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Let $G={\rm GL}(n,\mathbb{F}_q)$ and let $N$ be the nilpotent radical of the standard Borel subgroup, i.e., $N$ is the subgroup of $G$ consisting of upper triangular matrices with 1’s on the diagonal. Given a non-trivial character $\psi: \mathbb{F}_q\rightarrow \mathbb{C}^\times$, we define a character $\psi_N$ of $N$ by

$\psi_N( \begin{pmatrix} 1 & x_{12} & x_{13} & \cdots & x_{1n}\\ & 1 & x_{23} & \cdots & x_{2n}\\ & & 1 & \cdots & \vdots \\ & & & \ddots & \vdots\\ & & & & 1\\ \end{pmatrix}) = \psi(x_{12}+x_{23}+\cdots+x_{n-1, n}).$

This defines a one-dimensional representation of $N$. One can show that the induced representation $\mathcal{G}={\rm Ind}_N^G(\psi_N)$ is multiplicity-free, i.e., when we decompose $\mathcal{G}={\rm Ind}_N^G(\psi_N)$ into a sum of irreducible constituents,

$\mathcal{G}={\rm Ind}_N^G(\psi_N)=\oplus a_i\pi_i,$

we have $a_1=1$ for all $i$. Each constituent $\pi_i$ is called a generic irreducible representation. There are many reasons for calling them generic. One is that no matter which non-trivial character $\psi$ we choose, we obtain the same induced representation up to equivalence. One can see this by looking at the induced character values on each conjugacy class. These values will be polynomials in $q$ and, once simplified, do not depend on the character $\psi$. Moreover, this Gelfand-Graev representation $\mathcal{G}$ contains most irreducible representations of the group.

To see this, we can compare the number of irreducible constituents of $\mathcal{G}$ with the number of irreducible representations of $G$. By a standard result in the representation theory of finite groups, the number of irreducible representations of $G$ is equal to its number of conjugacy classes.