# A subgroup of $\operatorname{sp}(4,\mathbb{Z})$

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?

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the subgroup generated by $M_2$ is certainly free; perhaps you'd like to refine the question – user8268 Nov 22 '11 at 22:39
I think a group generated by an element is cyclic! – MTG Nov 23 '11 at 0:36
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? – Geoff Robinson Jan 22 '12 at 7:49