Properties of subgroups not containing their centralizers

In this paper, we give a generalization of Baer Theorem on the injective property of divisible abelian groups. As consequences of the obtained result we find a sufficient condition for a group G to express as semi-direct product of a divisible subgroup D and some subgroup H. We also apply the main Theorem to the pgroups with center of index p2, for some prime p. For these groups we compute Nc(G) the number of conjugacy classes and Na the number of abelian maximal subgroups and Nna the number of nonabelian maximal subgroups.


Introduction
We shall recall some definitions: If H is a subgroup of a group G, a subgroup K is called a complement of H in G if G = HK and H ∩ K = {1}.Therefore if H G and K G, then G is said to be the direct product of H and K, in symbols, G = H K.
If H G, then G is said to be the semi-direct product of H and K, in symbols, G = H K.
An abelian group D is called divisible if for every x ∈ D and every positive integer n there is a y ∈ D so that x = ny, i.e., each element of D is divisible by every positive integer.The main property of divisible groups is that they satisfy the following "injectivity" condition: Theorem 1.1 (Baer Theorem [3]).If D is a divisible group, then any homomorphism f : A → D = from any abelian group A into D extends to any abelian group G which contains A i.e., there exists a homomorphism f : G → D so that f |A = f .The purpose of this paper is to generalize this result to the nonabelian groups.To this end, we introduce the property "N" in subgroups: Let H be a subgroup of an arbitrary group G.

This is equivalent to saying that C G (H)
H where C G (H) is the centralizer of H in G which is defined to be the set of all g in G such that hg = gh for all h in H, it is clearly a subgroup of G.
By the definition of the condition "N" we deduce that 1) If G is abelian, then every proper subgroup H satisfies the condition "N".
2) If G is a nonabelian nilpotent group, then every maximal normal abelian subgroup H of G does not satisfy the condition "N" because C G (H) = H, [3].
3) There exist a nonabelian groups G whose a subgroup H satisfies the condition"N", for example let G = Q 8 × Z/2Z where H = Q 8 is the quaternion group of order 8, [2].

Main results and proofs
The set S is partially ordered by the relation ≤.We aim to apply Zorn's Lemma to S and to this end we consider a chain (H i , f i ) i∈I .It has an upper bound ∪ i ∈I H i , f where Since every subgroup of an abelian group satisfies the condition"N", we can apply Theorem 2.1 to abelian groups, also we deduce the known result, [3]:

Corollary 2.2. Any divisible subgroup D of an abelian group G splits, i.e., D has a complement
If G is a group (not necessarily abelian), we write: Corollary 2.3.Let G be a group and let D be a divisible subgroup of G such that every subgroup H of G which contains D, satisfies the condition "N".Then D has a complement H so that G = H D.
Proof.We consider the identity map: id D : D → D, by Theorem 2.1, id D extends to the group G, i.e., there exists a homomorphism f : Since the center does not have a prime index and |G : H| = p, Z(H) = H, consequently H is abelian.Hence H is a maximal normal abelian subgroup of the nilpotent group G, so C G (H) = H, [3], and H does not satisfy the condition "N". "iii By the same way we deduce that H is abelian.
Proof of Corollary 2.4.Let H be a subgroup of G so that H ⊂ H , since H is nonabelian, by Lemma 2.5, there is g ∈ G − H so that [g, H] = 1.Then [g, H ] = 1 by maximality of H. Thus the conditions of Theorem 2.1 are satisfied, so we obtain Corollary 2.4.

Subgroups satisfying the condition "N"
If A is finitely generated abelian group, the rank of A is defined by rk(A) the minimum number of generators of A.
We denote us by x G the conjugacy class of x in an arbitrary group G and C G (x) the centralizer of x in G and N c (G) the number of the conjugacy classes.
If G is a finite p-group of class c, then from [4], we know that Let G be a finite p-group of order p n such that its center has index p 2 .In this section, we compute the number N c (G) and N 0 the number of maximal subgroups in G satisfying the condition "N".Theorem 3.1.Let G be a finite p-group of order p n such that its center has index p 2 , then 1) G has precisely p + 1 abelian maximal subgroups.
2) The number of maximal subgroups satisfying the condition "N" equals

and each nontrivial conjugacy class has p elements.
The proof of Theorem 3.1 results from the following Lemmas.Lemma 3.2.Let G be an abelian finite p− group, then the number of subgroups of order p equals (p r − 1)/(p − 1) where r is the rank of G.
If g is an element of order p in G, then g = g 1 g 2 . . .g r such that g i has order p or 1 and g = 1.The number of such elements g equals Since a group of order p has p − 1 elements of order p, the number of subgroups of order p is (p r − 1)(p − 1).

Proof. By induction on m.
Proof of Theorem 3.1.1) Let H be an abelian maximal subgroup of G, then H does not satisfy the condition "N", so Z(G) ⊂ H. Consequently H/Z(G) is a subgroup of order p of the elementary p− group G/Z(G) Z/pZ×Z/pZ by Lemma 3.2, there is . By using the known result of Steinitz [5]: The number of subgroups of order p k equals the number of subgroups of order p n−k in a finite abelian group of order p n , we conclude that the number of the maximal subgroups H is equal to the number of subgroups of order p of G/G .If rk(G/G ) = r, then the number of maximal subgroups satisfying the condition "N" is Proof.This follows easily from Theorem 3.1.

Examples
The following examples illustrate some applications of the previous results.
Example 4.1.Let G be a p−group of order p 3 , then 1) The number N of maximal subgroups is given in the following table To prove the result 1) we consider the two cases a) If G is abelian, we apply Lemma 2.
5. b) If G is nonabelian, |G : Z(G)| = p 2 because the index of center does not equal to a prime, so G = Z(G) and G is extra-special, [3].Now G ⊂ Frat(G) ⊂ G implies that |Frat(G)| = p 2 or p, the first case is impossible because, by Burnside Basis Theorem, [3], |G : Frat(G)| = p implies that G is generated by one element, that is, G is cyclic.Hence G = Z(G) = Frat(G).If G/Gis not elementary p−group, then by Theorem 3.1 G has one maximal subgroup, so |Frat(G)| = p 2 , a contradiction.Thus G/G is elementary p−group and the result is an immediate consequence of Corollary 3.4.Since the index of center does not equal to a prime, Z(G) = Z(M ).If y ∈ M , yx = xy so x ∈ Z(M ) and we reach the contradiction x ∈ Z(G).Hence M is abelian.Conversely, let M be an abelian maximal subgroup of G and x

Lemma 4 . 5 . 2 =
and |C G (x)| = p 3 , by this contradiction we obtain x G = p.Let G be a p−group of order p 4 such that |G : Z(G)| = p 3 and let M be a maximal subgroup of G. Then 1) If M is abelian, M contains exactly p 2 −1 nontrivial conjugacy classes which has p elements.2) If M is not abelian, M −G contains exactly p−1 nontrivial conjugacy classes which has p 2 elements.Proof of Lemma 4.5. 1) Assume that M is abelian.Let x ∈ M − Z(G), since M G, x G ⊂ M .By Lemma 13, x G = p, consequently G has p 3 −p p = p 2 − 1 nontrivial conjugacy classes which has p elements.2) If M is not abelian, let x ∈ M − G .From Lemma 13 it follows that x G = p 2 , so M −G has exactly p 3 −p 2 p p−1 nontrivial conjugacy classes which has p 2 elements.We will prove the last result as following.If M 1 and M 2 are two maximal subgroups in G, it is clear that M 1 ∩ M 2 = G .We denote by k a ( respectively k na ) the number of abelian (respectively nonabelian) maximal subgroups in G.If x ∈ G − Z(G), x G ⊂ G , so x G = p and |C G (x)| = p 3 , we have shown in the proof of Lemma 4.4 that C G (x) is abelian, consequently k a = 0. Let M 1 , M 2 be two abelian maximal subgroups of G. Let x ∈ G −Z(G), then G ⊂ M 1 G and by Lemma 4.4, x G = p.Let x ∈ M 1 .Since M 1 is abelian, M 1 ⊂ C G (x) G, so M 1 = C G (x).By the same way we obtainM 2 = C G (x). Hence M 1 = M 2 and k a = 1.Each maximal subgroup M satisfy G ⊂ M ⊂ G, so MG is a subgroup of order p of the group G/G Z/pZ × Z/pZ.By Lemma 3.2, G has p + 1 maximal subgroups, so k na = p.2) By using Lemma 4.4 and Lemma 4.5 and the first assertion of Example 4.2, we obtain N c (G) = p + (p 2 − 1) + p(p − 1) = 2p 2 − 1.
Then any homomorphism f : H → D from H into divisible group D extends to the group G. Proof.Let us consider the set S of all pairs (H i , f i ) where H i is a subgroup of G containing H and f i : Theorem 2.1.Let G be a group and let H be a subgroup of G such that each proper subgroup H of G which contains H, satisfies the condition "N".
Let n the smallest positive integer so that g n Corollary 2.4.Let G be a finite p-group with center of index p 2 .If H is a nonabelian maximal subgroup of G then any homomorphism f : H → D from H into divisible group D extends to the group G. Let G be a finite p-group such that its center Z(G) has index p 2 .If H is a maximal subgroup of G, then the following properties are equivalent.
i)H is abelian ii) Z(G) ⊂ H iii) H does not satisfy the condition "N".Proof of Lemma 2.5."i)⇒ ii)".Let us assume that H is abelian.If Z(G)H, there exists g