# Show that $SO_3(\mathbb{Z}) \simeq SO_3(\mathbb{Z}/3\mathbb{Z})$

I have read the surprising fact that $SO_3(\mathbb{Z}) \simeq SO_3(\mathbb{Z}/3\mathbb{Z})$. At first I could only come up with diagonal elements of $SO_3$ such as:

$$\left[ \begin{array}{rrr} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array} \right]$$

However, then I was able to come up with rotations:

$$\left[ \begin{array}{rrr} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right]$$

This rotation group needs to preserve the cube $\{ -1, 1\}^3$, and the are 24 symmetries of the cube, so $|SO_3(\mathbb{Z})| \leq 24$.

How can I show this is an isomorphism?

• There is a surjective homomorphism from ${\rm SO}_3({\mathbb Z})$ to ${\rm SO}_3({\mathbb Z}/3{\mathbb Z}))$ defined by taking all matrix entries modulo $3$. Since both groups have the same order, it must be an isomorphism. – Derek Holt Apr 5 '16 at 8:10

For a simple explanation : a matrix in $O_n(\mathbb{Z})$ can only have $0$ and $\pm 1$ as coefficients, so reducing modulo $3$ is injective.