EQUIVALENT CONDITIONS FOR P-NILPOTENCE

In the first part of this paper we prove without using the transfer or characters the equivalence of some conditions, each of which would imply p-nilpotence of a finite group G. The implication of p-nilpotence also can be deduced without the transfer or characters if the group is p-constrained. For p-constrained groups we also prove an equivalent condition so that O ′ (G)P should be p-nilpotent. We show an example that this result is not true for some non-p-constrained groups. In the second part of the paper we prove a generalization of a theorem of Itô with the help of the knowledge of the irreducible characters of the minimal non-nilpotent groups.


Definitions and known results
We know the remarkable theorems of Frobenius which tell that in Theorem 2.1 (i) and (iv) both imply that the finite group G has a normal p-complement.All existing proofs of them use the transfer homomorhism or characters.
The well-studied minimal non-nilpotent groups, i.e. non-nilpotent groups, each of whose subgroups are nilpotent, sometimes are called Schmidt groups or (p, q)-groups.They can be described without using the transfer, see 5.1 Satz and 5.2 Satz in pp.280-281 of [6].Let G be a minimal non-nilpotent group.Then it can be proved without using the transfer or characters, that 1. G is solvable.
6.If Q ∈ Syl q (G), then Q is cyclic; and if Q = x , then x q ≤ Z(G).
7. If P is abelian, then Q acts irreducibly on P ; if P is nonabelian, then Q acts irreducibly on P/Z(P ).If P is abelian, then P is of exponent p; and if P is nonabelian, then P/Z(P ) is also of exponent p. So, they can be considered as vector spaces over GF (p).Their dimension is o(p)(mod(q)), which is even in the nonabelian case.
8. G is generated by its Sylow q-subgroups.
These groups are non-p-nilpotent.It can also be proved using the transfer or characters that a group is p-nilpotent if and only if, it does not contain such a subgroup.
We shall use the following: Notation 11.We shall write (p, q) ≤ G if the group G does not contain a (p, q)-group, otherwise we write (p, q) ≤ G.
Let us recall the definition of Thompson-ordering: Definition 12. Let G be a finite group.Let P be a property of subgroups of G.

Main results
The aim of this paper is to prove that the equivalence of the following four conditions can be proved without the use of transfer or characters: (iii) (p, q) ≤ G for every prime q = p; As a corollary we get: Theorem 22.Let G be a p-constrained group, P ∈ Syl p (G).If any of the conditions of the above theorem holds for G, then one can deduce without using the transfer that G has a normal p-complement.
As an application of Theorem 2.1 we prove also the following: Theorem 23.Let G be a finite group, let p = q be primes with p, q ∈ π(G).
Let P ∈ Syl p (G) and let O q (G) denote the subgroup of G generated by the q-elements of G.If G is p-constrained then the following are equivalent: (ii) O q (G)P has a normal p-complement.
Another application of Theorem 2.2 is to prove without using the transfer the following generalization of a theorem of Itô: Theorem 24.Let G be a finite p-constrained group, let P ∈ Syl p (G).Let us suppose that χ is a character of G satisfying the following conditions:

Then one of the following two possibilities holds: (i) P is abelian and P is normal in G;
(ii) p is a Fermat-prime and one of the constituents of χ has degree at least p − 1.The inequality in α) is sharp.There is a solvable group G 0 having a character χ 0 ∈ Char(G 0 ) satisfying β) and γ) with degree χ 0 (1) = 2p − 1, such that for this pair the assertion of the Theorem does not hold.

Preliminary lemmas
In the proof of Theorem 2.1 we will need the following lemma, which is Lemma 2 in [4].
Lemma 31.Let G be a group with H ∈ Hall π (G) and with the property that every π-subgroup Y of G can be conjugated into H.Let K be a class of elements of H, which is closed under conjugation inside H with elements of G such that if two elements of K are conjugate in G then they are already conjugate in H. Then if G 1 G and |G : G 1 | = q, where q ∈ π, then for For the proof of Theorem 2.1 we will also need the following lemma, which generalizes both Lemma 5 in [4] and Lemma 3.2 in [3].
Lemma 32.Let q ∈ π(G)\{p} be a fixed prime, P ∈ Syl p (G), U < P abelian and strongly closed in P .Then if (p, q) ≤ N G (U ) then (p, q) ≤ G, as well.P roof.Let G be a counterexample of minimal order.
First we prove that we may assume that O p (G) = 1.
Let O p (G) > 1 and let B G be a p-subgroup.Let G = G/B, and the images of U and P in this factor group let U and P , respectively.Then P ∈ Syl p (G) and the triple (G, P , U ) satisfies the conditions set for (G, P, U ).To see this we have to show only that (p, q) ≤ N G (U ) implies (p, q) ≤ N G (U ).Let M be the inverse image of N G (U ) in G.Here M < G, since if U G then U G ≤ P , and as U ≤ P is strongly closed, U G = U would follow.This would imply (p, q) ≤ N G (U ) = G, which is a contradiction.So M < G.But P ≤ M and N G (U ) = N M (U ).The triple (M, P, U ) satisfies the conditions of the Lemma.By induction (p, q) ≤ M .By [2], (p, q) ≤ M = N G (U ), as well.Hence the conditions of the Lemma are satisfied by the triple (G, P , U ) and by induction (p, q) ≤ G.
Let V be a (p, q)-group in G. Then V = V p ∈ Syl p (V ) and by the above result, its image V in G is nilpotent.Hence V p ≤ B. There are two cases: Equivalent conditions for p-nilpotence 133 Ad (i): We know that U P and we may assume that V p ≤ P , by replacing V with a suitable conjugate of it.Hence [U, V p ] ≤ U .On the other hand as Hence in both cases (i) and (ii with the property U ≤ S, then the triple (N, S, U ) satisfies the conditions of the Lemma.So by induction (p, q) ≤ N .This contradicts the fact that

End of the proof: Let
Let K be a maximal element of A for the Thompson-ordering.Then |N G (K)| p is maximal and among those with this property K is also of maximal order.As A = ∅, so such K exists.Then K ≤ P 1 for a suitable Sylow satisfies the conditions of the Lemma.When we prove this, then (p, q) ≤ N G (K xt ) follows, contradicting our assumption.It is enough to prove that (p, q) ≤ N G (R xt ∩U ).If |R| = |P | then we have that the triple (N G (K xt ), P, U ) satisfies the conditions of the Lemma, and since N G (K xt ) < G, by induction we get that (p, q) ≤ N G (K xt ), contradicting the choice of K.
, by the maximality of K in the Thompson ordering.The proof is complete.
For the proof of Theorem 2.4 we will need the description of irreducible characters of minimal non-nilpotent groups.As we did not find any reference to it in the literature, for the sake of selfcontainedness we include it here.
Lemma 33.Let G be a (p, q)-group, P ∈ Syl p (G), Q ∈ Syl q (G), |Q| = q n .Then G has exactly q n linear characters.
(i) If P is abelian, then all other characters in Irr(G) are of degree q.
They are induced from nontrivial characters of the unique index q subgroup of G.There are (|P | − 1)q n−2 such characters. (ii , where 2m ≡ o(p)(mod(q)).P/Z(P )Q is a (p, q)-group of type (i).So it has (p 2m − 1)q n−2 irreducible characters of degree q.The p − 1 irreducible characters of degree p m of P can be extended to G giving (p − 1)q n irreducible characters of degree p m .
(iii) If P is special and nonabelian, then if |Z(P )| = p k then Z(P ) has p k −1 p−1 maximal subgroups.By factoring with one of them we get a (p, q)group of type (ii).The union of inverse images of these characters give Irr(G).
P roof.As G = P , |G : G | = q n , so G has exactly q n linear characters.Let H = P x q .Then |G : H| = q and H is normal in G.
Ad (i): If P is abelian, then so is H, so if χ ∈Irr(G) nonlinear, then χ H = σ 1 + ... + σ q and χ = σ G i for i = 1, ..., q.So χ(1) = q and χ is induced from exactly q linear characters of H.As |G| = |P |q n = q n + q 2 (|P | − 1)q n−2 , we get that each nontrivial character of H that does not contain P in its kernel is induced to Irr(G).Ad (ii): If P is extraspecial, then |P | = p 2m+1 .As Q acts irreducibly on P/Z(P ), by Lemma 3.10 in Chapter II. of [6] we get that 2m = o(p)(mod(q)).The p − 1 faithful irreducible characters of P are of degree p m , they are 0 outside Z(P ), so they are G-invariant, and as (|P |, |G : P |) = 1, they can be extended to G. By Gallagher's theorem, se e.g.[7], they can be extended in q n ways.This way we get (p−1)q n irreducible characters of G.By taking into consideration those of degree 1 and q the sum of squares of the degrees gives: so we determined all Irr(G).
Ad (iii): We calculate the sum of squares of the irreducible characters we produced so far: q n +q 2 (p 2m −1)q n−2 +p 2m p k −1 p−1 (p−1)q n = q n p 2m+k = |G|, so we produced the whole Irr(G).

Proofs of the main results
Now we prove Theorem 2.1: (ii) → (iii): We use induction on |G|.The following argument is similar to one in the last part of the proof of Theorem 1 in [4].For the sake of selfcontainedness, we repeat it here.
Let A P be an abelian normal subgroup in P such that exp(A) ≤ p if p > 2, and exp(A) ≤ 4 if p = 2, and A is maximal with these properties.
(a) If A ≤ Z(P ): then according to Alperin's theorem [1], Ω j (P ) ≤ Z(P ), where j = 1 if p > 2 and j = 2 if p = 2. Then A is strongly closed in P , as if we take two elements a and a x of order p or of order 4 in A and P , then they are conjugate in P , and as A is normal in P , we get that a x ∈ A, too.Let N = N G (A).If N < G, then as P ∈ Syl p (N ), by induction we get that (p, q) ≤ N, and by Lemma 3.2 , (p, q) ≤ G.So we may assume that N G (A) = G.Then P ≤ C G (A) G.As for each a ∈ A # for the conjugacy class K G (a) of a in G and for the conjugacy class K P (a) in P it holds that K G (a) = K G (a) ∩ P = K P (a) = a, as A G and A ≤ Z(P ), so |G : C G (a)| = |P : C P (a)| = 1, and we get that A ≤ Z(G).Thus, if p > 2, then Ω 1 (P ) ≤ A ≤ Z(G); if p = 2, then Ω 2 (P ) ≤ Z(G); and this means that G ≥ (p, q) in cases p > 2 and p = 2, either, for every prime divisor q = p of G.

(b) If A ≤ Z(P ):
then A ∩ Z(P ) < A. Thus A/A ∩ Z(P ) contains a central subgroup of P/A ∩ Z(P ) of order p.Let A 1 be its inverse image in A. According to our assumption, A 1 = A ∩ Z(P ), x , where o(x) = p or o(x) = 4 and x ∈ Z(P ).A 1 is strongly closed in P as if a 1 ∈ A 1 and a u 1 ∈ P, then by assumption a u 1 is conjugate to a 1 in P .But as A 1 P , a u 1 ∈ A 1 .If ) and P ∩ C P .Then the triple (L, P ∩ L, U ) satisfies the conditions of the Lemma.Hence if L < G then by induction (p, q) ≤ L, which is impossible as V ≤ L. Thus G = L = CQ.Since P = P ∩ L = P ∩ C, P ≤ C and as C G, so by the Frattini argument we have that G = CN G (P ).As U is a strongly closed subgroup of P , N G (P ) ≤ N G (U ) and thus (p, q) ≤ N G (P ).Since V p ≤ C G (P ) ≤ N G (P ) and V p G, hence V p N G (P ).Thus |N G (P ) : C ∩N G (P )| ≡ 0 (q).But then, since |N G (P ) : C ∩ N G (P )| = |CN G (P ) : C| = |G : C| = |CQ : C|