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Given a pointed topological space $(X,x_0)$, let $p\colon (\tilde{X}, \tilde{x}_0)\to (X,x_0)$ be a covering of that space. Write $p^{-1}(x_0)= \{\tilde{x}_0, \tilde{x}_1,\ldots,\tilde{x}_n\}$. I'd like to show that there is a deck transformation $T$ s.t. $T\tilde{x}_k=\tilde{x}_{k+1}$ if and only if $p_*(\pi_1(X,x_0))$ is normal and the group of deck transformations is cyclic.

If $p_*(\pi_1(\tilde{X},\tilde{x}_0))$ is normal and the group of deck transformations is cyclic, must we have $p_*(\tilde{X},\tilde{x}_0)=deck(X,p)$?

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  • $\begingroup$ You mean $p_* (\pi_1(\bar{X},\bar{x}_0)$? $\endgroup$ – Espen Nielsen Jan 13 '15 at 11:38
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Let us denote $G=\pi_1(X,x_0)$ and $H=p_*(\pi_1(\bar{X},\bar{x}_0)$.

The first statement, about the cyclicity of the group $\text{deck}(X,p)$ of deck transformations, follows when you observe that $\text{deck}(X,p)$ acts transitively on $p^{-1}(x_0)$ if and only if $H$ is normal in $G$. (It is important to note where a subgroup is normal, as we will see!)

As for your second question, note that $\text{deck}(X,p)$ is isomorphic to $N(H)/H$, where $N(H)$ is the normalizer of $H$ in $G$, defined as the largest intermediate subgroup $H\subseteq N(H)\subseteq G$ such that $H$ is normal in $N(H)$. There is an explicit formula for $N(H)$ here. Therefore, if $H$ is already normal in $G$, we have $N(H)=G$, and $\text{deck}(X,p)(\bar{X})=G/H$. You can see that your second statement is false in general. For example, let $G=S_3$, the symmetric group on 3 variables, and let $H$ be $\mathbb{Z}/3\mathbb{Z}$. $H$ is then a normal subgroup, and the group of deck transoformations is $G/H=\mathbb{Z}/2\mathbb{Z}$ which is cyclic.

If order for the condition $H\simeq \text{deck}(X,p)$ to happen, you need $N(H)/H\simeq H$, which can happen, for example when $G$ is either a product $G\simeq H\times H$, or a semidirect product $G\simeq H \rtimes H$.

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