Let $\mathbb{F}_q$ be a finite field with $q$ elements and let $G$ be the group of inhomogeneous linear transformations over $\mathbb{F}_q$, that is $x\mapsto ax+b$ for some $a\in \mathbb{F}_q^*$ and $b\in\mathbb{F}_q$. Now I have to find all the irreducible representations of $G$. The hint in the book is that we should look at $V=\{f\colon \mathbb{F}_q\rightarrow \mathbb{C}\colon \sum_{x\in \mathbb{F}_q}f(x)=0\}$.

The dimension of this vector space is $q-1$ since we can choose the first $q-1$ 'coordinates' freely. First I need to show that this is indeed a representation, thus if $f\in V$ then also $\rho(g)f\in V$ where the action is defined by $\rho(g) f(x)=f(g(x))=f(ax+b)$. So I need to show that $\sum_{x\in \mathbb{F}_q}f(ax+b)=0$. Since the function $g\colon x\mapsto ax+b$ is bijective, this equality indeed holds. Now I have to show that this is irreducible. The order of $G$ is $q(q-1)$ and $(q-1)^2=q^2-2q+1$, so after this we have to find other representations such that the sum of squares formula is fulfilled. Can you give me hints on showing that this representation is irreducible? Thanks.

  • $\begingroup$ Have you figured out the conjugacy classes so that you know how many irreducible reps you are supposed to find? Did you notice that $G$ acts on $\Bbb{F}_q$ doubly transitively? Have you covered the theorem explaining that whenever a finite group $G$ acts on a set $S$ doubly transitively, then you get an irreducible rep of dimension $|S|-1$ by extracting the trivial rep out of $F(S)=\{f:S\to\Bbb{C}\}$? Last but not least you can calculate the character $\chi_V$, and check that $\langle \chi_V,\chi_V\rangle_G=1$. $\endgroup$ Nov 11, 2014 at 22:37
  • $\begingroup$ Oh, and for calculating the conjugacy classes the technique described here may help you. $\endgroup$ Nov 11, 2014 at 22:38


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