Conjugacy classes of PGL(3,Z)

Question

We know that every $2\times 2$ matrix in $PGL(2, \mathbb{Z})$ of order $3$ is conjugate to the matrix $$ \left( \begin{array}{cc} 1 & -1 \\ 1 & 0 \end{array} \right) $$.

I am interested in finding out to what extent this holds for $3\times 3$ integer invertible matrices.

In other words how many conjugacy classes of order 3 matrices in $PGL(3, \mathbb{Z})$ are there?

Answer 1

I will work in ${\rm GL}$ instead of ${\rm PGL}$.

The corresponding question over ${\rm GL}_3(\mathbb{Z})$ is essentially$^1$ equivalent to asking how many faithful $\mathbb{Z}[G]$-modules, free of rank 3 over $\mathbb{Z}$ there are up to isomorphism, where $G$ is the cyclic group of order 3. Any such module is a direct sum of indecomposable modules, and those have been classified in I. Reiner, Integral representations of cyclic groups of prime order, Proc. Amer. Math. Soc. 8 (1957), 142–146. There are three indecomposable $\mathbb{Z}[G]$-modules:

• the trivial module of $\mathbb{Z}$-rank 1, $\Gamma_1$,
• the augmentation ideal of $\mathbb{Z}[G]$, which has $\mathbb{Z}$-rank 2, $\Gamma_2$,
• the regular module $\Gamma_3=\mathbb{Z}[G]$ itself.

So there are two isomorphism classes of faithful modules of rank 3:

• $\Gamma_1\oplus \Gamma_2$,
• $\Gamma_3$.

One should beware, that, in general, the Krull-Schmidt Theorem fails for $\mathbb{Z}[G]$-modules, but in this case it is easy to see that the two guys are not isomorphic, since e.g. in one of them the trivial isotypical component is a direct summand, and in the other one it isn't. Alternatively, the two are not isomorphic over $\mathbb{Z}_3$, and Krull-Schmidt does hold over local rings.

${}^1$ To remove the word "essentially", one needs to check that any matrix obtained in this way is conjugate to its inverse. This is true because each of the indecomposable modules listed above can be extended to a module under the symmetric group $S_3$, as is easy to check.

Answer 2

The finite subgroups of $GL_3(\mathbb{Z})$ are known in the literature:

$\qquad$ Tahara: On the finite subgroups of $GL(3,\mathbb{Z})$. Nagoya Math. J. 41(1971), 169-209.

In particular Proposition 3 states that there are exactly two non-conjugate subgroups of order three. Representants are $$U_1=\langle \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & -1 \end{pmatrix}\rangle, \qquad U_2=\langle\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\rangle$$

Added: In $PGL_3(\mathbb{Z})=GL_3(\mathbb{Z})/\langle -I\rangle$ there are also exactly two conjugacy classes of subgroups of order 3 and representants are $\bar{U}_1,\bar{U}_2$.

For, let $V_i=\langle x_i\rangle \le G := GL_3(\mathbb{Z}),i=1,2$ be subgroups of order 3. If $\bar{V}_i$ are conjugate in $\bar{G} :=PGL_3(\mathbb{Z})$ then there is $g \in G$ s.t. $x_2=(\pm I)gx_1^kg^{-1} (k=1,2)$ and hence $(\pm I)^3=I$ and $x_2=gx_1^kg^{-1}$, i.e. the $V_i$ are conjugated in $G$.

Conversely, let $\bar{V}$ be a subgroup of $\bar{G}$ of order three. It's preimage in $G$ has order 6. Hence there is $V \le G$ of order three that maps to $\bar{V}$ and by the above $V$ is conjugated to some $U_i$.