is there any way to find the order of the group generated by $x=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ and $y=\begin{pmatrix}0&i\\i&0\end{pmatrix}$ under matrix multiplication ,other than calculating self product of these matrices and product of these matrcies etc? I have calculated though $x^4=I$

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    $\begingroup$ Well, if you compute $xy$ and $yx$, you may notice it's not so much work to brute-force it. $\endgroup$ – Daniel Fischer Aug 27 '13 at 13:14
  • $\begingroup$ Another way is to give it to some software like Maple that has packages for answering this sort of question. But then I guess the question becomes, how does Maple do it? $\endgroup$ – Gerry Myerson Aug 27 '13 at 13:15
  • $\begingroup$ I have calculated that but what can be concluded from that? $\endgroup$ – Marso Aug 27 '13 at 13:18

Lets denote $\langle x,y \rangle$ by $G$. Note, that $y^{-1}xy = x^{-1}$, so $\langle x \rangle$ is a normal subgroup of $G$ hence $G = \langle x \rangle \cdot \langle y \rangle$ (this means the complex product of the sets $\langle x \rangle$ and $\langle y \rangle$). Now consider the well known equation $|UV| = \frac{|U| |V|}{|U \cap V|}$ for any subgroups $U,V$ of any group.

  • $\begingroup$ $U\cap V=I$ so it will be order $16$ $\endgroup$ – Marso Aug 27 '13 at 13:26
  • $\begingroup$ No, $U \cap V = \{ \pm I \}$, so the order is 8. (It's $Q_8$.) $\endgroup$ – Derek Holt Aug 27 '13 at 13:31
  • $\begingroup$ @DerekHolt: So, mine is completely wrong. ? $\endgroup$ – mrs Aug 27 '13 at 13:32
  • $\begingroup$ The group does satisfy $x^4=y^4=(xy)^4=1$, and those relations alone would indeed define an infinite group, but the group here also satisfies $y^{-1}xy=x^{-1}$ and $x^2=y^2$. $\endgroup$ – Derek Holt Aug 27 '13 at 13:52

It can be found that $$x^k\neq I_{2\times 2},~y^k\neq I_{2\times 2},~(xy)^k\neq I_{2\times 2},~~~k=2,3, $$ where $xy=\begin{pmatrix} i & 0\\ 0 & -i \end{pmatrix}$ and so


  • $\begingroup$ so it is a group of order $10$? $\{x,x^2,x^3,y,y^2,y^3,xy,(xy)^2,(xy)^3,I\}$? $\endgroup$ – Marso Aug 27 '13 at 13:23
  • $\begingroup$ sorry it will be more I missed some product $\endgroup$ – Marso Aug 27 '13 at 13:23
  • $\begingroup$ @SenoreBunuel: Regarding to the relations, the group satisfies Von Dyck group relations for some powers of $x$, $y$ and $xy$. It seems that the group is infinite. $\endgroup$ – mrs Aug 27 '13 at 13:27

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