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# On Fundamental Groups of Galois Closures of Generic Projections

On Fundamental Groups of Galois Closures of Generic Projections

## So we obtain the

So we obtain the following commutative diagramK d,n → Kn,d∗↓ ↓ ∼ =K(F d , n) ↩→ F dnWe already know that the map from K d,n to K(F d , n) is surjective. To show thatit is also injective it is enough to prove that the homomorphism from K d,n to Kd,n∗is injective.To achieve this we define a series of groups lying in between K d,n and Kd,n ∗ :We define K ≤kd,n to be the group generated by K d,n and the elements t a with a ≤ ksubject to the relations (†1) and (†2). Of course, only those relations that involvet a ’s and t b ’s with a, b ≤ k are imposed. Hence we obtain the following groupsand homomorphismsK n,d = K ≤0n,d → K≤1 n,d→ ... → K≤dn,d = K∗ n,d .By Lemma 5.8 each of these homomorphisms is injective and so the compositehomomorphism from K n,d to Kn,d ∗ is injective.□Lemma 5.7 The mapµ : K ∗ n,d → F dndefined in the proof of Proposition 5.6 is an isomorphism of groups.It is clear that µ defines a surjective homomorphism.We define a map ˆµ viaPROOF.ˆµ : F dn→ K ∗ n,df an↦→ t af ai↦→ t a · f ainIf we can show that ˆµ defines a homomorphism of groups it will be the inverse ofµ and it follows that µ is an isomorphism.The group F d n is generated by the elements f a i with a = 1, ..., d and i =1, ..., n subject to the commutator relations [f a i , f b j ] = 1 for all i ≠ j .First we establish two further sets of relations that hold true inside K ∗ d,n :in f a · t b · (f in a ) −1 =−1 t a · t b · t a i ≠ n (†3)in jn f a · f b · (f in a ) −1 =−1 jn t a · f b · t a i ≠ j, and i, j ≠ n (†4)The relation (†3) can be seen by applying (†2) to the right hand side of (†1) withj = n. The relation (†4) is only a reformulation of (†1).48

First, assume that i = n. ThenNow assume that i ≠ n. Thenˆµ([f a i , f b j ])ˆµ([f a n , f b j ])jn −1 = t a · t b f b · t a · (t b f jn b ) −1jn −1 = t a · t b f b · t a · (f jn b ) −1} {{ }apply (†3)= t a t b · t −1 b t −1 −1a t b · t b= 1t b−1in jn = t a f a · t b f b · (t a f in a ) −1 · (t b f jn b ) −1= t a f in a t b · ((f in a ) −1 f} {{ }in a ) · f jn b (f in a ) −1 t −1 a (f jn b ) −1 · (t a t} {{ }−1 −1a ) · t bapply (†3)apply (†4)= t a · t −1 a t b t a · f in a f jn b (f in a ) −1 · f in a (f jn b ) −1 (f in a ) −1 · t −1 −1 a t b= 1Hence ˆµ defines a homomorphism and so we are done.□Lemma 5.8 Keeping the notations introduced in the proof of Proposition 5.6there is an isomorphismK ≤kn,d∼ =K≤k−1n,d⋊ ¡where the infinite cyclic ¡ group is generated by t k . In particular, the map fromto K ≤kn,dconsidered in the proof of Proposition 5.6 is injective.K ≤k−1n,dPROOF.We want to define a map from K ≤k−1n,dto itself viaϑ : K ≤k−1n,d→ K ≤k−1n,df aij↦→ f knm · f aij · (f k nm ) −1 m ≠ i, j, nt a ↦→ f knm · t a · (f k nm ) −1 m ≠ nFirst we have to show that ϑ does not depend on the choice of m in the definition ofϑ: For the definition of ϑ(f a ij ) this means we have to check that for m, m ′ ≠ i, j, nf knm · f aij · (f k nm ) −1 = f knm ′ · f aij · (f knm ′ ) −1holds true. If i, j ≠ n then both expressions are equal to f ij a by relation (∗4). Ifnmi = n then we conjugate this expression with f ′k and after applying (∗2) wemmare done since f ′ k and f ij a commute using (∗4).49

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