Let $G$ be generated by $r$ elements and choose $h\geq\max\{2,r\}$. The <fundamental group of a closed orientable surface> has the presentation
$$
\pi_1(S_h)=
\langle a_1,b_1,\ldots,a_h,b_h\mid
\textstyle\prod_{i=1}^h[a_i,b_i]=1\rangle.
$$
Mapping $a_1,\ldots,a_r$ to generators of $G$ and every $b_i$ and remaining $a_i$ to the identity defines a <surjective group homomorphism> $\pi_1(S_h)\twoheadrightarrow G$.

Its kernel determines a connected <regular covering> $S\to S_h$ with <deck transformation group> $G$. The action is free and orientation preserving. The covering surface has genus at least two, and the lifted hyperbolic metric makes every deck transformation an isometry. Part a now gives $G\hookrightarrow\operatorname{Mod}(S)$. This is the <realization of a finite group as a surface deck group>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 156 / 1 / b / Solution

{wiki=Covering_space#Regular_covering}

A connected covering is regular when its deck transformation group acts transitively on every fibre. Under the classification of coverings, this is equivalent to the corresponding subgroup of the fundamental group being normal.


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