Periodic matrices in SL(3,Z) will be conjugated to product of periodic matrices in SL(2,Z) by +- indentity on a third integer direction. Is this true?
Sorry, following your comments, maybe something I said is misleading. I state the original question: Consider a periodic automorphism $\phi$ on $Z^3$, can we find a coordinate on $Z^3$, such that $\phi$ is either (1,0;0,A) or (-1,0;0,A).
ok, let me expand Geoff's suggestion. Let $A\in SL(3,\mathbb{Z})$ be such that $A^n=Id$ for some positive integer $n$. Since the characteristic polynomial of $A$ is a cubic polynomial of the form $-t^3+\cdots +1$, it has a positive real root; and since all roots of $A$ are roots of the unit, 1 is an eigenvalue of $A$. Moreover, since $A^n=Id$, $A$ is semisimple over $\overline{\mathbb{Q}}$ and so its minimal polynomial over $\mathbb{Q}$ is of the form $(t-1)p(t)$ with $p(t)\in \mathbb{Z}[t]$ a cyclotomic polynomial of degree at most 2 with $p(1)\neq 0$. This leaves only the following possibilities: $1$, $t+1$, $t^2+t+1$, $t^2+1$ or $t^2-t+1$, corresponding to $A$ having period $1,2,3,4$ or $6$ respectively. The period 1 case is trivial. For the period 6 case one can work as follows: we have a splitting of $\mathbb{Q}^3$ as $V\oplus W$, where $V$ and $W$ are $A$-stable $\mathbb{Q}$-vector spaces, with $\dim_{\mathbb{Q}}(V)\geq 1$ and with $A$ acting as the identity on $V$. Moreover, since $p(1)=1$, we can find polynomials $a(t)$ and $b(t)$ in $\mathbb{Z}[t]$ such that $a(t)p(t)+b(t)(t-1)=1$. This implies that $\mathbb{Z}^3=(\mathbb{Z}^3\cap V)\oplus (\mathbb{Z}^3\cap W)$. The abelian subgroup $\mathbb{Z}^3\cap V$ of $\mathbb{Z}^3$ is free since subgroups of free abelian groups are free, and has rank $\dim_{\mathbb{Q}}V$, since $V$ is a $\mathbb{Q}$-subspace of $\mathbb{Q}^3$. So we can find a $\mathbb{Z}$-basis $B_V$ for $\mathbb{Z}^3\cap V$ consisting of $\dim_{\mathbb{Q}}V$ elements. The same for $\mathbb{Z}^3\cap W$. The two basis $B_V$ and $B_W$ together are a $\mathbb{Z}$-basis of $\mathbb{Z}^3$ and up to multiplying by $-1$ one of the vectors in this basis we may assume that the change of basis matrix $P$ from the standard basis of $\mathbb{Z}^3$ to the basis $B_V\cup B_W$ is an element of $SL(3;\mathbb{Z})$. By construction, $PAP^{-1}$ is a block-diagonal matrix in $SL(3;\mathbb{Z})$, with an upper $1\times 1$ block $(1)$ and a lower $2\times 2$ block in $SL(2;\mathbb{Z})$.
This method, however, clearly does not apply to the remaining three cases.
The matrix
$0\ 1\ 0$
$1\ 0\ 0$
$0\ 0 \-1$
in $SL_3(\mathbb Z)$ is not conjugate to any block sum of an $SL_2(\mathbb Z)$ matrix and $+1$. And of course it is not conjugate to any block sum of an $SL_2(\mathbb Z)$ matrix and $-1$, either.
For $GL_3$ and $GL_2$ the answer is yes.
EDIT This last statement is wrong; see Geoff Robinson's comments.
