Hello,
I've just learnt the notion of deficiency of a group but I don't know how work with it.
I want to construct a group with large negative deficiency ; naively I think that $(Z\_2)^n$ will work because we need $n$ generators and $n$ relations for the square of the elements being $1$, plus $n(n-1)/2$ relations of commutation ; but maybe we can do better ...
Another problem is to control the deficiency of a direct product. Still naively we take generators and relations in both groups and add relations of commutation but it certainly does not work.
So my two questions :


* Does there exist finite groups with arbitrarily large negative deficiency ?
* If I fix a finitely presentable group G and consider the product of G with groups of arbitrarily large negative deficiency, can I obtain products with arbitraly large negative deficiency ?


Thank you,
mister\_jones



Groups with large negative deficiency
One way to bound the deficiency of a group is via the Morse inequalities. If $n$ is the number of generators and $m$ is the number of relations, then


$$1 - n+m \geq b\_0(G) - b\_1(G) + b\_2(G)$$


where $b\_i(G)$ denote the $i$-th Betti number of $G$ (with coefficients in some field $k$). This is due to the fact that there is a projective resolution of $\mathbb Z$ by projective $\mathbb Z G$-modules which starts like
$$0 \leftarrow \mathbb Z \leftarrow \mathbb ZG \leftarrow \mathbb ZG^{\oplus n} \leftarrow \mathbb ZG^{\oplus m} \leftarrow \cdots.$$


Hence, one gets:


$${\rm def}(G) \leq b\_1(G) - b\_2(G).$$


In your example, i.e. $G=(\mathbb Z/ 2\mathbb Z)^n$, one has $b\_1(G;\mathbb F\_2) = n$ and $b\_2(G; \mathbb F\_2)= \binom{n}{2}+n$. The first computation is clear and the second follows from the Kuenneth theorem:
$$b\_2( (\mathbb Z/ 2\mathbb Z)^n) = \sum\_{p\_1+ \cdots +p\_n = 2}\prod\_{i=1}^n b\_{p\_i}(\mathbb Z/ 2\mathbb Z).$$ Just note that all Betti numbers of $\mathbb Z/ 2\mathbb Z$ with coefficients in $\mathbb F\_2$ are equal to one and hence you just have to count the number of summands. There are precisely $n$ with one $2$ and $\binom{n}{2}$ with two $1$'s.


Hence, ${\rm def}((\mathbb Z/ 2\mathbb Z)^n) \leq - \binom{n}{2}$. Your concrete presentation shows that this is sharp.


