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Research Article

Groups Satisfying Some Normalizer Conditions

A. Liggonah
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The aim of the study is to characterize all finite groups that satisfy the normalizer conditions stated in this manuscript. Group and character theoretic methods are used in the study and it is proved that such groups are not simple. Specifically, the following result is established. Let a finite group G have a maximal subgroup H satisfying: (I)H = XP<t>, where P = <x, y: x3 = y3 = [x, y] = 1> and t2 = (zt)2 = 1 for all z in H. (II) X = RxKxT, where R has odd order and y acts fixed-point-free on X; K and T are 2-groups, xy centralizes K and acts fixed-pint-free on T; x centralizes T and acts fixed-point-free on K. T≠1, K≠1. (III) H is the only maximal subgroup of G containing XP and |Ω1(Z(KxT))|>4. Then G is not a simple group.

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A. Liggonah , 2008. Groups Satisfying Some Normalizer Conditions. Asian Journal of Algebra, 1: 25-28.

DOI: 10.3923/aja.2008.25.28



The problem originated from the study of alternating group A7 on 7 letters, by Liggonah (1977). It is the generalization of the proposition in the proof of the main result established by Liggonah (1977).

The notations used are standard, as used and defined in Gorenstein (1968). Conditions (i) through (iii) are conditions on the subgroup H of G, Giving the structure of H. These are the normalizer conditions referred to in the theorem.

The proof of the theorem is by contradiction. We suppose that G is simple and satisfies the conditions of the theorem and aim at arriving at a contradiction.The method used in the proof involves studying the fusion of involutions in H, By using character and group theoretic methods.

We divide the proof through a series of lemmas. From now on, G is assumed to be simple and satisfies conditions (i) through (iii).


Lemma 1
For each class L of involutions of H, either there is an element l ε L such that l-1yl = y-1 or L⊂O3`(H).

Conditions (i) to (iii) imply that NG(X) = H = NG(KxT). Since KxT admits <y><t> and y acts fixed-point-free, KxT is unique in (KxT)<t> and of class ≤ 2 by Stephen and Tyrer (1973a, b). Hence (KxT)<t> is a Sylow 2-subgroup of its own normalizer, so it is a Sylow 2-subgroup of G. Consider H1=(RxKxT)<y><t>. The principal 3-block B0(H1) of H1 has the form:

e y L
1 1 1
d 1 δ
d+1 -1 δ+1

If l-1yl = y-1is false for all l εL, then #; (l`l` = sy) = 0 for all 3`-elements s commuting with y, by Higman (1968); where #(l`l` = sy) is the number of conjugates of l with product equal to sy. Applying Higman (1968) results, we

#(l`l` = sy) = |H1|/|CH1(y)|Σ χ (l)2 χ(y) = 0

where, summation is over all characters χ in B0(H1). Then this implies that

Σ χ(y)χ(l)2 = 0 giving 1+δ2/d - (δ+1)2/(d+1) = 0.

That is, (δ–d)2 = 0, giving δ = d. Then l lies in the kernel of every character in the principal 3-block of H1. By Brauer (1964a, b), l lies in O3`(H1) = O3`(H) and O3`(H) = RxKxT and hence L⊂O3`(H) as required.

Hence we have that a class L of involutions in H, either there is an element lεL such that l-1yl = y-1 or L⊂O3`(H).

The involutions of H not in X = RxKxT are all conjugate (even in H).

The extended centralizer of y in H1 is C*H1(y) = <y><t>, so all involutions of H1 not in X must be in <y><t> D6 by Lemma 1 (otherwise they will lie in O3`(H1) = X). Using the fact that all involutions of D6 are conjugate and C*H(y) = P<t>, all involutions of H not in O3`(H) = O3`(H1) = X are conjugate to t as required.

We have seen in the proof of Lemma 1 that KxT is weakly closed in (KxT)<t>, a Sylow 2-subgroup of G and of class 2 and any involution of H is conjugate in H to t or lies in KxT. Since Ω1(Z(KxT)) is characteristic in KxT, elements of Ω1(Z(KxT)) are conjugate in G only if they are conjugate in NG1(Z(KxT))).

By maximality of H and the supposition that G is simple, we must have that NG1(Z(KxT))) = H. In particular, if k is an element of Ω1(Z(KxT)), its conjugates in Ω1(Z(KxT)) are k, ky, ky2. Since |Ω1(Z(KxT))|>4, we can always pick k so as not to be conjugate to t in H. From now on, we assume that such a k has been picked.

Lemma 2
The only conjugates of k in G lying in (KT)<t> are k, ky, ky2.

By the corollary and the underlying assumption, any further conjugates kg lies in KxT, whence kg, k, ky, ky2 generate an abelian group. Let A be a subgroup of KxT chosen such that:

A contains the greatest possible number of conjugates of k
A is as large as possiblee as possible

We first show that the conjugates of k lying in A are already conjugate in NG(A). Indeed, let kA. Then kgAg and so≤ CG(k). Since (KxT)<t> is a Sylow 2-subgroup of G and of CG(k), we can assume ≤ (KxT)<t>. An element of (KxT)<t> not in KxT transforms ky to ky2 since it must involve t, so that an abelian subgroup of (KxT)<t> not in KxT can not contain ky or ky2.

Furthermore, the conjugates of k in lie in KxT. Thus, if is not contained in KxT, then ( (KxT)) <ky> is an abelian subgroup of KxT containing more conjugates of k than A, contrary to the choice of A. Hence ≤ KxT.

By choice of A, each A and are maximal abelian subgroups of KxT and since KxT is of class at most 2, (KxT)` = Z(KxT) = A and because [KxT, A] = (KxT)` = A and applying theorem 2.1 in Gorenstein (1968), page 18, gives A, are normal in KxT. By weak closure of KxT in (KxT)<t>, it means that A and are conjugate in NG(KxT) which is H. Since y acts fixed-point-free, <A,> is abelian by Stephen and Taylor, so it implies A = by maximality of A. Thus the conjugates of A in H are A and At. Replacing g by tg if necessary, we can assume A = Ag, which implies g εNG(A). Hence conjugates of k lying in A are already conjugate in NG(A). A is normalized by P so NG(A) = XP and hence NG(A) = H or NG(A) = XP. This implies the only conjugates of k in A are k, ky, ky2 as required.

Lemma 3
The element t is not conjugate to any element of KxT in CG(k).

By Lemma 2, the conjugates of k are k, ky, ky2 and kkyky2 = 1 because y fixes kkyky2 and y acts fixed-point-free. This implies the only conjugates of ky<k> in CG(k)/<k> is ky<k>. By Glaubermann (1966), it implies that ky<k> ε Z(CG(k)/<k>). This means that for some normal subgroup M of CG(k) of odd order, M<k, ky> is normal in CG(k). By Frattini argument, this gives CG(k) = M NCG(k)(<k, ky>). Since <k, ky> ≤ Ω1(Z(KxT)) and because of the fact that the conjugates of k in H lying in (KxT)<t> are k, ky, ky2, then <k, ky> = H. Hence by maximality of H, we have NG(<k, ky>) = H. Hence CG(k) = MX<t> or CG(k) = MX<x><t>. Since KxT ≤ X, we have t is not conjugate to any element of KxT from the structure of CG(k) as required.

We could stop here, at this stage, by quoting Goldschmidt`s results, presented by Thompson (1968), concerning strongly closed abelian 2-subgroups because we have shown that <k, ky> is a strongly closed abelian 2-subgroup of CG(k). But we can also finish more concisely.

Lemma 4
The element t is conjugate in CG(k) to some element of KxT.

By Thompson`s Transfer Theorem given by Thompson (1968), t is conjugate in G to some element of KxT, say tgKxT. then both kg and <k, ky> centralize tg, so we can chose h in CG(tg) so that kgh lies in the same Sylow 2-subgroup of CG(tg) as <k, ky>. By Lemma 3, a Sylow 2-subgroup of G contains only three conjugates of k, so this implies kgh = ky2 for some i = 1,2,3. Since hCG(tg), tghy-1 lies in KxT and ghy-i CG(k), so t is conjugate in CG(k) to some element of KxT, proving the lemma.

The contradiction between Lemma 3 and Lemma 4 completes the proof of the theorem. That is, G is not simple.

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