# Automorphism group of $UT(n,p)$, the group of unitriangular matrices over the field $\mathbb{F}_p$

I am interested in understanding (at least roughly, if no such a description exists) the group of automorphisms for the group $UT(n,p)$, of unitriangular matrices over the field $\mathbb{F}_p$ on $p$-elements. Unfortunately the online searching I've carried out wasn't quite helpful. If no description exists, maybe some information how the automorphisms behave on the set of subgroups of it may be helpful too.

For instance when $n=3$, I know we have that any $\phi \in Aut(UT(3,p))$, permutes the pairs of non-commuting elements of $UT(3,p)$ (which in that case happens always to be a generating set) and the subgroups of order $p^2$.

So am wondering if something similar exists (of course modified somehow) in the general case too.

If the above sounds quite general, the case where $n=4$, is of great importance for me either. Any comment might be useful, and of course references too.

Thanks!

• I haven't given this any thought at all, but computer calculations with $3 \le n \le 8$ strongly suggest that the order of the automorphism group of ${\rm UT}(n,p)$ is $p^3(p-1)^2(p+1)$ when $n=3$, and $2p^e(p-1)^{n-1}$ with $e = (n^2+n-4)/2$, when $n > 3$. So, for $n=4$, this would give $2p^8(p-1)^3$. – Derek Holt Sep 11 '17 at 12:04
• Derek thank you very much for your reply. For $n=3$ I know that this is the case indeed, since I can write down explicitly what's going on. For $n=4$ though the situation becomes really hard and what you wrote is helpful indeed. What I am trying to understand is mainly how the automoprhisms permute the subgroups of order $p^3$ (for $n=4$ at least) and because I don't have access to any computational system at the moment is difficult to do any computation by hand. – mayer_vietoris Sep 11 '17 at 12:11
• I did some calculations on the orbits of the automorphism group of ${\rm UT}(4,p)$ on subgroups of order $p^3$ for $p=3,5,7$. At this stage, computations are starting to take longer, but I could go a bit further if necessary. In summary, for $p=3,5,7$, there are respectively $109$, $391$, and $953$ conjugacy classes of subgroups of order $p^3$, and the automorphism group has respectively $19$, $21$ and $23$ orbits on these subgroups. – Derek Holt Sep 11 '17 at 12:56
• Thank you very much for all the effort you have put on this. One question only, when you mean "...and the automorphism group has respectively 19, 21 and 23 orbits on these subgroups.", you mean that the $Aut(UT(4,p))$-action on the set of subgroups of order $p^3$ yields these 19,21 and 23 orbits respectively or something different? Because it confuses me a little bit. – mayer_vietoris Sep 11 '17 at 13:30
• Yes that is what I meant. I did the calculation in Magma by computing orbits of the outer automorphism group on the conjugacy classes of subgroups of order $p^3$. I can provide more detailed information on this necessary. – Derek Holt Sep 11 '17 at 13:35

• over $F_2$, any size: Maginnis J. S., (1993/11)."Outer Automorphisms of Upper Triangular Matrices." Journal of Algebra 161(2): 267-270. Abstract: The outer automorphism group of the upper triangular matrices over the field of two elements is calculated. A. J. Weir (Proc. Amer. Math. Soc.6 (1955), 454-464) performed a similar calculation for Fields of odd characteristic, and we borrow the term extremal auto ....