#  "Is there a group of order 48 in the set of" \ \ 3 xx 3 \ \ "matrices of integers ?"   "If so, can you exhibit one ? If not, prove its impossibility."

Feb 17, 2018

Yes, for example the group generated by:

$\left(\begin{matrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ - 1 & 0 & 0\end{matrix}\right)$, $\left(\begin{matrix}- 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & - 1 & 0\end{matrix}\right)$, $\left(\begin{matrix}- 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1\end{matrix}\right)$

#### Explanation:

Consider the subgroup of $G L \left(3 , \mathbb{Z}\right)$ generated by the $2$ matrices:

$A = \left(\begin{matrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ - 1 & 0 & 0\end{matrix}\right)$

$B = \left(\begin{matrix}- 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & - 1 & 0\end{matrix}\right)$

Notice that $A$ represents a rotation of $\frac{\pi}{2}$ about the $y$-axis, while $B$ represents a rotation of $\pi$ about the line $x = y + z = 0$.

Note that $\det \left(A\right) = \det \left(B\right) = 1$, so they are elements of $S L \left(3 , \mathbb{Z}\right)$.

Between them, these two geometrical operations generate a subgroup of $S L \left(3 , \mathbb{Z}\right)$ of order $24$, isomorphic to ${S}_{4}$, and the rotational symmetry group of the cube.

If we then add a third generator:

$C = \left(\begin{matrix}- 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1\end{matrix}\right)$

with $\det \left(C\right) = - 1$

we get a subgroup of $G L \left(3 , \mathbb{Z}\right)$ of order $48$, which is the full symmetry group of the cube.

For an in depth analysis of the finite subgroups of $G L \left(3 , \mathbb{Z}\right)$ you may like to read the paper at https://projecteuclid.org/download/pdf_1/euclid.nmj/1118798212