# Irreps of $S_3=GL(2,2)$. Who is cuspidal?

$S_3 = GL(2, F_2)$. It has $3$ irreps - trivial, sign, standard 2d.

Can one help me to understand general words about Irreps of $G(F_q)$ ("principal series", "parabolically induced", "cuspidal", "complimentrary series", "Steinberg irrep") in this particular example ? I mean who of these irreps is who ?

Notes on the subject I am looking:

Paul Garrett: http://www.math.umn.edu/~garrett/m/v/toy_GL2.pdf

Etingof&Students: http://arxiv.org/abs/0901.0827

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Side remark: Have a look at Weintraub - "Representation theory of finite group" in the section Mackey Machine math.stackexchange.com/questions/38571/… What is the relationship between Mackey's theorem in character theory and Mackey's theorem in transfer theory? math.stackexchange.com/questions/189430/… –  Alexander Chervov Sep 26 '12 at 10:49
math.stackexchange.com/questions/153381/… Estimates on conjugacy classes of a finite group. Theorem: Let A be a normal subgroup of G such that A is the centralizer of every non-trivial element in A. If further G/A is abelian, then G has |G:A| linear characters, and (|A|−1)/|G:A| non-linear irreducible characters of degree =|G:A| which vanish off A. math.stackexchange.com/questions/117500/… –  Alexander Chervov Sep 26 '12 at 10:50
Let be a normal subgroup of a finite group . Let belonging to be a conjugacy class of elements in , and assume that belongs to . Prove that is a union of conjugacy classes in , all having the same cardinality, where equals the index of the group generated by and the centralizer in of and element belonging to . math.stackexchange.com/questions/5614/… –  Alexander Chervov Sep 26 '12 at 10:54
Here, the "parabolically-induced" (="principal series") repn's standard model is just left $P$-invariant ($P$=upper-triangular) functions on the group $G$, since there is only the trivial character on diagonal things $M=\{1\}$. This repn is $|G|/|P|$-dimensional, that is, $(3\cdot 2)/2=3$, contains the trivial repn, obviously, and the 2-dimensional complement is the "Steinberg", by conventional naming. Evidently, the single "supercuspidal" (meaning sums to $0$ over the unipotent radical $N$ of $P$) is the "sign" repn. A peculiar outcome!