3

I believe that I've read a paper or two that have fully classified groups of prime power order acting irreducibly on elementary abelian $p$-groups. Or, at least, cyclic groups of prime power order acting irreducibly on elementary abelian $p$-groups. However, I cannot find the reference. Can someone please point me in the right direction? Thank you.

Shaun
  • 47,747

1 Answers1

4

The general situation you are asking about has a $q$-group $Q$ acting irreducibly on $k^n$ where $k$ is the prime field of order $p.$ When $q=p$ there is only the trivial action with $n=1,$ but when $q\neq p$ this situation effectively includes the whole representation theory of $Q$ in coprime characteristic. In particular, we can choose $p=1$ mod $|Q|$ and then $\mathbb F_p$ will be a splitting field for $Q,$ which means the representation theory of $Q$ over $k=\mathbb F_p$ is essentially the same as over $\mathbb C.$ There is definitely no classification, though of course many important facts about this general situation are known.

In the cyclic case, there certainly is a classification. I have to correct my earlier answer in that I can't seem to locate an explicit treatment in any textbook, but there is a clear statement and proof on StackExchange here (that answer covers arbitrary $|Q|$). In summary, the irreducible faithful representations of cyclic $Q$ over $k$ have dimension $f$ where $f$ is the order of $p$ mod $|Q|,$ in other words $f$ is minimal such that $Q$ is isomorphic to a subgroup of $\mathbb F_{p^f}.$ There are $\phi(|Q|)/f$ inequivalent representations, in each of which a generator of $Q$ acts like multiplication by a primitive $|Q|$-th root of unity on $\mathbb F_{p^f}$ regarded as an $f$-dimensional vector space over $k.$ The representations are irreducible because by choice of $f,$ $\mathbb F_{p^f}=k[\zeta]$ where $\zeta$ is any primitive $p^f$-th root of unity. Galois conjugate roots of unity give equivalent representations, hence the count of inequivalent representations just given. The sum of the dimensions of these representations is of course $\phi(|Q|),$ so considering all the representations, we get a total dimension sum of $$\phi(1)+\phi(q)+\cdots+\phi(|Q|)=|Q|.$$ In coprime characteristic, all irreducibles are direct summands of the group algebra, so it follows that these are all the irreducible representations of $Q.$

Tom WIlde
  • 504