# Poincaré duality and Mayer–Vietoris sequence

In his article, Davis states that if a $$n$$-dimensional Poincaré duality group $$G$$ splits along a subgroup $$C$$ then the cohomological dimension of $$C$$ must be $$n-1$$. I am struggling to understand why this must be the case. The idea is to use Mayer–Vietoris sequence. Suppose $$\operatorname{cd}(C). I could only get as far as concluding that $$H_{n}(G)\simeq H_{n}(A)\oplus H_{n}(B)$$. But I don't see how this leads to a contradiction. I would greatly appreciate any help with regards to this.

• What article do you mean to link? Your link does not resolve. Commented Jul 19 at 14:01
• Do you know how to prove that an infinite index subgroup of a PD(n) group has $cd<n$? Commented Jul 19 at 14:37
• The result that @MoisheKohan refers to is sometimes called Strebel's theorem. See R. Strebel. A remark on subgroups of infinite index in Poincaré duality groups. Comment. Math. Helv., 52(3):317–324, 1977.
– HJRW
Commented Jul 19 at 16:15
• If I remember correctly, it is also in Ken Brown's book. Commented Jul 19 at 16:39
• @LSpice It should open now. Commented Jul 19 at 20:12

$$\DeclareMathOperator\PD{PD}\DeclareMathOperator\cd{cd}$$You need two ingredients which Davis assumes the reader is aware of:
1. Strebel's theorem: If $$G$$ is a $$\PD(n)$$ group, then every infinite index subgroup $$H< G$$ satisfies $$\cd(H). Ken Brown in his book "Cohomology of groups" gives this theorem as an exercise with a detailed hint (chapter 8, section 10). You can also read the original Strebel's paper "A remark on subgroups of infinite index in Poincaré duality groups".
2. Suppose that a group $$G$$ admits a nontrivial graph of groups decomposition. Then every vertex group of this decomposition has infinite index in $$G$$. This is an easy consequence of Bass–Serre theory of group actions on trees (if a finite index subgroup in $$G$$ has a fixed vertex in the Bass–Serre tree $$T$$, then $$G$$ has a finite orbit in $$T$$, hence, has a fixed vertex in $$T$$, contradicting nontriviality of the graph of groups).
Now, combine these two facts with the MV sequence; you will see that if $$G=A*_C B$$ is a nontrivial amalgam, and $$\cd(C) and $$G$$ is a $$\PD(n)$$ group, then for every $$\mathbb ZG$$-module $$M$$ we have $$H^n(G, M)=0$$, which is a contradiction with the assumption that $$G$$ is a $$\PD(n)$$ group.
• Ah, of course. I confused myself by applying the theorem to $C$ instead of to $A$ and $B$. Commented Jul 20 at 0:11