Let us that the group Gis the free product of two finite groups G1 and G2, for which we use the notationG=G1∗G2.
B.6.1 Proposition. The sequence
0→R[G]−→α R[G/G1]⊕R[G/G2]−→² R→0 withα(g) = (gG1,−gG2)and²(gG1,0) = 1 =²(0, gG2)is exact.
For the proof, which is completely elementary, we follow [Bieri]. LetGbe a group and M an R[G]-module. Recall that a mapd:G→M is called a derivation if
d(xy) =d(x) +xd(y)
(these are precisely the1-cochains of the standard resolution!). Denote by²the mapR[G]→Rgiven byg7→1. Then it is easy to check thatdextends to anR-linear map
d:R[G]→M, satisfyingd(ab) =d(a)²(b) +ad(b).
LetFbe a free group on generators{fi}. Since there are no relations, it is clear that the assignment suppose thatπ :F ³ Gis a free presentation of the group Gby the free groupF discussed so far.
We denoteπ(fi)bygi and extendπlinearly toπ :R[F]→R[G]. From the above, we immediately
Proof of Proposition B.6.1. Suppose thatG1 is generated by the (minimal) set{xi}andG2 by {yj}. Let F be the free group on symbols{xi, yj}so thatπ : F ³ G is given byxi 7→ xi and yi7→yi.
Clearly,²is surjective and also²◦α= 0. Next we compute exactness at the centre. The image of Equation 2.20 inR[G/G1] =R[G]/(R[G](1−h)|h∈G1)is and hence the exactness at the centre.
It remains to prove thatαis injective. Note that we have not yet used the freeness of the product;
the discussion above would remain valid if there were additional relations inG. Now we do use the freeness, namely as follows. Every element16=g∈Ghas a unique representation as products of the formg = a1b2a3b4· · ·ak−1bk, org = a1b2a3b4· · ·bk−1akwhere either all theai ∈ G1− {1}and allbj ∈G2− {1}or all theai ∈G2− {1}and allbj ∈G1− {1}. The integerkis defined to be the length ofg, denoted byl(g). We letl(1) = 0. For cosetsgG1 ∈G/G1we letl(gG1) :=l(g)wheng is represented by a product as above ending in an element ofG2. We definel(gG2)similarly.
Let λ = P
waww ∈ R[G] be an element in the kernel of α. Hence, P
wawwG1 = 0 = P
wawwG2. Let us assume that λ 6= 0. It is clear thatλcannot just be a multiple of 1 ∈ G, as otherwise it would not be in the kernel ofα. Now pick theg∈Gwithag6= 0having maximal length l(g) (among all thel(w)withaw 6= 0). It follows thatl(g) >0. Assume without loss of generality that the representation ofgends in a non-zero element of G1. Thenl(gG2) = l(g). Further, since ag 6= 0and0 = P
wawwG2, there must be anh ∈ Gwithg 6=h, gG2 =hG2 andah 6= 0. Asg does not end inG2, we must haveh=gyfor some06=y∈G2. Thus,l(h)> l(g), contradicting the
maximality and proving the proposition. 2
B.6.2 Proposition. (Mayer-Vietoris) Let M be a left R[G]-module. Then the Mayer-Vietoris se-quence gives the exact sese-quences
0→MG →MG1 ⊕MG2 →M →H1(G, M)→H1(G1, M)⊕H1(G2, M)→0.
and for alli≥2an isomorphism
Hi(G, M)∼=Hi(G1, M)⊕Hi(G2, M).
Proof. We see that all terms in the exact sequence of Proposition B.6.1 are freeR-modules. We now apply the functorHomR(·, M) to this exact sequence and obtain the exact sequence of R[G]-modules
0→M →HomR[G1](R[G], M)⊕HomR[G2](R[G], M)→HomR(R[G], M)→0.
The central terms, as well as the term on the right, can be identified with coinduced modules. Hence, the statements on cohomology follow by taking the long exact sequence of cohomology and invoking
Shapiro’s Lemma B.4.1. 2
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Accepted for publication in J. für die reine und angewandte Mathematik.
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[W2] G. Wiese. On modular symbols and the cohomology of Hecke triangle surfaces.
Preprint.arXiv:math.NT/0511113.
Gabor Wiese
NWF I - Mathematik Universität Regensburg D-93040 Regensburg Germany
E-mail:gabor@pratum.net
Webpage:http://maths.pratum.net/