-b)$ are equivalent. 
(Consider the automorphism of $G$: $(x,0)\mathbb{m}apsto (x,1)$, $(y,0)\mathbb{m}apsto (yx,0)$, $(e,1)\mathbb{m}apsto (e,1)$)
\end{remark}

\noindent
Authors' Address: 

\noindent
Binru Li, Sascha Weigl\\
Lehrstuhl Mathematik VIII, Universit\"at Bayreuth\\
Universit\"atsstra\ss e  30, D-95447 Bayreuth\\
E-mail address:\\
binru.li@uni-bayreuth.de\\
sascha.weigl@uni-bayreuth.de

\end{document}