Complex Representations of GL(2,K) for Finite Fields K by Ilya Piatetski-Shapiro

By Ilya Piatetski-Shapiro

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We have (xy)2 = xyxy = xxyy (G is abelian) = x2 y 2 (x, y ∈ H) = ee = e. Therefore, xy ∈ H. ) Let x ∈ H. We have x−1 2 = x−2 = x =e (law of exponents) 2 −1 −1 (law of exponents) (x ∈ H) = e. Therefore, x−1 ∈ H. By the Subgroup Theorem, H is a subgroup of G. 3 Cyclic subgroup Let G be a group and let a be a fixed element of G. Put a = {am | m ∈ Z}. 1 Theorem. a is a subgroup of G. Proof. 2). ) We have e = a0 ∈ a . ) Let x, y ∈ a . Then x = am and y = an for some m, n ∈ Z. So xy = am an = am+n ∈ a .

G2) We claim that (e1 , e2 ) is an identity element, where ei is the identity element of Gi (i = 1, 2). For (x1 , x2 ) ∈ G1 × G2 , we have (e1 , e2 )(x1 , x2 ) = (e1 x1 , e2 x2 ) = (x1 , x2 ) and similarly (x1 , x2 )(e1 , e2 ) = (x1 , x2 ). Therefore, (e1 , e2 ) is an identity element. −1 (G3) Let (x1 , x2 ) ∈ G1 × G2 . We claim that (x−1 1 , x2 ) is an inverse of (x1 , x2 ). We have −1 −1 −1 (x−1 1 , x2 )(x1 , x2 ) = (x1 x1 , x2 x2 ) = (e1 , e2 ) −1 −1 −1 and similarly (x1 , x2 )(x−1 1 , x2 ) = (e1 , e2 ).

For instance: (1, 5, 2, 4) = (4, 1, 5, 2) = (2, 4, 1, 5) = (5, 2, 4, 1). ” If the numbers are arranged in order around a circle, then a cyclic permutation corresponds to a rotation of the circle. 59 The inverse of a cycle is obtained by writing the entries in reverse order. For example, (1, 5, 2, 4)−1 = (4, 2, 5, 1). A transposition is a 2-cycle. The transposition (i1 , i2 ) transposes (interchanges) the two numbers i1 and i2 and fixes every other number. A 1-cycle (i1 ) is the identity since it fixes i1 as well as every other number.

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