# Minimum number of relations that must be added to make a group abelian

Let $$G$$ be a finitely generated group and let $$c(G)$$ denote the minimal number of relations that we must add to a presentation fro $$G$$ in order to make $$G$$ abelian. I would like to find examples of groups where $$c(G)$$ is arbitrarily large but I do not know how to get lower bounds on $$c(G)$$. Do you know of such groups? There is no algorithm to compute $$c(G)$$ since $$c(G) = 0$$ implies $$G$$ is abelian - but I just want some examples where $$c(G)$$ is big.

Actually, for an application I have in mind, I would like to find 3-manifold groups (or in particular knot complement groups) $$G_i$$ where $$c(G_i) \to \infty$$.

I imagine there is a relation between $$c(G)$$ and some other invariants of groups and I am hoping someone can point these out to me. Maybe $$c(G)$$ has a name that I am totally unaware of...

• But as a normal subgroup, it is generated by only $1$ element. – BS. Aug 26 '19 at 11:25
• This was a point of confusion for me also - namely, $c(G)$ is not the minimum number of generators of $[G,G]$ or the minimal number of normal generators, since a bigger subgroup might require fewer generators... – user101010 Aug 26 '19 at 11:30
• Oh, I see. But don't the free groups actually give the desired examples?(even in the sense of the minimal number of generators as a normal subgroup) Any abelian group has a non-positive deficiency(the minimal difference between numbers of generators and relations) so to get an abelian quotient of $F_n$ one must impose at least $n$ relations. – SashaP Aug 26 '19 at 13:44
• You only need $n-1$ relations to get an abelian quotient of $F_n$, i.e. $n-1$ of the generators. – Derek Holt Aug 26 '19 at 16:13
• I am guessing you want to preserve the abelianization, because otherwise for every knot group $c(G)=1$ as introducing the meridian as a relation kills the whole group. – Neil Hoffman Aug 28 '19 at 14:30

## 1 Answer

The question has been answered in the comments by SashaP and Derek Holt, taking the following definition of $$c(G)$$:

Definition 1. Let $$G$$ be a finitely generated group. We denote by $$c(G)$$ the minimal number of elements of $$G$$ required to normally generate a subgroup of $$G$$ that contains $$G'$$, the derived subgroup of $$G$$.

The invariant $$c(G)$$ is related to both the weight $$w(G)$$ and the deficiency $$\text{def}(G)$$ of $$G$$.

The weight $$w(G)$$ of a finitely generated group $$G$$ is the minimal number of elements required to generate $$G$$ as a normal subgroup. Clearly, we have $$d(G/G') \le w(G) \le d(G)$$ where $$d(G)$$ denotes the minimal number of generators of $$G$$.

Neil Hoffman has suggested another possible definition of OP's invariant.

Definition 2. For $$G$$ a finitely generated group, we define $$c_{ab}(G)$$ as the minimal number of elements required to normally generate $$G'$$ as a normal subgroup of $$G$$.

I will address Definition 1 first by wrapping up the hints provided in the comments attached to the question. Definition 2 will be addressed later in Claim 3.

The following is a combination of Derek Holt's and Neil Hoffman's observations.

Claim 1. Let $$G$$ be a finitely generated group. Then we have $$c(G) \le \min(w(G), d(G) - 1).$$ In particular, $$c(G) = 1$$ if $$G$$ is a non-Abelian knot group.

The deficiency $$\text{def}(P)$$ of a finite group presentation $$P = \langle x_1, \dots , x_n \vert \, r_1, \dots, r_m \rangle$$ on $$n$$ generators and $$m$$ relators is the integer $$n - m \in \mathbb{Z}$$. The deficiency $$\text{def}(G)$$ of a finitely presented group $$G$$ is the maximum of the deficiencies of its finite presentations.

The following summarizes SashaP's observations.

Claim 2. If $$G$$ is a finitely generated Abelian group, then $$\text{def}(G) \le 1$$. In addition, we have

• $$\text{def}(G) = 1$$ if and only if $$G \simeq \mathbb{Z}^r$$ for some $$r \in \{1, 2\}$$.
• $$\text{def}(G) = 0$$ if and only if $$G \simeq \mathbb{Z}/n\mathbb{Z}, \mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$$ with $$n > 0$$ or $$\mathbb{Z}^3$$.

Proof. Let $$r = \dim(G \otimes_{\mathbb{Z}} \mathbb{Q})$$. Therefore we can write $$G = \mathbb{Z}^r \times H$$. By [2, Lemma 1.2], we have $$\text{def}(G) \le \dim\left(H_1(G, \mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{Q}\right) - d(H_2(G, \mathbb{Z}))$$. Actually, equality holds since $$G$$ is efficient by [2, Lemma 1.3]. As $$d(H_2(G, \mathbb{Z})) = d(\Lambda^2(G)) = d(\Lambda^2(\mathbb{Z}^r \times H)) \ge \frac{r(r-1)}{2}$$, we deduce that $$\text{def}(G) \le r - \frac{r(r-1)}{2} \le 1$$. The remaining statements are easily inferred from the efficiency of finitely generated Abelian groups.

We are now in position to answer the main question, that is, to provide a sequence of groups $$(G_n)$$ with arbitrarily large values of $$c(G_n)$$.

Corollary 1. Let $$G$$ be a finitely presented group. Then $$c(G) \ge \text{def}(G) - 1$$.

Proof. Consider a presentation $$P$$ of $$G$$ with deficiency $$\text{def}(G)$$. By adding $$c(G)$$ relators to $$P$$ we obtain the presentation $$P'$$ of an Abelian group. Thus $$\text{def}(P') = \text{def}(G) - c(G) \le 1$$ by Claim 2. The result immediately follows.

Note that the free group on $$n$$ generator is the fundamental group of a 3-dimensional handlebody of genus $$n$$. It is also the fundamental group of the connected sum of $$n$$ copies of $$S^1 \times S^2$$.

Corollary 2. Let $$F_n$$ be the free group on $$n$$ generators. Then $$c(F_n) = n - 1$$.

Proof. By Claim 1, we have $$c(F_n) \le n - 1$$. It follows from Corollary 1 that $$c(F_n) \ge n - 1$$, hence the result.

In order to address Definition 2, further definitions are required. If $$M$$ is a finitely generated module over a unital ring $$R$$, we denote by $$d_R(M)$$ the minimal number of generators of $$M$$. Let $$G$$ be a group. Then the conjugation action of $$G$$ on the Abelian group $$M \Doteq G'/G''$$ induces an action of $$G_{ab}$$ on $$M$$ so that $$M$$ is naturally a $$\mathbb{Z}[G_{ab}]$$-module.

Claim 3. Let $$G$$ be a finitely generated group and let $$n = d(G)$$. Then the following holds $$d_{\mathbb{Z}[G_{ab}]}(G'/G'') \le c_{ab}(G) \le c_{ab}(F_n).$$ In addition, we have $$c_{ab}(F_n) = \frac{n(n - 1)}{2}.$$

Proof. The first two inequalities follow easily from the definitions. Clearly, we have $$c_{ab}(F_n) \le \frac{n(n-1)}{2}$$. The reverse inequality follows from [3, Theorem 3]. Indeed, set $$M \Doteq F_n'/F_n''$$, $$(F_n)_{ab} = \mathbb{Z}^n$$ and let $$I$$ be the augmentation ideal of $$\mathbb{Z}[(F_n)_{ab}]$$. Then by [3, Theorem 3] the group $$M$$ is a $$\mathbb{Z}[(F_n)_{ab}]$$-module generated by $$\frac{n(n-1)}{2}$$ elements and we have $$M/IM \simeq \mathbb{Z}^{\frac{n(n-1)}{2}}$$. As $$(F_n)_{ab}$$ acts trivially on $$M$$, this module cannot be generated by less than $$\frac{n(n - 1)}{2}$$ elements.

A similar reasoning applies to the integral Alexander module $$A(K)$$ of a classical $$1$$-knot $$K \subset \mathbb{S}^3$$. Recall that the Alexander module $$A(K) \Doteq G'/G''$$ is a module over $$\mathbb{Z}[t^{\pm 1}] \simeq \mathbb{Z}[G_{ab}]$$ where $$G = \pi_1(\mathbb{S}^3 \setminus K)$$.

Claim 4. Let $$K$$ be the knot sum of $$n$$ copies of an oriented knot with a non-zero Alexander module. Let $$G = \pi_1(\mathbb{S}^3 \setminus K)$$. Then we have $$c_{ab}(G) \ge n.$$

Claim 4 is an easy consequence of

Lemma 1. Let $$K_1$$ and $$K_2$$ be two oriented $$1$$-knots. Then $$A(K_1 \# K_2) = A(K_1) \times A(K_2)$$.

Proof. This is a straightforward consequence of the two following facts:

• If $$V$$ is a Seifert matrix of a $$1$$-knot $$K$$, then $$V - tV^{T}$$ is a presentation matrix of $$A(K)$$ where $$V^T$$ denotes the transpose of $$V$$.
• If two $$1$$-knots $$K_1$$ and $$K_2$$ admit $$V_1$$, resp. $$V_2$$, as a Seifert matrix, then the block-diagonal matrix $$\text{diag}(V_1, V_2)$$ is a Seifert matrix of $$K_1 \# K_2$$.

We are now in position to prove Claim 4.

Proof of Claim 4. Write $$K = K_0 \# K_0 \# \cdots \# K_0$$ with $$A(K_0) \neq \{0\}$$. By Lemma 1, the Alexander module $$A(K)$$ of $$K$$ is the direct product of $$n$$ copies of $$A(K_0)$$. Considering a maximal ideal $$\mathfrak{m}$$ of $$\mathbb{Z}[t^{\pm 1}]$$ such that $$A(K_0)$$ surjects onto the field $$\mathbb{Z}[t^{\pm 1}]/\mathfrak{m}$$, we see that $$A(K)$$ cannot be generated by less than $$n$$ elements.

 E. Rapaport, "On the commutator subgroup of a knot group", 1960.
 D. Epstein, "Finite presentations of groups and 3-manifolds", 1961.
 S. Bachmuth, "Automorphisms of free metabelian groups", 1965.

• Even if implicit, at least in the free group case it seems that the lower bound on $c(G)$ uses the 2-nilpotent quotient $G/[G,[G,G]]$. That is, stated differently, obviously $c(G)\ge c(G/[G,[G,G]])$ while $c(G/[G,[G,G]])$ can be bounded below by hand. Hence I'm curious of examples for which $c(G)$ is large while $c(G/[G,[G,G]])$ is bounded (e.g., $[G,G]$ is perfect). – YCor Sep 7 '19 at 9:02
• @LucGuyot Sorry for the late comments, I was just looking at this again and I had two questions. Why is $d_{\mathbb{Z}[G_{ab}]}(G'/G'') \leq c_{ab}(G)$? Additionally, why do we have a map $A(K) \to \mathbb{Z}[t^{\pm 1}]/\mathfrak{m}$ for all knots? I know we've got that if $A(K)$ is a sum of cyclic modules, which is good enough for the question. – user101010 Dec 10 '19 at 7:22
• @user101010 If $G'$ can be generated by $d = c_{ab}(G)$ elements $g_1,\dots, g_d$ as a normal subgroup of $G$, then any element $g \in G'$ is a product of some conjugates of $g_1, \dots, g_d$. In the module $G'/G''$ this translates in $g$ being a linear combination of $g_1,\dots,g_d$ with coefficients in $\mathbb{Z}[G_{ab}]$ (remind that commutators commute there). – Luc Guyot Dec 10 '19 at 12:40
• @user101010 Regarding your second question, this is a general fact about finitely generated modules. If $M$ is any non-zero finitely generated module over a commutative and unital ring $R$, then $M$ surjects onto a non-zero cyclic $R$-module (and hence onto a field). Proof: $M$ has minimal generating set with $0 < d < \infty$ elements, so we can take the quotient of $M$ by the submodule generated by the first $d - 1$ elements of this set. – Luc Guyot Dec 10 '19 at 12:55