Just realised that i had my safe group option on the whole time. When i generated a group it would check the axioms and also check for albelian.... turned it off much faster however the cayley table for $S_7$ has $7!^2$ elements so its gonna be a while
"it is common for people first starting to grapple with computers to make large-scale computations of things they might have done on a smaller scale by hand. They might print out a table of the first 10,000 primes, only to find that their printout isn't something they really wanted after all. They discover by this kind of experience that what they really wanted is usually not some collection of "answers" - what they want is understanding."
@Hippalectryon Give your computer a rest do some work. Go outside. Eat a bagel. Chase cats. Run around like a maniac. Talk to random people. Watch TV. Annoy your neighbors. Sleep. The list goes on.
there are certain combinatorial moves one can do to young tableaux or something, and they were looking at ways to realize these as paths between orthogonal matrices
i am seeing what you're doing, @alexander. you're looking for a way to find a "right" smoothness for some representation of a discrete group on GL2(C).
doesn't have to be optimal or anything, just seems like there ought to be some obvious choice, like how you can connect $A$ and $B$ by $A+(B-A)t$ from $t=0$ to $1$
@MikeM I have always wondered whether homeomorphism and homeopathy are diffeomorphic.
@PedroTamaroff I want to prove that nonabelian groups of order $216$ are not simple. I am trying a pretty convoluted way by counting elts of sylow subgroups and intersections, but i guess it's not the right way to go about this. can you hint (not revealing the answer?)
@AlexanderGruber looks like you want to "continuify" a (wlog unitary) linear transformation A. To do this, diagonalize as $A=U{\rm diag}(e^{i\theta_1},\cdots,e^{i\theta_n})U^{-1}$, then you can write $A_t:=U{\rm diag}(e^{i\theta_1 t},\cdots,e^{i\theta_n t})U^{-1}$
@BalarkaSen Err... what about this. Suppose there are $4$ Sylow subgroups. Then we have a representation $\eta:G\to S_4$, that is nontrivial, so $\ker \eta\lhd G$ must be trivial, so $G$ embeds into a group of order $4!=24$ which is impossible?
@Ted Hmm... No, I'd rather like to buy them from you firsthand. Anyway, he must have sold it to some reseller, because I got it from some small group that specializes in out-of-print texts
@TedShifrin if you're talking about the problem, I have a poorly written up solution that they gave me in the class. They reasoned that the solution was an exponential, but they didn't really prove why.
Whoever invented this, he must be not aware of democracy. I want $\epsilon$ to be large not small. Also, why $\theta$ is an angle?
