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Suppose $a,b \in GL(n,\mathbb{C})$, and $\langle a,b\rangle$ is a free group of rank $2$.

Is there a way to choose a $c$ to guarantee that $\langle a,b,c\rangle$ is a free group of rank $3$?

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Yes. Note first that it suffices to answer the question in $\textit{SL}(2,\mathbb{C})$. In particular, if $A,B\in\textit{GL}(2,\mathbb{C})$, let $\hat{A},\hat{B}$ be scalar multiples of $A,B$ that lie in $\textit{SL}(2,\mathbb{C})$, and suppose that there exists a $C\in\textit{SL}(2,\mathbb{C})$ so that $\langle\hat{A},\hat{B},C\rangle$ is a free group of rank $3$. Then for any word $w(x_1,x_2,x_3)$ in the free group $\langle x_1,x_2,x_3\rangle$, if $w(A,B,C) = I$, then $w(\hat{A},\hat{B},C)$ would have to be a scalar matrix, and hence would lie in the center of $\langle\hat{A},\hat{B},C\rangle$, which is impossible. We conclude that $w(A,B,C) \ne I$ for all $w$, and hence $\langle A,B,C\rangle$ is free as well.

So suppose $A,B\in\textit{SL}(2,\mathbb{C})$ and $\langle A,B\rangle$ is free of rank $2$. Consider the group $\textit{SL}(2,R)$, where $R$ is the ring $\mathbb{C}[t_1,t_2,t_3,t_4]/(t_1t_4-t_2t_3-1)$, and let $$ T \;=\; \begin{bmatrix}t_1 & t_2 \\ t_3 & t_4\end{bmatrix}. $$ Note that $T$ is invertible over $R$, with $$ T^{-1} \;=\; \begin{bmatrix}t_4 & -t_2 \\ -t_3 & t_1\end{bmatrix}, $$ and hence $T \in \textit{SL}(2,R)$. Then $\langle A,B,T\rangle$ must be free, since any element of $\langle A,B\rangle$ can be substituted for $T$, and the free group $\langle A,B\rangle$ has no non-trivial laws.

Now, if $w = w(x_1,x_2,x_3)$ is any non-trivial word in the free group $\langle x_1,x_2,x_3\rangle$, the equation $$ w(A,B,T) \;=\; I $$ is equivalent to a nontrivial system of polynomial equations in $t_1,t_2,t_3,t_4$, which define a proper subvariety $V_w$ of $\textit{SL}(2,\mathbb{C})$. But $\textit{SL}(2,\mathbb{C})$ cannot be expressed as a countable union of proper subvarieties, and hence there exists a matrix $C\in\textit{SL}(2,\mathbb{C})$ that does not lie in any $V_w$. Then $\langle A,B,C\rangle$ is free.

Practically speaking, the way to choose $C$ is to choose a generic matrix, i.e. a matrix whose entries do not satisfy any polynomial equations over the field generated by the entries of $A$ and $B$.

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  • $\begingroup$ thanks for your answer. So i guess the same argument also works when you have $n$ free generators and you want to choose a $(n+1)$'th element such that they form a free group of rank $n+1$ $\endgroup$ – FunctionOfX Jun 11 '15 at 14:47

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