maximal order of elements in GL(n,p) - MathOverflow most recent 30 from http://mathoverflow.net2013-06-18T06:53:35Zhttp://mathoverflow.net/feeds/question/109483http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/109483/maximal-order-of-elements-in-gln-pmaximal order of elements in GL(n,p)idanps2012-10-12T19:26:08Z2012-10-13T07:25:43Z
<p>I am looking for a formula for the maximal order of an element in GL(n,p), where p is prime. </p>
<p>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. </p>
http://mathoverflow.net/questions/109483/maximal-order-of-elements-in-gln-p/109488#109488Answer by Ilya Bogdanov for maximal order of elements in GL(n,p)Ilya Bogdanov2012-10-12T20:13:53Z2012-10-13T07:25:43Z<p>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 $p^n-1$.</p>
<p>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.</p>
<p><b>ADDENDUM.</b> 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 $A^d-I=(A-I)^d=0$.</p>
<p>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 $A^d=I+J^d=I$.</p>
<p><b>NB.</b> 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.</p>