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 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).
PROOF OF THE THEOREM
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
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 NG(Ω1(Z(KxT))).
By maximality of H and the supposition that G is simple, we must have
that NG(Ω1(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.
The only conjugates of k in G lying in (KT)<t> are k, ky,
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
||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.
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 ≤
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
are conjugate in NG(KxT) which is H. Since y acts fixed-point-free,
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.
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.
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.