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.

  • $\begingroup$ 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

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

I tested the following small example:


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

A := AlgebraAsQuiverAlgebra(KG)[1];
R := RelationsOfAlgebra(A);
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  • $\begingroup$ 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
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    $\begingroup$ 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

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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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