TITLE: Groups with large negative deficiency
QUESTION [6 upvotes]: 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

REPLY [4 votes]: 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 $E$.