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 $n1$. I am struggling to understand why this must be the case. The idea is to use Mayer–Vietoris sequence. Suppose $\operatorname{cd}(C)<n1$. 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.

1$\begingroup$ What article do you mean to link? Your link does not resolve. $\endgroup$– LSpiceCommented Jul 19 at 14:01

2$\begingroup$ Do you know how to prove that an infinite index subgroup of a PD(n) group has $cd<n$? $\endgroup$– Moishe KohanCommented Jul 19 at 14:37

3$\begingroup$ 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. $\endgroup$– HJRWCommented Jul 19 at 16:15

1$\begingroup$ If I remember correctly, it is also in Ken Brown's book. $\endgroup$– Moishe KohanCommented Jul 19 at 16:39

$\begingroup$ @LSpice It should open now. $\endgroup$– Harsh PatilCommented Jul 19 at 20:12
1 Answer
$\DeclareMathOperator\PD{PD}\DeclareMathOperator\cd{cd}$You need two ingredients which Davis assumes the reader is aware of:
Strebel's theorem: If $G$ is a $\PD(n)$ group, then every infinite index subgroup $H< G$ satisfies $\cd(H)<n$. 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".
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 $\cd(C)<n1$ 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.

1$\begingroup$ Ah, of course. I confused myself by applying the theorem to $C$ instead of to $A$ and $B$. $\endgroup$ Commented Jul 20 at 0:11
