Burnside group $B(2, 3)$ has $27$ elements, isomorphic to unitringular matrix group $\text{UT}(3, 3)$?

I realized my question here 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 $x,y,z\in\mathbb{F}_3$?

• There's a nice exercise on p.80 (exercise 13) of Hatcher's Algebraic Topology, which shows how to get that $B(2,3)$ has 27 elements. – Jim Conant Oct 1 '16 at 4:09
• I voted to close because this is clearly not a research level problem, but I have noticed that the typical level of questions about group theory on Math StackExchange is decreasing and there and more very elementary questions. This results in fewer and fewer people being available to answer questions like this on MSE, and it is not clear that there is an ideal place to ask them. – Derek Holt Oct 1 '16 at 11:48
• I do think that this question is closer to research level than many group theory questions that are allowed to stand, and that it is not completely elementary. I haven't read the paper that Igor Rivin cites, but I know that a standard trick is to note that if $\langle x,y \rangle$ is a non-cyclic group of exponent $3$, then $x^{-1}y$ and $yx^{-1}$ commute ( since $(xy)^{3} = 1$, so that $yxy = (xyx)^{-1} = x^{-1} y^{-1}x^{-1}$). Hence $\langle x^{-1}y, yx^{-1} \rangle$ is an elementary Abelian group of order $9$, of index (dividing) $3$ in $\langle x,y \rangle$. I can't leave it as an answer. – Geoff Robinson Oct 1 '16 at 14:42
• @GeoffRobinson: the question is a simple exercise, one of whose easy answers is given in the linked MathSE original question posted 1 hour before the MO question (math.stackexchange.com/questions/1948807). – YCor Oct 1 '16 at 18:12
• @YCor: I had not seen your MSE answer, and I don't really want to get into a long discussion about whether the question is research level- this is a subjective judgement, depending on how much prior knowledge one has, or assumes others have. My main point was that this question was better than a lot of group theory questions which are allowed to stand unchallenged on MO. – Geoff Robinson Oct 1 '16 at 18:28

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.
• I am somewhat surprised that no one has mentioned Chapter 18 of Marshall Hall's book, The Theory of Groups. He does cite the Levi-van der Waerden paper and proves that $B(r,3)$ has order $3^{m(r)}$ where $m(r) = r$ $+$ ${r}\choose 2$ + ${r}\choose 3$. Hall also includes information about the commutator power structure for elements of $B(r,3)$. – David A. Jackson Oct 1 '16 at 15:16
• The notes of G. Traustason, which can be found at people.bath.ac.uk/gt223/paper27.pdf , make it clear that Burnside knew that the Burnside Problem was true for groups of exponent $3$, and that any two conjugate elements commute in a group of exponent $3$, from which it follows directly that a two-generator group of exponent $3$ has a commutator group which is central and cyclic. So I am now persuaded that he was aware of the structure of $B(2,3)$ in particular. – Geoff Robinson Oct 4 '16 at 13:48