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*](https://en.wikipedia.org/wiki/Zappa-Sz%C3%A9p_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?



Cohomological dimension of knit products
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 <https://math.stackexchange.com/questions/107781/has-this-generalized-semidirect-product-been-studied/107788#107788>). They always fit in a short exact sequence
$$ 0 \rightarrow \mathbb{Z}^2 \rightarrow \mathbb{Z} \bowtie \mathbb{Z} \rightarrow \mathbb{Z}\_2 \oplus \mathbb{Z}\_2 \rightarrow 0. $$


