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.

normalsubgroup, it is generated by only $1$ element. $\endgroup$ – BS. Aug 26 '19 at 11:256more comments