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A group $G$ is generated by $\begin{pmatrix}1&n\\0&1\end{pmatrix}$ and $\begin{pmatrix}1&0\\n&1\end{pmatrix}$, then we know $G\cong \mathbb{F}_2$ which is a free group generated by two elements. Now I consider the representation: $G\to GL(2,\mathbb{R})$, it is necessary that the image of $\begin{pmatrix}1&n\\0&1\end{pmatrix}$ is a triangular matrix under conjugation? Thanks in advance.

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  • $\begingroup$ Do you require the representation to be faithful? $\endgroup$ – user1729 Jun 4 '14 at 12:10
  • $\begingroup$ @user1729PhD Of course. $\endgroup$ – user151938 Jun 4 '14 at 12:17
  • $\begingroup$ Just checking! Interesting question. What about mapping $a\mapsto [a, b]$ and $b\mapsto b$ ($a$ is the first matrix, the upper triangular one, while $b$ is the second). Then so long as $\langle a, b\rangle$ is non-trivial in the abelianisation of $GL_2(\mathbb{R})$ then this works. However, I have no idea what the abelinisation of $GL_2(\mathbb{R})$ looks like! EDIT: nope, doesn't work. $\endgroup$ – user1729 Jun 4 '14 at 12:22
  • $\begingroup$ (Works over $\mathbb{Z}$ though!) $\endgroup$ – user1729 Jun 4 '14 at 12:40
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No, it is not necessary. For instance the matrices $$A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}^{100} $$ and $$B = \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix}^{100} $$ generate a free group of rank 2 (I just put the exponent $100$ to be safe; any sufficiently large exponent will work). Those matrices have determinant one and they have trace of absolute value $>2$---let me call this a "hyperbolic" matrix---so they are not conjugate to upper triangular matrices with $1$'s on the diagonal. Furthermore, every element of the group they generate is a hyperbolic matrix (except for the identity).

Hyperbolic matrices have distinct eigenvalues and a distinct pair of eigenlines.

The general theorem in this regard is that if $A,B \in SL(2,\mathbb{R})$ are two hyperbolic matrices, and if they do not have a common eigenline, then there exists $M>0$ such that if $m,n \ge M$ then $A^m,B^n$ freely generate a rank 2 free group, every element of which is a hyperbolic matrix (except for the identity).

This is part of the whole "ping-pong" construction, although I see that the Wikipedia page on the ping-pong construction seems to omit mention of hyperbolic matrices.

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  • $\begingroup$ He wanted the matrices to be not conjugate to any upper triangular matrix, not just not conjugate to one with $1$'s on the diagonal. But your matrices have complex eigenvalues, so they still work. $\endgroup$ – Derek Holt Jun 5 '14 at 8:57

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