5
$\begingroup$

Given a finite field $K$ with $p$ elements and a finite $p$-group $G$, is there a way to obtain the quiver and relations of $KG$ with GAP (and its package QPA)?

Since $KG$ is local, the quiver should easily be obtainable via the radical $J$ as $J/J^2$. But I do not know how to obtain the relations and whether this can be done in a quick way with a computer.

$\endgroup$
  • $\begingroup$ My guess is that if $G$ is a finite p-group the quiver has one vertex and $\log_p |G/\Phi(G)|$ loops where $\Phi(G)$ is the Frattini subgroup. $\endgroup$ – Benjamin Steinberg Jan 30 '19 at 16:43
1
$\begingroup$

Perhaps the QPA function AlgebraAsQuiverAlgebra can be used to obtain what you want?

I tested the following small example:

LoadPackage("qpa");

p  := 3;
K  := GF(p);
G  := CyclicGroup(p);
KG := GroupRing(K, G);

A := AlgebraAsQuiverAlgebra(KG)[1];
R := RelationsOfAlgebra(A);
| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'll delete the answer if you don't mind. The GAP output suggests $A = K[x]/(x(x^3+1))$, which is of course not isomorphic to $KG$. Perhaps I misinterpreted what some of the functions are supposed to do... $\endgroup$ – Jabby Feb 8 '19 at 14:40
  • 1
    $\begingroup$ I now use AlgebraAsQuiverAlgebra instead of IsomorphismFpAlgebra. Can you check if the result is now correct? $\endgroup$ – Jabby Feb 8 '19 at 15:05
  • $\begingroup$ I just saw (after reading your answer) that the command AlgebraAsQuiverAlgebra is even in the QPA manual and does exactly what is needed. So this solves this problem. Thanks. $\endgroup$ – Mare Feb 8 '19 at 15:09
4
$\begingroup$

Here is a computation of the quiver. I don't know how to get the relations. Let $G$ be a finite $p$-group and $K$ the $p$-element field. Note that the trivial module is the unique simple module and the radical $J$ is the augmentation ideal. It is well known that $J$ has basis the elements of the form $g-1$ with $g\in G\setminus\{1\}$. Note that

$(g-1)(h-1)= gh-1-(g-1)-(h-1)$

and so

$gh-1+J^2=(g-1)+J^2+ (h-1)+J^2$.

Thus $J/J^2$ is a $K$-vector space with generators $[g]$ with $g\in G$ and relations $[g]+[h]=[gh]$. In other words, $J/J^2$ is the universal elementary abelian $p$-group image of $G$, which is $G/\Phi(G)$ with $\Phi(G)$ the Frattini subgroup of $G$ (the intersection of all maximal subgroups $=[G,G]G^p$).

Thus $\dim J/J^2 = \log_p |G/\Phi(G)|$ and so you have one vertex and that number of loops. I would guess GAP could find a set of generators for the Frattini quotient and then your path algebra would map the loop corresponding to $x\Phi(G)$ to $x-1$ and I assume there are algorithms to get the kernel.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thanks, that can be also found in the book "Representations and cohomology volume 1" by Benson as proposition 3.14.2. The main problem is to find a good/quick way to obtain the relations. It might be possible that there is already an easy way with GAP commands that I am not aware of. $\endgroup$ – Mare Jan 30 '19 at 18:04
  • $\begingroup$ I didn't know it was in to the book. $\endgroup$ – Benjamin Steinberg Jan 30 '19 at 18:16
  • 1
    $\begingroup$ The way the question was worded made it sound like the quiver was not known either. $\endgroup$ – Benjamin Steinberg Jan 30 '19 at 18:21
  • $\begingroup$ No problem, I probably should have added that in the question. But your answer gives a nice quick proof so it is still useful. $\endgroup$ – Mare Jan 30 '19 at 18:48
  • 2
    $\begingroup$ I see. He does it via cohomology. In fact his proof is very close to mine. The augmentation ideal represents derivations so $J/J^2\cong Hom_{KG}(J/J^2,K)\cong Hom_{KG}(J,K)\cong Der(K)$. But since $K$ is the trivial module a derivation is a homomorphism from $G$ to $K,+$, which is the same thing as a homomorphism $G/\Phi(G)$ to $K,+$. $\endgroup$ – Benjamin Steinberg Jan 30 '19 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.