Minimal number of generators of a group and direct product with an infinite cyclic group

Question

Is it true that the direct product of any group with the minimal number of generators $k$ with an infinite cyclic group will have the minimal number of generators $k + 1$? If yes, any reference or a proper argument? If it is not true, is it at least true that the minimal number of generators of a product is not smaller than $k$?

Answer 1

Let $G$ be a finitely generated group. Let us denote by $d(G)$ the minimal number of generators of $G$. Then we always have $d(G \times \mathbb{Z}) \le d(G) + 1$ and also $d(G \times \mathbb{Z}) \ge d(G)$ since the former group surjects onto the latter.

As indicated by YCor in his comment, it is not true in general that $d(G \times \mathbb{Z}) = d(G) + 1$. Here is a workable necessary and sufficient condition.

Claim. Let $G$ be a finitely generated group. Then the equality $d(G \times \mathbb{Z}) = d(G) + 1$ holds true if and only if $d(G/[G, G]) = d(G)$ where $[G, G]$ stands for the commutator subgroup of $G$.

The proof relies on the invariant decomposition theorem of finitely generated Abelian groups:

Theorem. Let $G$ be a finitely generated Abelian group. Then there are positive integers $1 < d_1, d_2, \dots, d_k$ such that $d_i$ divides $d_{i + 1}$ for every $i$ and an integer $r \ge 0$ such that $G$ is ismorphic to $\mathbb{Z} /d_1 \mathbb{Z} \times \cdots \times \mathbb{Z} /d_k \times \mathbb{Z}^r$. In addition, $d(G) = k + r$.

Let us prove the claim.

Proof of the claim. Let us assume first that $d(G/[G, G]) = d(G)$ holds true. Since we have $d(G \times \mathbb{Z}) \ge d(G/[G, G]\times \mathbb{Z}) = d(G/[G, G]) + 1$ by the above theorem and obviously $d(G \times \mathbb{Z}) \le d(G) + 1$, we deduce that $d(G \times \mathbb{Z}) = d(G) + 1$.

In order to prove the converse, let us assume now that $d(G/[G, G]) < d(G)$ holds true. Consider a $d$-tuple $\mathbb{g} = (g_1, \dots, g_d)$ of elements generating $G$ with $d = d(G)$. The image $\overline{\mathbb{g}}$ of $\mathbb{g}$ by the natural map $G \twoheadrightarrow G/[G, G]$ is a generating $d$-tuple of $G/[G, G]$. Since $d > d(G/[G, G])$, it follows from By [1, Theorem 1.1] that we can find a Nielsen transformation $\psi \in \text{Aut}(F_d)$, with $F_d$ the free group on $d$ generators, such that the first coordinate of $\overline{\mathbb{g}} \cdot \psi$ is the trivial element. Hence, replacing $\mathbb{g}$ by $\mathbb{g} \cdot \psi$, we can assume that $g_1 \in [G, G]$. Hence we can find $w \in [F_d, F_d]$ such that $g_1 = w(g_1, g_2, \dots, g_d)$. Let $h_1 = (g_1, 1)$ where $1$ is the canonical generator of $\mathbb{Z}$ and let $h_i = (g_i, 0)$ for $2 \le i \le d$. Since $(g_1, 0) = w(h_1, \dots, h_d)$, the $d$-tuple $\mathbb{h} \Doteq (h_1,\dots, h_d)$ generates $G \times \mathbb{Z}$. As a result, $d(G\times \mathbb{Z}) = d$, which settles the claim.

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[1] D. Oancea, "A note on Nielsen equivalence in finitely generated abelian groups ", 2011.