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Let $M_1$ and $M_2$ be two matrices as follows:

$ M_1= \left( \begin{array}{cccc} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 5 & 5 & 1 & 0 \\ 0 & -5 & -1 & 1\\ \end{array} \right), \quad M_2= \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\\ \end{array} \right). $

One may easily check that $$(M_1M_2)^5=I, $$ where $I$ is the identity matrix. A well known problem is the following:

Is the relation $(M_1M_2)^5=I $, the only relation between $M_1$ and $M_2$?

Now my question is the following:

Does the group generated by $M_1$ and $M_2$ contain a free subgroup?

Is it known?

  • 2
    the subgroup generated by $M_2$ is certainly free; perhaps you'd like to refine the question2011-11-22
  • 0
    I think a group generated by an element is cyclic!2011-11-23
  • 0
    Yes, but I think user8268 is making the point that an infinite cyclic group is free of rank 1. Presumably your intention was to ask if there was a free subgroup of rank at least two. Have you heard of the Tits alternative?2012-01-22

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