Let $q$ be a prime power and let $n\geq2$ be an integer. Is it known what is the largest order of a unipotent uppertriangular $n\times n$ matrix over the ring $\mathbb{Z}/q\mathbb{Z}$? I am mostly interested in upper bounds.

1$\begingroup$ Just checking: you really mean the integers modulo $q$, and not the field of $q$ elements, right? $\endgroup$– Gerry MyersonMar 2, 2018 at 11:49

4$\begingroup$ @GerryMyerson  Yes, the ring. $\endgroup$– user05811Mar 2, 2018 at 11:55
2 Answers
This works for any unipotent matrix whose characteristic polynomial is $(x1)^n$ (including some uppertriangular matrices without ones on the diagonal). By CayleyHamilton, the matrix $U$ satisfies $(U1)^{n}=0$, so we can write $U = 1 + N$ with $N^n=0$.
Then $$ U^{p^k} = (1+N)^{p^k} = \sum_{i=0}^{n1} {p^k \choose i} N^i$$ so it is sufficient to find $k$ such that ${p^k \choose i }=0\mod q$ for all $i$ from $0$ to $n1$.
Consider the group action of $\mathbb Z/p^k$ by translation on subsets of $\mathbb Z/p^k$ of size $i$. The order of the stabilizer of any point divides $\gcd ( i,p^k)$, so by the orbitstabilizer theorem, the binomial coefficient is a multiple of $p^k/\gcd(i,p^k)$. The largest possible value of $\gcd(i,p^k)$ is the largest power of $p$ less than $n$, and we want $p^k$ to be $q$ times it, so it suffices to take
$$p^k = q p^ { \lfloor \log_p (n1) \rfloor }$$
and the order of the matrix is at most this.
One can check this is sharp by taking a single unipotent Jordan block, and checking that the lower bound on the $p$adic valuation of binomial coefficients is sharp. (At this point one needs to use the fact that the order is necessarily a power of $p$ to reduce to the case of a $p^k$th power.)

1$\begingroup$ Thank you for the beautiful argument. Doesn't the argument in fact lead to the formula $p^k=qp^{\lceil\log_pn\rceil1}$? $\endgroup$ Mar 2, 2018 at 15:21

3$\begingroup$ In the first displayed formula, is $U^N$ supposed to be $U^{p^k}$? If so, should one start by mentioning that the order of the element must be a power of $p$, since the order of the group is a power of $p$? $\endgroup$ Mar 2, 2018 at 15:42

$\begingroup$ As Joe Silverman points out, the notation $U^N$ in the display doesn't make sense. Please edit. $\endgroup$ Mar 2, 2018 at 16:02

$\begingroup$ @user05811 You're welcome. Yes. I actually meant to write a slightly different formula, which is equivalent. $\endgroup$ Mar 2, 2018 at 16:57

$\begingroup$ @JoeSilverman Yes, that's correct. The argument doesn't actually need to use independently the fact that the order is a power of $p$  we show explicitly that the $p^k$th power of the element is $1$, from which the fact about the order follows. $\endgroup$ Mar 2, 2018 at 16:59
Let $q=p^r$. The eigenvalues of a unipotent matrix are all 1 and so we may conjugate it into Jordan canonical form without extending the field. This yields a unipotent upper triangular matrix with entries in ${\Bbb Z}/p{\Bbb Z}$. Thus the answer is independent of $r$.
Consider a Jordan block $A$ of size $m$. Computing its powers we see that its order is $p^t$ where $t=\lceil m / p \rceil$. Thus your answer is $p^{\lceil n / p \rceil}$.

5$\begingroup$ What is a reference for 'Jordan canonical form' over rings? Please note that $\mathbb{Z}/p^k\mathbb{Z}$ is not a field if $k>1$. $\endgroup$ Mar 2, 2018 at 10:56

1$\begingroup$ To second @PeterHeinig's question, in what sense can you guarantee that you can conjugate a matrix over $\mathbb Z/q\mathbb Z$ so that it not only is upper triangular, but has its entries in $\mathbb Z/p\mathbb Z$? For that matter, what do you mean by $\mathbb Z/p\mathbb Z$? Since it's implicitly inside $\mathbb Z/q\mathbb Z$, I guess you mean $q'\mathbb Z/q\mathbb Z$ (with $q' = q/p$), but that seems to be quite a strong statement. $\endgroup$– LSpiceMar 2, 2018 at 11:13

1$\begingroup$ Note that when $n=2$, the group of $2\times 2$ uppertriangular unipotent matrices over $\mathbb{Z}/q\mathbb{Z}$ is cyclic of order $q$. $\endgroup$ Mar 2, 2018 at 11:29

4$\begingroup$ I think that this is a good and correct answer for the field $\mathbb{F}_q$ of order $q$, but the question was about $\mathbb{Z}/q\mathbb{Z}$, which is of course different. $\endgroup$ Mar 2, 2018 at 11:34

2$\begingroup$ This answer is not right even over the field $\mathbb F_p$. You mean $p ^ { \lceil \log_p n \rceil }$. $\endgroup$ Mar 2, 2018 at 12:49