# Minimal presentations and (co)homology groups

I wonder whether there exists a link between the number of generators and relations of a presentation for a given group $G$ and the ranks of its (co)homology groups $H_1(G,\mathbb{Z})$ and $H_2(G, \mathbb{Z})$. For example, if $\langle X \mid R \rangle$ is a presentation of a group $G$ such that $R$ has as less relations as possible provided that $|X|= \mathrm{rank}(G)$, can we say that $\mathrm{rank}~H_2(G,\mathbb{Z}) = |R|$? It seems to be true at least for some groups.

I am not really familiar with (co)homology of groups, so I don't know what would be a pertinent formulation. Instead, I conclude with a pretty vague question:

Is it possible to link minimal presentations and (co)homology groups?

## 2 Answers

Let$G=\langle X \mid R \rangle$, so $G=F/N$, with $F$ free on $X$ and $N=\langle R^F \rangle$.

Then the Schur Multiplier $M(G) = H_2(G,{\mathbb Z})$ of $G$ is isomorphic to $N \cap [F,F]/[F,N]$.

Note that $N/(N \cap [F,F]) \cong N[F,F]/[F,F]$ is free abelian of rank equal to $|X| - r_1$, where $r_1$ ia the torsion-free rank of $F/N[F,F] \cong G/[G,G] \cong H_1(G,{\mathbb Z})$. So, since $N/[F,N]$ can be generated by the images of the elements of $R$, $m(G)$ can be generated by at most $|R|-|X| + r_1$ elements. The presentation is called efficient if this is the minimal number of generators of $M(G)$.

There has been a lot of work done on search for efficient presentations of finite groups (where $r_1=0$). There are known examples that do not have efficient presentations, but I think it is still an open problem whether every finite $p$-group has one. So an efficient presentation of a group with $M(G)=0$ would be a balanced presentation with $|X|=|R|$. For example

$Q_8 = \langle x,y \mid x^2=y^2,y^{-1}xy=y^{-1} \rangle$.

Note also that $|X|-|R|$ is called the deficiency of the presentation.

Brown mentions the same relation in the fifth exercice of chapter II.5 in his book Cohomology of groups. He suggests a more topological approach, and here is an attempt to do it:

Let $\langle X \mid R \rangle$ be a presentation of a group $G$ with $|X|=n$ and $|R|=m$. Let also $r$ denote $\mathrm{rk}_{\mathbb{Z}} G^{ab}$. We want to prove that $\mathrm{rk}_{\mathbb{Z}} H_2(G) \leq r-n+m$.

Let $Y$ be the CW-complex associated to the given presentation. It is known that it is possible to add $n$-cells to $Y$, with $n \geq 3$, in order to "kill" higher homotopy groups. More precisely, the new CW-complex $Y^{\mathrm{aug}}$ satisfies $\pi_n ( Y^{\mathrm{aug}}) = 0$ for all $n\geq 2$. Moreover, because we did not modify the $2$-squeleton of $Y$, $\pi_1(Y) \simeq \pi_1( Y^{\mathrm{aug}})$. Therefore, $Y^{\mathrm{aug}}$ is a $K(G,1)$, hence $H_{\ast}(Y^{\mathrm{aug}}) \simeq H_{\ast} (G)$.

Notice that $1-n+m = \chi(Y)= 1- \mathrm{rk}_{\mathbb{Z}} H_1(Y) + \mathrm{rk}_{\mathbb{Z}} H_2(Y)$, hence $\mathrm{rk}_{\mathbb{Z}} H_2(Y)=r-n+m$.

On the other hand, we clearly have an epimorphism $H_2(Y) \twoheadrightarrow H_2(Y^{\mathrm{aug}})$, therefore:

$$\mathrm{rk}_{\mathbb{Z}} H_2(G)= \mathrm{rk}_{\mathbb{Z}} H_2( Y^{\mathrm{aug}}) \leq \mathrm{rk}_{\mathbb{Z}} H_2(Y)=r-n+m.$$