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...

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    $\begingroup$ But as a normal subgroup, it is generated by only $1$ element. $\endgroup$
    – BS.
    Aug 26, 2019 at 11:25
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    $\begingroup$ 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... $\endgroup$
    – user101010
    Aug 26, 2019 at 11:30
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    $\begingroup$ 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. $\endgroup$
    – SashaP
    Aug 26, 2019 at 13:44
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    $\begingroup$ You only need $n-1$ relations to get an abelian quotient of $F_n$, i.e. $ n-1$ of the generators. $\endgroup$
    – Derek Holt
    Aug 26, 2019 at 16:13
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    $\begingroup$ 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. $\endgroup$ Aug 28, 2019 at 14:30

1 Answer 1


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.

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

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    $\begingroup$ 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). $\endgroup$
    – YCor
    Sep 7, 2019 at 9:02
  • $\begingroup$ @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. $\endgroup$
    – user101010
    Dec 10, 2019 at 7:22
  • $\begingroup$ @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). $\endgroup$
    – Luc Guyot
    Dec 10, 2019 at 12:40
  • $\begingroup$ @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. $\endgroup$
    – Luc Guyot
    Dec 10, 2019 at 12:55

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