Order of $A \in GL(n,\mathbb Z_p)$ cannot exceed $p^n-1$ ?

Question

If $A \in GL(n,\mathbb Z_p)$ then is it true that order of $A$ cannot exceed $p^n-1$ ?

Answer 1

If $A$ is a matrix in your group, its eigenvalues (which are not zero) are zeroes of a polynomial of degree $n$. It follows that they belong to a field extension $E$ of $\mathbb Z_p$ of degree at most $n$.

Since the multiplicative group of $E$ has order $p^n-1$, we see that every eigenvaluee has order dividing $p^n-1$. If $A$ is diagonalizable, this gives what you want.

Can you extend this to the non-diagonalizable case? (the rational canonical form can serve as a good replacement of diagonalizability)

Answer 2

More generally, the multiplicative order of any $A\in GL(n,\Bbb F_q)$ is at most $q^n-1$, if $n>0$. (Here $\Bbb F_q$ is a finite field of order $q$; for any prime$~p$ one has $\Bbb F_p=\Bbb Z/p\Bbb Z$.)

Every such $A$ has a minimal polynomial $\mu_A\in\Bbb F_q[X]$ of degree$~d\leq n$ (by the Cayley-Hamilton theorem, if you like). This means that the vector subspace of $M_n(\Bbb F_q)$ spanned by all powers of$~A$ is already spanned by $A^0,A^1,\ldots,A^{d-1}$ (see this recent question) and has dimension$~d$. Since all powers of $A$ are nonzero elements of this subspace, there are at most $q^d-1\leq q^n-1$ distinct such powers.

The condition $n>0$ was necessary, because in the trivial ring $0$ is invertible. Also the given bound is attained, because the multiplicative group of $\Bbb F_{q^n}$ is cyclic, and viewing a generator as a linear operator on the $\Bbb F_q$-vector space $\Bbb F_{q^n}$ provides (after choosing a basis) an example of an $A$ for which the bound is attained.

More about orders of elements of $GL(n,\Bbb F_q)$ in this answer.