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 $x,y,z\in\mathbb{F}_3$?


1 Answer 1


That $u^3=1$ for all $u$ in this matrix group is an immediate calculation, and moreover this matrix group is generated by the matrices $e_{12}(1)$ and $e_{23}(1)$. So this matrix group is a quotient of $B(2,3)$.

Therefore it is enough to show that $B(2,3)$ has order $\le 27$. Let $x,y$ be its generators, and write $X=x^{-1}$, $Y=y^{-1}$, and $[u,v]=uvu^{-1}v^{-1}$.

\begin{align*}[[x,y],y]= & xyXYyyxYXY\\ = & xyXyxxyx(XY)^3\\ = & xyXyXyx \\ = & xx(Xy)^3x=xxx=1\end{align*}

Similarly $[[x,y],x]=1$. So $[x,y]$ is central. It follows that every element can be written as $x^ay^b[x,y]^c$; moreover by the exponent condition, $a,b,c$ can be chosen in $\{0,1,2\}$. This yields at most 27 elements and concludes the proof.


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