# Groups with large negative deficiency

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

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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.

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There even exist 2-generator finite groups with arbitrarily large negative deficiency.

Let $G$ be the standard wreath product of a cyclic group of order 2 with a cyclic group of order $n$. I don't know the exact negative deficiency of $G$, but I can show that it is at least $\lfloor n/2 \rfloor$.

It is a standard result that the negative deficiency of a finite group is at least equal to the rank of its Schur Multiplier $M(G) = H_2(G)$. I can show that this is at least $\lfloor n/2 \rfloor$ by constructing a central extension $E$ of $G$ by an elementary abelian 2-group $Z$ of rank $r := \lfloor n/2 \rfloor$ with $Z \le E'$.

In outline, to define $E$, let $y_1,\ldots, y_n$ be generators of the base group $B$ of $G$, let $Z$ have generators $z_1,\ldots, z_r$, and in $E$ we put $y_i^2=1$ for all $i$ and $[y_i,y_j] = z_r$, where $r$ is the distance of $i$ from $j$ when the integers $1,\ldots,n$ are arranged around a circle each at distance 1 from the next. Then the action of the cycle in $G$ of order $n$ on $B$ induces the trivial action on $Z$, so we can define the required extension $E$.

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