# Obtaining quiver and relations for finite p-groups

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.

• 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. – 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:

LoadPackage("qpa");

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

A := AlgebraAsQuiverAlgebra(KG)[1];
R := RelationsOfAlgebra(A);

• 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... – Jabby Feb 8 '19 at 14:40
• I now use AlgebraAsQuiverAlgebra instead of IsomorphismFpAlgebra. Can you check if the result is now correct? – Jabby Feb 8 '19 at 15:05
• 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. – 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.

• 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. – Mare Jan 30 '19 at 18:04
• I didn't know it was in to the book. – Benjamin Steinberg Jan 30 '19 at 18:16
• The way the question was worded made it sound like the quiver was not known either. – Benjamin Steinberg Jan 30 '19 at 18:21
• No problem, I probably should have added that in the question. But your answer gives a nice quick proof so it is still useful. – Mare Jan 30 '19 at 18:48
• 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 $K,+$. – Benjamin Steinberg Jan 30 '19 at 18:58