Return to Answer

These two matrices generate a free group: $$ A=\left( \begin{array}{ccc} \frac{1}{3} & \frac{2 \sqrt{2}}{3} & 0 \\ -\frac{2 \sqrt{2}}{3} & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{array} \right), B=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{3} & \frac{2 \sqrt{2}}{3} \\ 0 & -\frac{\sqrt{2}}{3} & \frac{1}{3} \end{array} \right). $$ To see that they do, note that the entries of $3A$ and $3B$ live in the ring $\mathbb{Z}[\sqrt{2}]$, which admits a surjective homomorphism to the field $\mathbb{F}_3(i)$. This map induces a map on the matrix rings, under which $A$ and $B$ become

$$ A' =\left( \begin{array}{ccc} 1 & - i & 0 \\ i & 1 & 0 \\ 0 & 0 & 0 \end{array} \right), B'=\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & - i \\ 0 & i & 1 \end{array} \right). $$ If there were some non-trivial reduced group word in $A$ and $B$ that gave the identity, then a similar word in $3A$, $3A^{\top}$, $3B$ , and $3B^{\top}$ would be a multiple of the identity matrix. However, one may check that any monoid word in the generators $A'$, $A'^{\top}$, $B'$, $B'^{\top}$ that evaluates to a multiple of the identity must contain the subword $A'A^{\top}$ or some similar forbidden subword.

These two matrices generate a free group: $$ A=\left( \begin{array}{ccc} \frac{1}{3} & \frac{2 \sqrt{2}}{3} & 0 \\ -\frac{2 \sqrt{2}}{3} & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{array} \right), B=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{3} & \frac{2 \sqrt{2}}{3} \\ 0 & -\frac{\sqrt{2}}{3} & \frac{1}{3} \end{array} \right). $$ To see that they do, note that the entries of $3A$ and $3B$ live in the ring $\mathbb{Z}[\sqrt{2}]$, which admits a surjective homomorphism to the field $\mathbb{F}_3(i)$. This map induces a map on the matrix rings, under which $3A$ and $3B$ become

$$ A' =\left( \begin{array}{ccc} 1 & - i & 0 \\ i & 1 & 0 \\ 0 & 0 & 0 \end{array} \right), B'=\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & - i \\ 0 & i & 1 \end{array} \right). $$ If there were some non-trivial reduced group word in $A$ and $B$ that gave the identity, then a similar word in $3A$, $3A^{\top}$, $3B$ , and $3B^{\top}$ would be a multiple of the identity matrix. However, one may check that any monoid word in the generators $A'$, $A'^{\top}$, $B'$, $B'^{\top}$ that evaluates to a multiple of the identity must contain the subword $A'A^{\top}$ or some similar forbidden subword.

These two matrices generate a free group: $$ \left( \begin{array}{ccc} \frac{1}{3} & \frac{2 \sqrt{2}}{3} & 0 \\ -\frac{2 \sqrt{2}}{3} & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{array} \right), \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{3} & \frac{2 \sqrt{2}}{3} \\ 0 & -\frac{\sqrt{2}}{3} & \frac{1}{3} \end{array} \right). $$ To see that they do, consider a set of corresponding matrices over the field $\mathbb{F}_3(i)$:

$$ \left( \begin{array}{ccc} 1 & \mp i & 0 \\ \pm i & 1 & 0 \\ 0 & 0 & 0 \end{array} \right), \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & \mp i \\ 0 & \pm i & 1 \end{array} \right). $$ Non-trivial reduced words in these matrices always have disagreement along the diagonal.

These two matrices generate a free group: $$ A=\left( \begin{array}{ccc} \frac{1}{3} & \frac{2 \sqrt{2}}{3} & 0 \\ -\frac{2 \sqrt{2}}{3} & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{array} \right), B=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{3} & \frac{2 \sqrt{2}}{3} \\ 0 & -\frac{\sqrt{2}}{3} & \frac{1}{3} \end{array} \right). $$ To see that they do, note that the entries of $3A$ and $3B$ live in the ring $\mathbb{Z}[\sqrt{2}]$, which admits a surjective homomorphism to the field $\mathbb{F}_3(i)$. This map induces a map on the matrix rings, under which $A$ and $B$ become

$$ A' =\left( \begin{array}{ccc} 1 & - i & 0 \\ i & 1 & 0 \\ 0 & 0 & 0 \end{array} \right), B'=\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & - i \\ 0 & i & 1 \end{array} \right). $$ If there were some non-trivial reduced group word in $A$ and $B$ that gave the identity, then a similar word in $3A$, $3A^{\top}$, $3B$ , and $3B^{\top}$ would be a multiple of the identity matrix. However, one may check that any monoid word in the generators $A'$, $A'^{\top}$, $B'$, $B'^{\top}$ that evaluates to a multiple of the identity must contain the subword $A'A^{\top}$ or some similar forbidden subword.

These two matrices generate a free group: $$ \left( \begin{array}{ccc} \frac{1}{3} & \frac{2 \sqrt{2}}{3} & 0 \\ -\frac{2 \sqrt{2}}{3} & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{array} \right), \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{3} & \frac{2 \sqrt{2}}{3} \\ 0 & -\frac{\sqrt{2}}{3} & \frac{1}{3} \end{array} \right). $$ To see that they do, consider a corresponding pair of matrices over the field $\mathbb{F}_3(i)$.

These two matrices generate a free group: $$ \left( \begin{array}{ccc} \frac{1}{3} & \frac{2 \sqrt{2}}{3} & 0 \\ -\frac{2 \sqrt{2}}{3} & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{array} \right), \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{3} & \frac{2 \sqrt{2}}{3} \\ 0 & -\frac{\sqrt{2}}{3} & \frac{1}{3} \end{array} \right). $$ To see that they do, consider a set of corresponding matrices over the field $\mathbb{F}_3(i)$:

$$ \left( \begin{array}{ccc} 1 & \mp i & 0 \\ \pm i & 1 & 0 \\ 0 & 0 & 0 \end{array} \right), \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & \mp i \\ 0 & \pm i & 1 \end{array} \right). $$ Non-trivial reduced words in these matrices always have disagreement along the diagonal.