Group Theory: The Simple Group of Order 168 - Part 2

We show that there are no nontrivial normal subgroups in SL(3,Z/2). Techniques include Jordan canonical forms and companion matrices.

Category:

Education

Tags:

• Mathematics
• algebra
• finite
• group
• theory
• matrix
• linear
• minimal
• polynomial
• characteristic
• Jordan
• form
• companion
• simple

[hdarjus]

awesome, I just don't understand why aren't there simple groups of order between 60 and 168

[MathDoctorBob]

@hdarjus Thanks! We worked out up to order 100 in grad school. The difficulty depends on your tools for the cases 36, 48, 72, 96. Burnside's Theorem covers these.

Beyond 100, the Sylow Theorems go far. BT handles 108, 144, and 160. Cases 120, 132, and 150 fall outside BT and are not immediately ruled out by Sylow.  I rushed through the calculations, so I'm sure I overlooked some hard cases. - Bob

[njwildberger]

Hi That was a nice lecture, thanks. It is certainly an interesting group. Do you know what the hypergroup structure of the conjugacy classes is? Roughly that is the same as the structure constants of the center of the group algebra, normalized as probabilities. I am also an avid maths video creator, more in the line of geometry these days.

[MathDoctorBob]

@njwildberger You're welcome, and thanks for the comment. It definitely needs to be reshot.

I haven't worked it out, but: If we realize the group algebra over C as L^2(G) with convolution, then the center is also spanned by the characters of the irreducible representations, and these characters are orthogonal idempotents (up to scalars). Fulton/Harris' Representation Theory has the character table around p. 70. I'll see if I can find a better reference.  - Bob

[njwildberger]

@MathDoctorBob Thanks for the reply. I was not however suggesting at all that your video needs to be reshot!

Yes the hypergroup structure can be found from the character table, though the other way around is perhaps more natural. Since you have good control of the conjugacy classes, I was thinking perhaps you could compute the hypergroup structure constants directly (if g1 is in class C1 and g2 is in class C2, what is the probability that g1g2 is in class C3?) Then the character table follows.

[MathDoctorBob]

@njwildberger I like that better. It's a nice way to build the irreducible representations from scratch.

I haven't much intuition for this group beyond what I have here. I stumbled across this problem while trying to find good normal form applications for seniors.  - Bob

[MathDoctorBob]

@bigfatAMERCANbastard Now I know what to put on my business cards. :) - Bob