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Let $p,q$ be primes s.t. $q=np+1$. Denote $m=p^2$. Then $\mathbb{Z}_p$ acts non-trivially on $\mathbb{Z}_q$, so we have a non-abelian semi-direct product $\mathbb{Z}_q\rtimes \mathbb{Z}_m$, with the action of $\mathbb{Z}_p$. This group has a faithful irreducible representation.

Over $\mathbb{C}$, every faithful irreducible representation must be of dimension $p$, because the dimension must divide $p^2q$, it cannot be 1 since the group is non-abelian, and it cannot be greater than $p$ because there is an abelian subgroup of order $pq$.

Is it possible to bound the dimensions of the faithful irreducible representations in characteristic $p,q$ as well?

Also, is there a simple way to check if the faithful irreducible representation I have is also irreducible in characteristic $p,q$?

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4  
But surely this group has a nontrivial normal $p$-subgroup and also a nontrivial normal $q$-subgroup, so there are no faithful irreducible representations in characteristic $p$ or $q$. –  Derek Holt Jan 10 '13 at 19:36
    
Well, clearly the center is $Z_p$, and $Z_q$ is isomorphic to a normal subgroup by construction. Why does this mean there are no faithful irreducible representations in characteristic $p$ or $q$? (I guess it must be elementry, but I don't understand). Thanks. –  A.B. Jan 10 '13 at 20:50
3  
Any irreducible representation of a finite group $G$ in characteristic $p$ has $O_p(G)$ in its kernel. –  Derek Holt Jan 11 '13 at 14:23

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