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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!

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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 ^_^

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    $\begingroup$ 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). $\endgroup$ Feb 25 at 0:54
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    $\begingroup$ What an elegant answer!Thank you,I appreciate it! ^_^ $\endgroup$ Feb 25 at 11:37

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