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

By Ilya Piatetski-Shapiro

Best abstract books

Groebner bases and commutative algebra

The center-piece of Grobner foundation conception is the Buchberger set of rules, the significance of that's defined, because it spans mathematical concept and computational purposes. This entire therapy turns out to be useful as a textual content and as a reference for mathematicians and machine scientists and calls for no necessities except the mathematical adulthood of a sophisticated undergraduate.

Group Rings and Class Groups

The 1st a part of the e-book facilities round the isomorphism challenge for finite teams; i. e. which homes of the finite team G could be decided by means of the vital team ring ZZG ? The authors have attempted to provide the consequences roughly selfcontained and in as a lot generality as attainable about the ring of coefficients.

Finite Classical Groups [Lecture notes]

(London Taught direction middle 2013)

Extra info for Complex Representations of GL(2,K) for Finite Fields K

Sample text

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.