# maximal order of elements in GL(n,p)

I am looking for a formula for the maximal order of an element in GL(n,p), where p is prime.

I recall seeing such a formula in a paper from the mid- or early 20th century, but could not find again this reference. I will be grateful for any hint.

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Well, by Hamilton--Cayley, each matrix $A\in {\rm GL}(n,p)$ generates an at most $n$-dimensional subalgebra ${\mathbb F}_p[A]\subseteq M(n,p)$ thus containing at most $p^n-1$ nonzero elements. Hence the order of $A$ cannot exceed $p^n-1$.

On the other hand, consider a degree $n$ monic polynomial $P_n$ whose root is a generator $\xi$ of ${\mathbb F}_{p^n}^*$. Then a matrix with $P_n$ as its characteristic polynomial has order at least $p^n-1$ since $\xi$ is its eigenvalue.

ADDENDUM. if you wish the order to be the power of $p$, then the answer is $d=p^{\lceil \log_p n\rceil}$. Since the order of $A$ is divisible by the multiplicative orders of its eigenvalues, all the eigenvalues should be $1$. Hence the characteristic polynomial is $(x-1)^n$, so $A^d-I=(A-I)^d=0$.

On the other hand, if $A=I+J$ is the Jordan cell of size $n$ (with eigenvalue 1), then $A^{d/p}=I^{d/p}+J^{d/p}\neq I$, but $A^d=I+J^d=I$.

NB. The subgroup of all (upper-)unitriangular matrices is a Sylow $p$-subgroup in ${\rm GL}(n,p)$. So you may concentrate on it when looking at the elements of this kind.

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Thank you very much for the elegant answer! A related question: what is the maximal p-power which is the order of an element of GL(n,p)? –  user27196 Oct 13 '12 at 6:41
I've added an answer to this question, too. –  Ilya Bogdanov Oct 13 '12 at 7:20
Thank you very much again - this was very helpful. –  user27196 Oct 13 '12 at 8:31
Beautiful trick to use Cayley-Hamilton here! –  Peter Mueller Oct 13 '12 at 9:35
I have meanwhile found the paper: Ivan Niven, Fermat theorem for matrices, Duke Math. J. 15 (1948), 823-826, which gives an elementary and explicit description of the possible orders of elements in GL(n,q), where q is a prime power. Thanks again. –  user27196 Oct 14 '12 at 7:26