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# On Fundamental Groups of Galois Closures of Generic Projections

On Fundamental Groups of Galois Closures of Generic Projections

## Question 2.14 Is it true

Question 2.14 Is it true that for distinct transpositions τ 1 and τ 2 every irreduciblecomponent of p −1 (R τ1 ) intersects every irreducible component of p −1 (R τ2 ) ?In this thesis we want to compute the fundamental groups π top1 (X gal ) andπ top1 (Xgal aff ). The main result (Theorem 6.2) is that there is always a surjectivehomomorphismπ toptop) ↠ ˜K(π1 (X aff ), n)1 (Xgal affwhere ˜K(−, n) is the group-theoretic construction defined in Section 5.3.Now, if Question 2.14 has an affirmative answer for all topological covers ofthen the kernel of this surjective homomorphism is trivial.X affgal10

3 Semidirect products by symmetric groups3.1 Definition of K( − ,n) and E( − ,n)Ein voller Becher Weins zur rechten ZeitIst mehr wert, als alle Reiche dieser Erde!Dunkel ist das Leben, ist der Tod.Let G be an arbitrary group and n ≥ 2 be a natural number. We denote by θ thepermutation representation of the symmetric group S n on G n given byθ : S n → Aut(G n )σ ↦→ ( θ(σ) : (g 1 , ..., g n ) ↦→ (g σ −1 (1), ..., g σ −1 (n)) )Then we form the split extension of groups with respect to θ1 → G n → G n ⋊ θ S n → S n → 1and denote by s : S n → G n ⋊ θ S n the associated splitting.We define the subgroup E(G, n) of G n ⋊ θ S n to be the group generated byall conjugates of s(S n ) and define K(G, n) to be the intersection G n ∩ E(G, n).Hence we get a split extension1 → K(G, n) → E(G, n) → S n → 1.We give another characterisation of these groups in Proposition 3.3.More generally, let S be a subgroup of S n . Then we defineE(G, n) S := 〈⃗gs(σ)⃗g −1 |⃗g ∈ G n , σ ∈ S〉 ≤ E(G, n)K(G, n) S := E(G, n) S ∩ K(G, n) K(G, n) .These subgroups remain the same when passing to a G n -conjugate splitting. Wewill therefore suppress s in future. Clearly, K(G, n) S is always a normal subgroupof G n and K(G, n).In the notation introduced in Definition 2.10 we have the following equalitiesand isomorphisms:E(G, n) Sn = E(G, n)E(G, n) (i) S n−1∼ = E(G, n − 1) for n ≥ 3and similarly for K(−, n).Later on we have to deal with subgroups of K(G, n) that are generated byK(G, n)-conjugates of a subgroup S of S n rather than G n -conjugates. Fortunately,we have the following11

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