For the definition of ϑ(t a ) we have to check that for m, m ′ ≠ nf knm · t a · (f k nm ) −1 = f knm ′ · t a · (f knm ′ ) −1nmholds true. We conjugate by f ′ mm k and then we are done since t a and f ′kcommute by (†1).Hence the definition of ϑ does not depend on the choice of the m’s occuring.Next we want to show that ϑ defines an endomorphism of K ≤k−1n,d. For this wehave to show that the relations are preserved by ϑ. If we pick a relation from (∗1)to (∗4), (†1) and (†2) then we can find an index m distinct from the i, j, k, n’s inthis particular relation since we assumed n ≥ 5. The action of ϑ is then given byconjugating every element occuring in this relation by f nm k . Since the relationsform a normal subgroup this means that ϑ preserves the relations of K ≤k−1n,dandso ϑ defines an endomorphism of this group.Clearly, ϑ defines an automorphism of K ≤k−1n,dfor we can just define its inverseby replacing f nm k by (fk nm ) −1 in the definition of ϑ.To obtain K ≤kn,dfrom K≤k−1n,d∗ 〈t k 〉 we only need the relations (†1) and (†2) .For (†1) it is enough to consider all relations with a = k and arbitrary b:t k f b ij t k−1= f k nm f b ij (f k nm ) −1 = ϑ(f b ij ).We have to impose one relation for every m ≠ i, j but we have already shownabove that all these elements define the same element ϑ(f ij b ) of K ≤k−1n,d.And for a = k and b < k the relation (†2) is equivalent to−1t k t b t k = f nm k f nj b (f nm k ) −1 (f nj b ) −1 ·t} {{ } bapply (†4)= f mn k t b (f mn k ) −1 = ϑ(t b )As we have shown above this element does not depend on the choice of m ≠ n.Hence we have shown thatK ≤kn,d∼ =K≤k−1n,d∗ 〈t k 〉/ ≪ t k xt k −1 = ϑ(x) ∀x ∈ K ≤k−1n,d≫and this is precisely the semidirect product of K ≤k−1n,dby 〈t k 〉. □5.3 Affine subgroups and the construction of ˜K( − ,n)We denote by F d be the free group of rank d ≥ 1. We embed K(F d , n) as usualinto F d n , cf. Section 3.1.Definition 5.9 A subgroup of K(F d , n) with n ≥ 3 is called an affine subgroupif it is normally generated by elements of the form (r, r −1 , 1, ..., 1), r ∈ F d andtheir S n -conjugates.50

We note that for affine subgroups normal generation with respect to K(F d , n) hasthe same effect as normal generation with respect to F d n : This follows since weassumed n ≥ 3 and so we compute for (f, 1, f −1 , 1, ...) ∈ K(F d , n) and r ∈ F d :(f, 1, 1, ..., 1) (r, r −1 , 1, ..., 1) (f, 1, 1, ..., 1) −1= (f, 1, f −1 , ..., 1) (r, r −1 , 1, ..., 1) (f, 1, f −1 , ..., 1) −1 .We let G be a group and n ≥ 3 be a natural number. We then choose a presentationF d /N ∼ = G of G. Then we define R :=≪ K(N, n) ≫. This is an affine subgroupof K(F d , n) since it is normally generated by the elements (s, s −1 , 1, ..., 1) withs ∈ N and their S n -conjugates. We define˜K(G, n) := K(F d , n)/R.Since R is S n -invariant the S n -action on K(F d , n) descends to an action on thequotient ˜K(G, n) and we defineẼ(G, n) := ˜K(G, n) ⋊ S n .with respect to this action. This is well-defined because ofTheorem 5.10 Let n ≥ 3 be a natural number. For every finitely generated groupG the construction of ˜K(G, n) and its S n -action do not depend on the choice of apresentation for G. Moreover, the construction of ˜K(−, n) is functorial in its firstargument.If we denote by H 2 (G) the second group homology of G with coefficients inthe integers then there is a central extension0 → H 2 (G) → ˜K(G, n) → K(G, n) → 1and the image of H 2 (G) lies inside the commutator subgroup of ˜K(G, n).PROOF. We embed K(F d , n) into F n nd . We denote by π the projection from F donto its last n − 1 factors. From Proposition 3.4 we know that ker π restricted toK(F d , n) equals the commutator subgroup [F d , F d ].We let f ∈ F d and s ∈ N. Then[(f, 1, f −1 , 1, ...), (s, s −1 , 1, ...)] = ([f, s], 1, 1, 1, ...)and this element lies in R. Thus [F d , N] is contained in R ∩ ker π.Conversely, R is generated by elements of the form (fsf −1 , s −1 , 1, ...) andtheir S n -conjugates where f runs through F d and s runs through N. From this itfollows that every element of R can be written as a product of the form∏(([fi , s i ], 1, ...) · (s i , 1, ..., s −1 i , 1, ...) ) .i(f i s i f i −1 , 1, ..., s i −1 , 1, ...) = ∏ i51