group of order 540 is simple~~

 

I get from sylow's theorem that r5 = 36, r3= 10, r2 = 45 ..I cannot get contracdiction with counting elements...

 

Any help would be appreciated~

 

Simon

And r5 etc are?

I'll guess the number of Sylow5 supgroups and so on.

 

Usually one looks at the action of G by conjugation on the sylow subgroups has no fixed points to show there is no normal Sylow subgroup (does that imply there are no normal subgroups at all?)

 

Perhaps you mean show it is not simple? Or that the non-commutative groups of order 540 are simple.