Rank of a free group times a free abelian group.

Question

I know that the rank (i.e. minimal number of generators) of the product $\mathbb{Z}\times F_2$, of the infinite cyclic and free group on two generators, is three, but the only argument I could quickly piece together uses quite advanced tools (name-bearing and quite recent results on deficiency of a group/presentation). Can anyone point me to a direct, elementary proof, if there is one? Presumably, this (i.e. that the rank is the sum of ranks) also holds for product of a free group of finite rank and any cyclic group/product of cyclic groups?

Answer 1

Let $G=\mathbb{Z}^n\times F_m$ where $F_m$ denotes the free group of rank $m$. Consider the abelianization of $G$, i.e., the surjective map $$ \pi:G\twoheadrightarrow G/[G,G]\simeq\mathbb{Z}^{n+m}. $$ Since this map is surjective, if $\{g_1,\dots,g_k\}$ generate $G$, then $\pi(g_1),\dots,\pi(g_k)$ generate $\mathbb{Z}^{n+m}$. Since the rank of $\mathbb{Z}^{n+m}$ is $n+m$, this implies that $k\geq n+m$.

On the other hand, we can write down a generating set of $G$ consisting of $n+m$ generators as follows: $\mathbb{Z}^n$ is generated by $n$ coordinate vectors $\{e_1,\dots,e_n\}$ and $F_m$ is, by definition, generated by $m$ elements $\{f_1,\dots,f_m\}$. Therefore, $G$ is generated by $\{e_1,\dots,e_n,f_1,\dots,f_m\}$.

Since we have shown that it is possible to generate $G$ with $n+m$ generators and, in addition, that any generating set for $G$ must have at least $n+m$ elements, it follows that exactly $n+m$ generators is the best possible, i.e., the rank is $n+m$.

For the further generalization, combining this argument with the fundamental theorem of finitely generated abelian groups should provide exactly what you're looking for.