I realized my question [here](https://math.stackexchange.com/questions/1948807/burnside-group-b2-3-how-to-see-has-27-elements-and-isomorphic-to-certain) might have been too hard for MSE, so I'm asking it here as well.


The Burnside group $B(d, n)$ is defined as the quotient of the free group on $d$ generators by the normal subgroup generated by all $n$th powers.


**Question.** How do I see that $B(2, 3)$ has $27$ elements and is isomorphic to the group of matrices of the form$$\begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$$for $x,y,z\in\mathbb{F}\_3$?



Burnside group $B(2, 3)$ has $27$ elements, isomorphic to unitringular matrix group $\text{UT}(3, 3)$?
Since the question has been reopened, I mention a familiar argument which certainly appears in papers of G. Higman. If $G = \langle x,y \rangle$ is a non-Abelian group of exponent $3$, then since $xyxyxy = 1,$ we have $yxy = (xyx)^{-1} = x^{-1}y^{-1}x^{-1}.$ Hence $yx^{-1}x^{-1}y = x^{-1}yyx^{-1}$ as $x$ and $y$ both have order $3$. Hence $H = \langle x^{-1}y,yx^{-1} \rangle$ is an Abelian subgroup (of order $9$) of $G$ which is easily checked to be normalized by $x$ ( and hence by $y$) so is an Abelian normal subgroup of $G$. Clearly $G = H \langle x \rangle,$ so that $G$ has order $27$ as $G$ is non-Abelian.


