# Is it possible to produce with $3$ elements the group $G=F_2 \ast (\mathbb{Z} \times \mathbb Z)$?

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!

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.
• And if you don't know about tensors, mod out $G^{\rm ab}$ by $2G^{\rm ab}$ to get $(\mathbf{F}_2)^4$, which cannot be generated by three elements (either through linear algebra, or because you can only get at most $8$ elements out of three in that finite group). Feb 25 at 0:54