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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.

etotheipi
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  • 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 $\langle \chi_V,\chi_V\rangle_G=1$. – Jyrki Lahtonen Nov 11 '14 at 22:37
  • Oh, and for calculating the conjugacy classes the technique described here may help you. – Jyrki Lahtonen Nov 11 '14 at 22:38

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