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I have given the presentation $ G = \langle a_1,a_2\ |\ [[a_i,a_j],a_k],\forall i,j,k\in{1,2},\ a_1^2 = a_2^2 \rangle $ of a torsionfree, nilpotent group which looks pretty similar to the Heisenberg group $ H = \langle x,y\ | \ [[x,y],x] = [[x,y],y] = 1 \rangle $ except for the relation $ a_1^2 = a_2^2$. This additional relation is already confusing to work with. How can two elements satisfy this relation without being inverse or equal to each other?

Second I figured out $G$ can be written as $G = \langle a_1,a_2\ |\ [a_1,a_2]=[a_2,a_1],\ a_1^2=a_2^2 \rangle $ using both relations. Now I'm stuck with embedding $G$ into the group of unitriangular matrices $ UT_n(\mathbb{Z})$. Basically because the second relation doesn't make sense to me. Here the Heisenberg group is embedded Group presentation for discrete Heisenberg group.

Unfortunatly I am not able to adapt. Maybe there is also a way to picture the group elements of $G$ in a different way, I'm not exactly sure how the words are generated. I'm not asking for a full solution but every hint in the right direction is highly appreciated.

Thanks a lot for taking the time to check out the question!

Arwid
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    The element $[a_1,a_2]$ has order $2$ (that follows from your relation $[a_1,a_2]=[a_2,a_1]$) , so $G$ is not torsion-free, and so it cannot be embedded into any ${\rm UT}_n({\mathbb Z})$. – Derek Holt Aug 27 '18 at 18:10
  • Yes, there are several maximal torsion-free subgroups, $\langle a_1 \rangle$, $\langle a_2 \rangle$, $\langle a_1[a_1,a_2] \rangle$, etc, and they are all infinite cyclic. – Derek Holt Aug 27 '18 at 19:56
  • I checked and the transformation is indeed correct and the group itself not torsionfree. Am I right that the biggest torsion free subgroup is $G_1 = \langle a_1,a_2 \ | \ [[a_i,a_j]a_k] ,\ i=j=k ,\ a_1^2 = a_2^2 \rangle = \langle a_1,a_2 \ | \ a_1 = a_2 \rangle = \langle a_1 \rangle = \mathbb{Z} $?

    This than could be embedded in $ UT_n(\mathbb{Z})$ through $G_1 \rightarrow UT_n(\mathbb{Z}), \ a_1 \begin{bmatrix}1&a_1\0&1\end{bmatrix}$.

     – Arwid Aug 27 '18 at 19:56

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