# Cohomological dimension of knit products

Let $G$ be a group with complementary subgroups $A$ and $B$ (meaning $A\cap B=1$ and $AB=G$).

If $A$ and $B$ are both normal in $G$, then $G\cong A\times B$ is a direct product. If $A$ is normal, then $G\cong A\rtimes_{\varphi} B$ is a semi-direct product. More generally, $G\cong A\bowtie B$ is a knit product, or Zappa-Szép product, or general product of $A$ and $B$.

My question concerns the cohomological dimension of such knit products. To ask a specific question:

Does anyone know of any examples where $\operatorname{cd}(A\bowtie B)<\operatorname{cd}(A\times B)$?

(Edit: Other than free products $A\ast B$.)

More generally I would be interested in any references concerning the cohomology of knit products. Is there any weaponry akin to the Lyndon-Hochschild-Serre spectral sequence?

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Is there even a reason why cd(A\bowtie B) should be bounded by cd(A×B)? –  Dieter May 31 '13 at 13:10
Mark: Maybe you meant the opposite inequality? –  Misha May 31 '13 at 16:50
@Misha: I was tacitly assuming that $\mathrm{cd}(A\bowtie B)\leq \mathrm{cd}(A) + \mathrm{cd}(B)$ and asking for examples where the inequality is strict. Now it seems from Dieter's update that this inequality is not true in general. –  Mark Grant Jun 17 '13 at 15:24

First of all, there are examples for semi-direct products, but maybe this is not what you want.

1) Let $F_m$ be the free group on $m$ generators. Given a surjection $\varphi: F_m \rightarrow F_n$, we obtain a short exact sequence $$1 \rightarrow N \rightarrow F_m \xrightarrow{\varphi} F_n \rightarrow 1.$$ Here, $N$ is free and $F_m \cong N\rtimes F_n$.

Hence, $\mathrm{cd}(N\rtimes F_n)=1$, but $\mathrm{cd}(N\times F_n)=2$.

2) Stallings example: Let $F_2$ be the free group on two generators and consider the short exact sequence $$1 \rightarrow H \rightarrow F_2 \times F_2 \rightarrow \mathbb{Z} \rightarrow 1$$ obtained by sending the generators of $F_2$ to 1, in both factors. Then $H$ is not a free group and $F_2 \times F_2\cong H\rtimes \mathbb{Z}$.

One has $\mathrm{cd}(H\rtimes \mathbb{Z})=2$, but $\mathrm{cd}(H\times \mathbb{Z})=3$.

Secondly, a knit product of two torsion-free groups does not have to be torsion-free. Hence, one cannot expect the inequality $\mathrm{cd}(A\bowtie B)\leq \mathrm{cd}(A)+\mathrm{cd}(B)$ to hold in general.

It is not difficult (but a little tedious) to construct an example of a group of the form $\mathbb{Z}\bowtie \mathbb{Z}$, with an element of order two. Keep in mind for this that the knit products $\mathbb{Z}\bowtie \mathbb{Z}$ have been classified (see this MathStack post for some references http://math.stackexchange.com/questions/107781/has-this-generalized-semidirect-product-been-studied/107788#107788). They always fit in a short exact sequence $$0 \rightarrow \mathbb{Z}^2 \rightarrow \mathbb{Z} \bowtie \mathbb{Z} \rightarrow \mathbb{Z}_2 \oplus \mathbb{Z}_2 \rightarrow 0.$$

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This is great, thanks. Do you know of any explicit examples for your last paragraph? –  Mark Grant Jun 17 '13 at 15:26

First of all, the free product of nontrivial groups is never a knit product of these groups.

Secondly, a similar question for short exact sequences of groups was already asked in this MO post, with examples similar to Dieter's.

The third remark is that, in view of the examples, the natural inequality to expect is subadditivity: $$cd(A\bowtie B) \le cd(A) + cd(B).$$ I do not know how to prove this inequality in general (apart from semidirect products). Something might follow from this paper where the related concept of "factorable group" is considered and the author did some homological computations (I did not look at the details though).

Suppose, however, that instead of cohomological dimension $cd(G)$ of groups, one considers the asymptotic dimension $asdim(G)$ (I assume from now on, that both groups $A$ and $B$ are finitely generated). Definitions and basic results can be found, for instance here. It then follows from the "Hurewicz theorem" for asymptotic dimension, using the product map $$A\times B \to A\bowtie B$$ that $$asdim(A\bowtie B) \le asdim(A) + asdim(B).$$

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