2 years ago

On Fundamental Groups of Galois Closures of Generic Projections

On Fundamental Groups of Galois Closures of Generic Projections

PROOF. If G is abelian

PROOF. If G is abelian (resp. finite, nilpotent, solvable) then so is K(G, n) beinga subgroup of G n . If G is perfect then K(G, n) ∼ = G n which is perfect.By the first exact sequence of Proposition 3.4 there exists a surjective homomorphismfrom K(G, n) onto G. So if K(G, n) is abelian (resp. finite, nilpotent,perfect, solvable) then so is G being a quotient of K(G, n).□We finally give some basic functoriality properties of our construction:Proposition 3.8 Let n ≥ 2 be a natural number and let G 1 , G 2 , G be arbitrarygroups.1. If G 1 → G 2 is an injection then so is K(G 1 , n) → K(G 2 , n).2. If G 1 → G 2 is a surjection then so is K(G 1 , n) → K(G 2 , n).3. If G is a semidirect product then so is K(G, n). However, the functorK(−, n) is not exact in the middle as Example 3.11 and Example 3.13 show.4. K(G 1 × G 2 , n) = K(G 1 , n) × K(G 2 , n).5. If G is an abelian group thenK(G, n) tors∼ = K(Gtors , n)K(G, n) ⊗¦§¥ ∼ = K(G ⊗¦§¥ , n)where − tors denotes the torsion subgroup of an abelian group.6. For n ≥ 3 the natural homomorphism from K(G, n) ab onto K(G ab , n) is anisomorphism. The assumption n ≥ 3 is needed as Example 3.12 shows.PROOF. We assume that 1 → K → G → Q → 1 is exact. Then also the inducedsequence 1 → K n → G n → Q n → 1 is exact. This induces homomorphisms(notation as in the beginning of this section)K n ⋊ θ S n → G n ⋊ θ S n → Q n ⋊ θ S nand induces injections E(K, n)↩→E(G, n) and K(K, n)↩→K(G, n). This provesthe first assertion (we do not need the normality of K in G in this step). Thegroup E(Q, n) is generated by S n and commutators [q, σ], q ∈ Q. Since G↠Qis surjective we see that E(G, n)↠E(Q, n) is surjective since we can lift elementsof S n and commutators. Similarly we see that K(G, n)↠K(Q, n) is surjective.If G is a semidirect product then there exists a split surjection G↠Q. Thismap induces a split surjection K(G, n)↠K(Q, n). Therefore also K(G, n) is asemidirect product.16

The assertions about the torsion and the free part of an abelian group followimmediately from Corollary 3.5.The surjection G↠G ab and the universal property of abelianisation imply thatthere is a natural surjective homomorphism K(G, n) ab ↠K(G ab , n):1 → K(G, n) → G n → G ab → 1↓ ↓ ab ||1 → K(G ab , n) → (G ab ) n → G ab → 1 .An element of the kernel K(G, n) → K(G ab , n) is also an element of the kernelof G n → (G ab ) n which is [G, G] n . Since we assumed n ≥ 3 every commutator(1, ..., 1, [h 1 , h 2 ], 1..., 1) lies not only in K(G, n) but is even a commutator of elementsof K(G, n), cf. the proof of Proposition 3.3. This implies that the kernelof K(G, n) → K(G, n) ab is the commutator subgroup of K(G, n). Hence thecanonical homomorphism from K(G, n) ab onto K(G ab , n) is an isomorphism forn ≥ 3.□3.3 UniversalityWe assume that we are given a group X and a homomorphism ϕ : S n → Aut(X)with n ≥ 3. Then we form the semidirect product1 → X → X ⋊ ϕ S n → S n → 1.We consider S n as a subgroup of the group in the middle via the associated splitting.For a subgroup S ≤ S n we denote [X, S] by X S . Again, X S is a normalsubgroup of X and does not change if we pass to an X-conjugate splitting.Proposition 3.9 Let ϕ : S n → Aut(X), n ≥ 3 be a homomorphism and let1 → X → X ⋊ ϕ S n → S n → 1be the split extension determined by ϕ. If we defineY := X Sn /X S(1)n−1then there exists a commutative diagram with exact rows1 → X Sn → X Sn ⋊ ϕ S n → S n → 1↓ ↓ ‖1 → K(Y, n) → K(Y, n) ⋊ θ S n → S n → 1where all homomorphisms downwards are surjective. Moreover, we have an exactsequence1 → ⋂ ni=1 X S (i)n−1→ X Sn → K(Y, n) → 1.17

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