Assume that \(X\) is a path-connected, locally simply-connected space with fundamental group isomorphic to \(\mathbb{Z}_2 \times \mathbb{Z}_2\). How many path-connected covering spaces of \(X\) are there, up to equivalence (isomorphism)?
Solution
1. Connected covering spaces of \(X\) up to equivalence correspond to conjugacy classes of subgroups of\[ \pi_1(X)\cong(\mathbb Z/2)^2. \]
This is the standard classification of connected coverings for path-connected, locally simply-connected spaces.
2. Since the group is abelian, conjugacy classes of subgroups are just subgroups themselves.
Every subgroup is fixed under conjugation in an abelian group.
3. The group \((\mathbb Z/2)^2\) has exactly five subgroups: the trivial subgroup, the whole group, and its three distinct subgroups of order \(2\).
There are three nonzero elements, and each generates a distinct order-\(2\) subgroup. There are no other possible subgroup orders by Lagrange’s theorem.
4. Hence\[ \boxed{5} \]path-connected covering spaces occur up to equivalence.
Apply <1>1–<1>3. The whole group corresponds to the identity cover and the trivial subgroup to the universal cover.