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Suppose I have a quotient $X \to S$ by a finite abelian group $G$ action (I have several cases, but in all of them the group $G$ and the action could be written explicitly), where $X,S$ are surfaces (smooth and path connected) over the complex numbers field. I also know $\pi_1(X)$, which is $\mathbb{Z}^4$ in my case and that the covering $X \to S$ is unramified. Is this data enough to compute $\pi_1(S)$? As far as I understand, it is known from general theory of coverings, that there is an injection $\pi_1(X) \to \pi_1(S)$, so I believe that $\pi_1(S)$ should be of the form $\pi_1(X) \rtimes G$, but I do not know what can I use to prove it (if this is correct).

Any input or advice is appreciated, thank you in advance.

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    $\begingroup$ Regular covers correspond to extensions, but not necessarily split ones (=semidirect products). If a group $G$ (not necessarily finite or abelian) acts freely and properly on a reasonable path connected space $X$ with quotient $Y$, then you have an extension $1 \rightarrow \pi_1(X) \rightarrow \pi_1(Y) \rightarrow G \rightarrow 1$. This can be found in chapter 1 of Hatcher. $\endgroup$
    – Andy Putman
    Commented Jan 12 at 4:54

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