Question: Is it possible to produce with $3$ elements the group $G=F_2 \ast (\mathbb{Z} \times \mathbb Z)$?
I figured that $G= \langle a,b,c,d \mid [c,d]=1 \rangle$. I also know that it is impossible to produce a free group of order $n$ with less than $n$ elements (even though I don't know if this proposition can help in this situation).
It is an exercise that i found in some old notes from a friend of mine. Any help would be greatly appreciated!
No.
Say you had three elements, $x,y,z$ which generated $G$. Then look at the abelianization $G^\text{ab} = G \big / [G,G]$. The images of $x,y,z$ in this quotient (if you like, $x[G,G], \ y[G,G], $ and $z[G,G]$) must generate $G^\text{ab} \cong \mathbb{Z}^4$, so we've found three elements which generate $\mathbb{Z}^4$.
If you know this isn't possible, then great! We can stop here. If you don't, then after tensoring with $\mathbb{Q}$ our three elements would have to span $\mathbb{Z}^4 \otimes \mathbb{Q} \cong \mathbb{Q}^4$, but obviously no three elements can span a $4$ dimensional vector space.
I hope this helps ^_^