## Results about the order of a group forcing a particular property.

Given a group of order $n$ where $n$ is either a specific number, or a number of a particular form, e.g. square-free, when does $n$ completely determine a particular group property among all groups of that order? Vipul's group theory wiki has several stubs on this topic, and in the language of his wiki, I will call this a $P$-forcing number, where $P$ is a particular group theoretic property.

We already have quite a few easy examples, for instance, orders $pq$, $pqr$, and $p^2q$ force solvability, and $p^2$ forces abelian. Then there are more specific results like 99 is an abelian-forcing number.

I am interested in general, in any results of this flavor beyond what would be considered a common result in a standard graduate-level group theory book.

-
See Pete Clark's answer at: mathoverflow.net/questions/11001/… – S. Carnahan Mar 8 2010 at 5:20

The numbers n such that every group of order n is cyclic, abelian, nilpotent, supersolvable, or solvable are known. Most are described in an easy to read survey:

Pakianathan, Jonathan; Shankar, Krishnan. "Nilpotent numbers." Amer. Math. Monthly 107 (2000), no. 7, 631-634. MR 1786236 DOI: 10.2307/2589118

If you want to go beyond results like this, you may have better luck looking at a slightly more refined version of the order: the isomorphism type of the Sylow subgroups. Sometimes a p-group P has the property that every group G containing it as a Sylow p-subgroup has a normal subgroup Q of order coprime to p such that G is the semi-direct product of P and Q. An easy version of this that does appear in many group theory texts is that if n=4k+2, then in each group G of order n there is a normal subgroup Q of order 2k+1 so that G is the semi-direct product of any of its Sylow 2-subgroups and Q.

Groups all of whose Sylow p-subgroups are cyclic have very nice properties, subsuming those of groups of square-free order. Groups all of whose Sylows are abelian have more flexibility, but are still basically under control.

-

$(n,\phi(n)) = 1$ forces $G$ cyclic. $p^nq^m$ forces $G$ solvable.

-

I haven't had time to read it yet, but The influence of conjugacy-class sizes on the structure of finite groups seems tailor made to answer these questions. It's a survey article, too.

-

Very famous one is the Feit-Thompson theorem: if n is odd, then G is solvable. Though I suppose this is stated (but not proved) in most modern algebra texts.

-