3

I'd like to know, in general, what is known about the following:

Given a pair $a\mid b$, does there exist a group $G$ of order $b$ with no subgroup of order $a$?

Sure, Sylow says $a$ can't take the form of prime power, and $6\mid 12$ and $20\mid 60$ constitute pairs with a positive answer. But how far has this been worked out? In particular, what about fixing $a$ so that $b=na$? What is known then about the set of $n$'s that give a positive answer?

Shaun
  • 44,997
David Feldman
  • 377
  • 1
    If $a$ is not a prime power, there exists a group $G$ with $a \mid |G|$ which has no subgroup of order $a$. I remember asking this a long time ago here: https://math.stackexchange.com/questions/114378/when-does-every-group-with-order-divisible-by-n-have-a-subgroup-of-order-n – Mikko Korhonen Apr 25 '23 at 12:44
  • 1
    See also this post for groups, where the converse of Lagrange holds. It also contains some information on the case, where it doesn't hold, e.g., direct products $A_4\times H$, with $|H|$ odd. – Dietrich Burde Apr 26 '23 at 18:23

0 Answers0