h4.pdf - Math 155 Designs and groups Handout#4 Simplicity of PSL2(F(|F | > 4 and PSLn(F(n > 3 — Outline 0 Let F be a finite field of q elements PSLn(F

# h4.pdf - Math 155 Designs and groups Handout#4 Simplicity...

• 2
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 1 - 2 out of 2 pages.

Math 155: Designs and groups Handout #4: Simplicity of PSL 2 ( F ) ( | F | 4) and PSL n ( F ) ( n 3) — Outline 0. Let F be a finite field of q elements. PSL n ( F ) is a normal subgroup [indeed the commutator subgroup, but we won’t need this] of PGL n ( F ) with index gcd( n, q - 1), and is generated by “transvections” because SL n ( F ) is; indeed even coordinate transvections suffice. (A coordinate transvection is a matrix with 1’s on the diagonal and a single nonzero off-diagonal entry. A linear transformation T : F n F n that is of that form for some choice of basis is a transvection; an equivalent coordinate-free criterion is: T - I has rank 1 and square zero.) When n = 2 the transvections in PSL 2 ( F ) are precisely the fractional linear transformations of P 1 ( F ) with exactly one fixed point; if that point is , the transformation is x x + c for some c F * . 1. Let G = PSL 2 ( F ) and assume H is a normal subgroup of G . If H contains a transvection then it contains all of them, and thus coincides with G . (The G -conjugates of x x + c

#### You've reached the end of your free preview.

Want to read both pages?

• Fall '11
• JacobLurie

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes