Given a prime number p and a natural number m not divisible by p, we propose the problem of finding the smallest number such that for , every group G of order has a non-trivial normal p-subgroup. We prove that we can explicitly calculate the number in the case where every group of order is solvable for all r, and we obtain the value of for a case where m is a product of two primes.
Throughout this note, p will be a fixed prime number. We use to denote the p-core of G, that is, its largest normal p-subgroup.
We propose the following optimization problem: Given a number m not divisible by p, find the smallest r0 such that every group having order n =prm, with , has a nontrivial p-core . Denote such number r0 by . In Theorem 2.1, we will prove that is well-defined for any prime p and number m (with ). In Theorem 2.3 we explicitly determine the value of in the case that all groups whose order have the form prm are solvable (for example, if m is prime or if both p and m are odd). Finally, in Section 3, we calculate Λ(2,15), a case that is not covered by the previous theorem.
We remark that the motivation for this research came from the search for examples of finite groups G such that the Brown complex of nontrivial p-subgroups of G (see for example  for the definition and properties) is connected but not contractible. It is known that is contractible when G has a nontrivial normal p-subgroup, and Quillen conjectured in  that the converse is also true.
Theorem 2.1. For any prime number p and natural number m such that , there is a number such that if , any group of order has a non-trivial p-core .
Proof. Let G be a group of order with . Let P be a Sylow p-subgroup of G. Since the kernel of the action of G on the set of cosets of P is precisely , we obtain that G embeds in Sm, and so pr divides (m-1)!. Hence, if is the largest power of p dividing ((m-1)!), we obtain that .
For t,q natural numbers, let be the product
(note that can also be defined as , where is the q-factorial of t), and if is a prime factorization of m, with the qi pairwise distinct and for each i, we let . We prove that if is the largest power of p dividing , then .
Theorem 2.2. Let where and . If , then there is a group of order n with .
Proof. Let K be the group , that is, a product of elementary abelian groups, where and are distinct primes and Cq denotes the cyclic group of order q. Then Γ(m) divides the order of , and hence so does ps. Let H be a subgroup of of order ps. For every S∈ H and k∈ K define the map by . Then is a subgroup of . If we identify H with the subgroup of maps of the form TS,0 and K with the subgroup of maps of the form , then G is just the semidirect product of K by H. Hence |G|=n. We have that G acts transitively on K in a natural fashion, and the stabilizer of 0∈ K is H, a p-Sylow subgroup of G. Hence the stabilizers of points in K are precisely the Sylow subgroups of G, so their intersection contains only the identity , as we wanted to prove.
The next theorem will show that the lower bound given by Theorem 2.2 is tight in some cases.
Theorem 2.3. Let , where . If G is a group of order n and ps does not divide Γ(m) then either:
1. , or
2. G is not solvable.
Proof. Let G be solvable with order and . Let F(G) be the Fitting subgroup of G. Consider the map , sending g to given by conjugation by g. The restriction of c to P, a p-Sylow subgroup of G, has kernel . Since (Theorem 7.67 from ), and F(G) does not contain elements of order p by our assumption on , we have P ∩ CG(F(G))=1 and so P acts faithfully on F(G). If is the prime factorization of m, we have that F(G) is the direct product of the for . Hence . Let such that the action induced by cg on , is the identity. Since cg acts on each factor as the identity, then by Theorem 5.1.4 from , we have that it acts as the identity on each . By the faithful action of P on F(G), we have that g=1. This implies that P acts faithfully on . But then |P| divides the order of the automorphism group of , which is a product of elementary abelian groups of respective orders with for all i. Hence divides
Corollary 2.4. Let ps be the largest power of p that divides . If m is prime, or if both p,m are odd, then .
Proof. By Burnside's p,q-theorem, and the Odd Order Theorem, we have that all groups that have order of the form for some r are solvable. Therefore, for all , by Theorem 2.3 we have that all groups of order have non-trivial p-core.
At this moment, we can prove that in some cases, the group constructed in 2.2 is unique.
Theorem 2.5. Let where and s>0. If , but for all proper divisors m' of m, then up to isomorphism, the group constructed in the proof of Theorem 2.2 is the only solvable group of order n with .
Proof. With the notation of the argument of the proof of 2.3, if G is a solvable group of order n with , we must have that and for all i in order to satisfy the divisibility conditions. Hence is elementary abelian and a qi-Sylow subgroup for all i, and so G is the semidirect product of a p-Sylow subgroup P of with F(G), where the action of P on F(G) by conjugation is faithful. Hence G is isomorphic to the group constructed in the proof of Theorem 2.3.
One case in that we may apply Theorem 2.5 is when n=864. There are 4725 groups of order , but only one of them has the property of having a trivial 2-core.
An example that cannot be tackled with the previous results is the case p=2, . In this case, . Not all groups with order of the form are solvable, however, we will prove that is actually 4. (The group S5 attests that .)
Theorem 3.1. Every group G of order for r ≥ 4 is such that .
Proof. Let G be a group of order for . Suppose that . From Theorem 2.3, we obtain that G is not solvable. We will prove then that . Suppose otherwise, and let T=O3(G). Then , and so G/T is solvable. Since , from Theorem 2.3, we have that . Let such that . Suppose . Since , and G/L is solvable, we have that divides , that is, . Now, L is also solvable and , hence if we had we would have , and G would have a non-trivial subnormal 2-subgroup, which contradicts our assumption that . Hence j=1. But then , which contradicts that . Hence . By a similar argument, we get that .
From  we obtain that G is not simple. Hence G has a proper minimal normal subgroup M. From the previous paragraph, we obtain that M is not abelian, since in that case we would have that . The only possibility is that . We have then a morphism sending g to cg, the conjugation by g. Since , and , in any case the kernel of c is a nontrivial normal 2-subgroup.