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When (n=2), we remark that the finite index condition can be omitted, since the finiteness of the (L^{2}) norm of the second fundamental form implies that M has finite index, which was proved by Bérard et al. [11].
Suppose M has finite index.
The idea iterates to cover all types of finite index.
Hence the iteration must stop at some finite index k.
Let J be a countable (or finite) index set.
Let I be a finite index set and (iin I).
where and are two finite index sets, and.
Let Λ = { 1, …, ℓ } be a finite index set.
Let Λ ^ 1 be a finite index subgroup of Λ1and Λ ^ 2 a finite index subgroup of Λ2.
Suppose Γ1is a finite index normal subgroup of Π1and Γ2is a finite index normal subgroup of Π2such that f× Γ1) ⊂ Γ2and g×(Γ1) ⊂ Γ2.
Let (I_{m} ) denote the finite index set (lbrace1,2,ldots,m rbrace).
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com