# Technique for showing a group is not free?

The specific case I present here is much less important than the general question. I have two matrices: $$p = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & -\frac{1}{2} \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$

$$q = \begin{pmatrix} 1 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$ and I would like to show that the group generated by them is (I think) not free. i.e. I would like to find some non-trivial identity satisfied by the two matrices. Is there a program that can help me search for such non-trivial identities? Is there a way to do it in Sage? (I'm afraid I'm very new to sage, so don't know much about it).

Ideally there would be some general way of dealing with this for general matrices. At the moment I don't have a better technique than simple trial and error.

EDIT:

I have gotten sage to run:

from sage.groups.free_group import is_FreeGroup

p = matrix(QQ, [[1, 0, 0, 0], [0, 1, 1, -1/2], [0, 0, 1, 1], [0, 0, 0, 1]])
q = matrix(QQ, [[1, -1, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1]])

G = MatrixGroup([p, q])

is_FreeGroup(G)


Which returns False. But I don't know how to have it tell me which non-trivial relation the generators satisfy. Is there a way to tell me which group it is isomorphic to? (I have a hunch it is isomorphic to the Heisenberg group: 3x3 upper triangular matrices with 1 on the diagonal)

• Hint: The group of strictly upper triangular matrices is nilpotent. Commented Jun 14 at 9:57
• It seems like Sage .is_isomorphic() does not work for infinite groups
– Jack
Commented Jun 14 at 10:00
• Just a side comment: Exhibiting a relation between some generators does not show that the group is not free. Only that these generators do not freely generate it Commented Jun 14 at 10:31
• @TimSeifert Thank you very much for the comment, I find this surprising. I will need to think some more
– Jack
Commented Jun 14 at 10:34
• It's really not hard to see. The set $\{2,3\}$ generates $\mathbb{Z}$ for instance (even minimally so). Commented Jun 14 at 10:40

Of course this example is easy because the matrices are upper unitriangular. Using Magma it was easy to determine that the group $$\langle p,q \rangle$$ s nilpotent of class $$3$$ with Hirsch length $$4$$, and polycyclic presentation
$$\langle p,q,r,s \mid [p,q]=r,[p,r]=s, [q,r]=[p,s],=[q,s]=[r,s]=1 \rangle.$$
(So it is not isomorphic to the Heisenberg group, which has class $$2$$ and Hirsch length $$3$$.)