Order of group $\langle x,y,z \mid x^2=y^2= z^2= e, xyz=yzx = zxy \rangle$

Question

Group $\langle x,y,z \mid x^2=y^2= z^2= e, xyz=yzx = zxy \rangle$ has order 16. I put $a = xy$ and we have $za=az=yzay$ but I can't continue it.

Answer 1

I would go about this as follows (leaving the details for you).

Let's denote the group by $G$. Let $b=xyz$. Justify the following claims:

• We have $xbx^{-1}=yby^{-1}=zbz^{-1}=b$, so $b$ is in the center $Z(G)$. Denote $H=\langle b\rangle$, so $H\unlhd G$.
• $b^2=(xy)^2=(yz)^2=(zx)^2$ implying that $G/H$ has exponent two and is thus elementary 2-abelian.
• $xyz\in H$, so $G/H$ is generated by the cosets $xH$ and $yH$ and therefore $|G/H|\le 4$.
• Because $b^2\in Z(G)$ we have $(xy)^2=b^2=(yx)^2$ implying that $b^4=1$.
• $|G|\le16$.
• Consider the following complex matrices and prove that $|G|\ge16$: $$ X=\left(\begin{array}{cccc}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{array}\right), $$ $$ Y=\left(\begin{array}{cccc}0&0&1&0\\0&0&0&-1\\1&0&0&0\\0&-1&0&0\end{array}\right), $$ $$ Z=\left(\begin{array}{cccc}0&0&0&i\\0&0&i&0\\0&-i&0&0\\-i&0&0&0\end{array}\right). $$ Hint $XYZ=i I_4$.