# Regular Covering Spaces

Recall that if $G$ is a group and $H$ is a subgroup then $H$ is said to be a normal subgroup of $G$ if for every $g \in G$ we have that:

(1)Where the sets $gH$ and $Hg$ are defined as:

(2)Also recall from The Fundamental Group of Spaces and Their Covers page that if $X$ is a topological space and $(\tilde{X}, p)$ is a cover of $X$, $x \in X$ and $\tilde{x} \in p^{-1}(x)$ then the induced map $p_* : \pi_1(\tilde{X}, \tilde{x}) \to \pi_1(X, x)$ is an injective homomorphism, and so $p_*(\pi_1(\tilde{X}, \tilde{x}))$ is a subgroup of $\pi_1(X, x)$. When this subgroup is normal we give it a special name.

Definition: Let $X$ be a topological space and let $(\tilde{X}, p)$ be a covering space of $X$. Let $x \in X$ and let $\tilde{x} \in p^{-1}(x)$. Then $(\tilde{X}, p)$ is said to be a Regular Cover of $X$ if $p_*(\pi_1(\tilde{X}, \tilde{x})$ is a normal subgroup of $p_*(\pi_1(X, x))$. |

For example, if $(\tilde{X}, p)$ is a universal cover of $X$ then $p_*(\pi_1(\tilde{X}, \tilde{x}))$ is the trivial group which is trivially a normal subgroup of $\pi_1(X, x)$. Hence, every universal cover is a regular cover.

The following theorem gives us a criterion for determining when a covering space is regular or not.

Theorem 1 (Regular Cover Criterion): Let $X$ be a topological space and let $(\tilde{X}, p)$ be a covering space of $X$. Then $(\tilde{X}, p)$ is a regular covering space of $X$ if and only if for every $x \in X$ and for every $\tilde{x_1}, \tilde{x_2} \in p^{-1}(x)$ there exists a covering transformation $f \in A(\tilde{X})$ for which $f(\tilde{x_1}) = \tilde{x_2}$. |

Theorem 1 gives us a nice characterization for regular covers that are easy to visualize. Theorem 1 implies that a cover space is regular if and only if for every $x \in X$ the covering space is symmetric locally for each point in $p^{-1}(x)$.

Proposition 1: Let $X$ be a topological space and let $(\tilde{X}, p)$ be a covering space of $X$. Let $x \in X$ and let $\tilde{x} \in p^{-1}(x)$. If $p_1(X, x)$ is an abelian group then then $(\tilde{X}, p)$ is a regular cover of $X$. |

**Proof:**Observe that every subgroup of an abelian group is a normal group. In particular, $p_*(\pi_1(\tilde{X}, \tilde{x}))$ is a normal subgroup of $\pi_1(X, x)$ and so $(\tilde{X}, p)$ is a regular cover of $X$. $\blacksquare$