Australian people can buy antibiotics in Australia online here: http://buyantibioticsaustralia.com/ No prescription required and cheap price!

## Spm.uem.br

Bol. Soc. Paran. Mat.

(3s.) v. 23 1-2 (2005):
On the index complex of a maximal subgroup and the group-theoretic
abstract: Let

*G *be a finite group,

*Sp*(

*G*)

*, *Φ (

*G*) and Φ1(

*G*) be generalizationsof the Frattini subgroup of

*G*. Based on these characteristic subgroups and usingDeskins index complex, this paper gets some necessary and sufficient conditions for

*G *to be a

*p*-solvable,

*π*-solvable, solvable, super-solvable and nilpotent group.

Key Words: : index complex; solvable groups; super-solvable groups; nilpo-
The relationship between the properties of maximal subgroups of a finite group
and its structure has been studied extensively. The concept of index complex(see[1]) associated with a maximal subgroup plays an important role in the study ofgroup theory.

Suppose that

*G *is a finite group, and

*M *is a maximal subgroup of

*G*. A
subgroup

*C *of

*G *is said to be a completion for

*M *in

*G *if

*C *is not contained in

*M *while every proper subgroup of

*C *which is normal in

*G *is contained in

*M *.

The set of all completions of

*M *, denote it by

*I*(

*M *), is called the index complex of

*M *in

*G*. Clearly

*I*(

*M *) contains a normal subgroup, and is a nonempty partiallyordered set by set inclusion relation. If

*C ∈ I*(

*M *) and

*C *is the maximal elementof

*I*(

*M *),

*C *is said to be a maximal completion for

*M *. If moreover

*C ✁ G*,

*C *thenis said to be a normal completion for

*M *. Clearly every normal completion of

*M*
*∗ *The project is supported by National Natural Science Foundation of China (10301004) and
Basic Research Foundation of Beijing Institute of Technology

*† *Correspondence should be addressed to Jiang Lining2000

*Mathematics Subject Classification: *20D10, 20F16
is a maximal completion of

*M *. Furthermore, by

*k*(

*C*) we denote the product ofall normal subgroups of

*G *which are also proper subgroups of

*C*,

*k*(

*C*) is a propernormal subgroup of

*C*.

In [2], Deskins studied the group-theoretic properties of the completions and its
influences on the solvability of a finite group. He also raised a conjecture concerningsuper-solvability of a finite group in the same paper. Deskins’s conjecture and otherinvestigations were continued by many successive works [3-5]. This paper will studythe structure of a finite group

*G*. Using the concept of index complex and applyingFrattini-Like subgroups such as

*Sp*(

*G*)

*, *Φ (

*G*) and Φ1(

*G*), the paper improves mainresults of [3-5] and obtains some necessary and sufficient conditions for the

*G *tobe a

*p*-solvable,

*π*-solvable, solvable, super-solvable and nilpotent group.

Throughout this paper,

*G *denotes a finite group. The terminologies and no-
tations agree with standard usage as in [6]. The notation

*M <· G *means

*M *is amaximal subgroup of

*G*, and

*N ✁ G *means that

*N *is a normal subgroup of

*G*. If

*p *is a prime, then

*p *denotes the complementary sets of primes and

*|G *:

*M |p *the

*p*-part of

*|G *:

*M |*.

For convenience, we give some notations and definitions firstly. Suppose that

*p*
*F c *=

*{M *:

*M <· G *and

*|G *:

*M | *is composite

*}*;

*F p *=

*{M *:

*M <· G *and

*M ≥ NG*(

*P *) for a

*P ∈ Sylp*(

*G*)

*}*;
Using subgroups above, one can define Frattini-Like subgroups of

*G *as follows.

*{M *:

*M ∈ F pc} if F pc is nonempty, otherwise Sp*(

*G*) =

*G*;

*{M *:

*M ∈ FG} if FG is nonempty, otherwise *Φ1(

*G*) =

*G*;

*{M *:

*M ∈ FG} if FG is nonempty, otherwise *Φ (

*G*) =

*G.*
We begin with a preliminary result which will be used frequently in connection
with induction arguments in the next section.

Lemma 2.1 Let

*M *be a maximal subgroup of a group

*G *and

*N *a normal subgroupof

*G*. If

*C ∈ I*(

*M *) and

*N ≤ k*(

*C*), then

*C/N ∈ I*(

*M/N *) and

*k*(

*C/N *) =

*k*(

*C*)

*/N *.

Proof. Since

*C ∈ I*(

*M *),

*C ≤ M *. Also

*C/N ≤ M/N *. And if

*A/N < C/N *,

*A/N ✁ G/N *, then

*A < C *and

*A ✁ G*. Since

*A ≤ M *,

*A/N ≤ M/N *, and

*C/N ∈*
On the index complex of a maximal subgroup

*I*(

*M/N *). Also

*C ≤ M *means

*k*(

*C*) =

*C*. Then

*k*(

*C/N *)

*≤ C/N *and moreover

*k*(

*C*)

*/N ≤ M/N *. So

*k*(

*C/N *)

*≤ k*(

*C*)

*/N *.

On the other hand, let

*k*(

*C/N *) =

*H/N *, then

*H ✁ G *and

*H/N < C/N *. Thus,

*H < C *and

*k*(

*C/N *) =

*H/N ≤ k*(

*C*)

*/N *. Therefore,

*k*(

*C/N *) =

*k*(

*C*)

*/N *.

Lemma 2.2[2] Let

*C *and

*D *be normal completions of a maximal subgroup

*M *of

*G*. Then

*C/k*(

*C*)

*∼*
The order of

*C/k*(

*C*), where

*C *is a normal completion of

*M *, is called the normal
index of

*M *in

*G*, denoted by

*η*(

*G *:

*M *).

Lemma 2.3[7] Φ1(

*G*) is a nilpotent group; Φ (

*G*) is a Sylow tower group.

Lemma 2.4 If

*G *is a group with a maximal core-free subgroup, the followings areequivalent:
(1) There exists a nontrivial solvable normal subgroup of

*G*.

(2) There exists a unique minimal normal subgroup

*N *of

*G *and the index of
all maximal subgroups of

*G *in

*FG *with core-free are powers of a unique prime.

Proof. Using Ref.[7], it suffices to prove that (2) implies (1). Indeed for every

*L ∈ FG *with core-free, let

*p *be the unique prime divisor of

*|G *:

*L|*. Since

*N ≤ L*,

*G *=

*LN *. Moreover

*|G *:

*L| |N |*, thus

*p |N |*. Let

*P ∈ Sylp*(

*N*). If

*P*
the Frattini argument we have

*G *=

*N · NG*(

*P *). Suppose that

*NG*(

*P *)

*≤ M <· G*,there exists

*Gp ∈ Sylp*(

*G*) satisfying

*NG*(

*P *)

*≥ NG*(

*Gp*). This means

*M ≥ NG*(

*Gp*)and therefore

*M ∈ FG*. But

*N ≤ M*, by the uniqueness of

*N *we get that

*M *iscore-free. By the hypothesis,

*p |G *:

*M |*. Since

*M ≥ NG*(

*Gp*),

*p*
leads to a contradiction. Thus

*P ✁ G *and

*P *=

*N *is a nontrivial solvable normalsubgroup of

*G*.

The following is the main result of the paper which gives a description of

*p*-
Theorem 3.1 Let

*p *be the largest prime divisor of the order of

*G*. The

*G *is

*p*-solvable if and only if for each non-nilpotent maximal subgroup

*M *of

*G *in

*F pc*,there exists a normal completion

*C *in

*I*(

*M *) such that

*C/k*(

*C*) is a

*p *-group.

Proof. It suffices to prove the sufficient condition. Suppose that the result is falseand let

*G *be a counterexample of minimal order, now we can claim that:
i)

*F pc *is not empty. Indeed if

*F pc *is empty, then

*Sp*(

*G*) =

*G*. Using [9, Lemma
2.2],

*Sp*(

*G*) is

*p*-closed. So

*P ∈ Sylp*(

*G*)

*✁ G *and

*G *is

*p*-solvable. This leads to acontradiction.

ii) Every maximal subgroup

*M *of

*G *in

*F pc *must be non-nilpotent. Indeed if
there exists a maximal subgroup

*M *in

*F pc *which is also nilpotent, then

*|G *:

*M |p *=
1 and

*G *is

*p*-solvable. It is a contradiction.

iii)

*G *has a unique minimal normal subgroup

*N *such that

*G/N *is

*p*-solvable.

Indeed if

*G *is simple, then for every

*M *of

*G *in

*F pc*,

*G *is the only normal completionin

*I*(

*M *) with

*k*(

*G*) = 1. By hypothesis,

*G *=

*G/k*(

*G*) is a

*p *-group. This contradictswith the fact that

*p *is the largest prime dividing

*|G|*, hence

*G *is not simple. Let

*N *be a minimal normal subgroup of

*G*, we will according to cases of

*N ≤ k*(

*C*) or

*N ≤ k*(

*C*) prove that

*G/N *satisfies the hypothesis of the theorem.

If

*N ≤ k*(

*C*), then

*N ≤ C *and

*C/N *is a normal completion for

*M/N *in

*G/N *.

By Lemma 2.1,

*C/N k*(

*C/N *) =

*C/N k*(

*C*)

*/N ∼*
=

*C/k*(

*C*). Again

*C/k*(

*C*) is a

*p *-group, so

*C/N k*(

*C/N *) is a

*p *-group.

If

*N ≤ k*(

*C*), then

*N ≤ C*. For otherwise, either

*N *=

*C *or

*N < C*, so either

*G *=

*M C *=

*M N *=

*M *or

*N < k*(

*C*). Each of which is a contradiction. Since

*N*is a minimal normal subgroup of

*G*, we have either

*C*
*N *=

*N *, then

*N ≤ C*. It is also a contradiction. So

*C*
is a normal completion for

*M/N *in

*G/N *. We are to show that

*C/N k*(

*C/N *) is a

*p *-group. Since

*k*(

*C*)

*< C *and

*C*
*N *= 1, it follows that

*k*(

*C*)

*N < CN *, and hence

*k*(

*C*)

*N/N < CN/N *. Also

*k*(

*C*)

*N/N ✁ G/N *, so we have

*k*(

*C*)

*N/N ≤ k*(

*CN/N *).

We define a map

*φ*:

*C/k*(

*C*)

*→ CN/N k*(

*CN/N *), by

*φ*(

*xk*(

*C*)) =

*xN k*(

*CN/N *)
for all

*xk*(

*C*)

*∈ C/k*(

*C*). Now

*xk*(

*C*) =

*yk*(

*C*) implies that

*x−*1

*y ∈ k*(

*C*), so(

*xN *)

*−*1(

*yN *) = (

*x−*1

*y*)

*N ∈ k*(

*C*)

*N/N ≤ k*(

*CN/N *) and
(

*xN *)

*k*(

*CN/N *) = (

*yN *)

*k*(

*CN/N *)

*.*
That is to say,

*φ*(

*xk*(

*C*)) =

*φ*(

*yk*(

*C*)). Hence the map is well defined. It can beverified that

*φ *is an epimorphism and

*CN/N k*(

*CN/N *) is an epimorphic image of
a

*p *-group. Thus

*G/N *satisfies the hypothesis of the theorem. By the minimalityof

*G*,

*G/N *is

*p*-solvable.

Similarly, it can be shown that

*G/N*1 is

*p*-solvable if

*N *is another minimal
normal subgroup

*N*1 of

*G*. Thus

*G *=

*G/N*
*N*1, which is isomorphic a subgroup
of the

*p*-solvable group

*G/N × G/N*1, is

*p*-solvable. So in the following supposethat

*N *is the unique minimal normal subgroup of

*G*.

*|N | *or

*N *is a

*p*-group, then

*N *is

*p*-solvable and so

*G *is

*p*-solvable. It is
a contradiction. Hence,

*|N |p *= 1 and

*N *=

*Np ∈ Sylp*(

*N*). Let

*M *be a maximalsubgroup of

*G *such that

*NG*(

*Np*)

*≤ M*. By the Frattini argument, we obtainthat

*G *=

*N · NG*(

*Np*). Using [7, lemma 5], there exists a

*Gp ∈ Sylp*(

*G*) with

*NG*(

*Np*)

*≥ NG*(

*Gp*), so

*M ∈ F p *and

*|G *:

*M|p *= 1. If

*|G *:

*M| *=

*q *be a prime lessthan

*p*, then

*|G| *divides

*q*!. This leads to another contradiction. Thus

*|G *:

*M | *is
On the index complex of a maximal subgroup
composite and

*M ∈ F pc*. By ii) and hypothesis, there exists a normal completion

*C *in

*I*(

*M *) such that

*C/k*(

*C*) is a

*p *-group. Obviously

*N *is a normal completionof

*M *. Combining with Lemma 2.2, we have

*C/k*(

*C*)

*∼*
=

*N/k*(

*N *) =

*N *. Thus

*N *is
a

*p *-group, which leads to the final contradiction. This completes the proof.

As we have known in [3], a group

*G *is

*π*-solvable if and only if for every maximal
subgroup

*M *of

*G *there exists a normal completion

*C *in

*I*(

*M *) such that

*C/k*(

*C*)is

*π*-solvable. We now extend this result by considering a smaller class of maximalsubgroups.

Theorem 3.2 Let

*G *be a finite group.

*G *is

*π*-solvable if and only if for everymaximal subgroup

*M *of

*G *in

*F *there exists a normal completion

*C *in

*I*(

*M *) such
that

*C/k*(

*C*)is

*π*-solvable.

Proof.

*⇐*) Let

*G *be a group satisfying the hypothesis of the theorem. If

*F*
empty then Φ (

*G*) =

*G*, and

*G *is solvable. Thus assume that

*F *is not empty. If

*G*
is simple, then for every

*M *in

*F *,

*G *is the only normal completion in

*I*(

*M *) with

*k*(

*G*) = 1 and thus

*G *=

*G/k*(

*G*) is

*π*-solvable. So suppose that

*G *is not simple.

Let

*N *be a minimal normal subgroup of

*G*. Without loss of generality, one cansuppose that

*F*
is not empty. We will use induction on the order of

*G*. For
, by [7, Lemma 3], it follows that

*M ∈ F *. So by hypothesis
there exists a normal completion

*C *in

*I*(

*M *) such that

*C/k*(

*C*) is

*π*-solvable.

Similar to the proof in Theorem 3.1,

*CN/N k*(

*CN/N *) is

*π*-solvable. Thus

*G/N *satisfies the hypothesis of the theorem. Using the induction we obtain that

*G/N *is

*π*-solvable. Furthermore, we can assume that

*N *is the unique minimalnormal subgroup of

*G*. By the same way,

*G/N *is still a

*π*-solvable group.

Now if

*N ≤ *Φ (

*G*), then from Lemma 2.3 Φ (

*G*) is solvable. Thus,

*N *is

*π*-
solvable, and furthermore

*G *is

*π*-solvable. If

*N ≤ *Φ (

*G*), there exists a maximalsubgroup

*M*0

*∈ F *with

*N ≤ M*
0. Then

*CoreGM*0 = 1 and

*G *=

*N M*0. So

*N *is a
normal completion in

*I*(

*M*0). By hypothesis there exists a normal completion

*C *in

*I*(

*M*0) such that

*C/k*(

*C*) is

*π*-solvable. By Lemma 2.2,

*N/k*(

*N*) =

*N ∼*
Again

*C/k*(

*C*) is

*π*-solvable, therefore

*N *is

*π*-solvable and moreover,

*G *is

*π*-solvable.

*⇒*) The converse is obvious.

The following theorem can be proved similarly as Theorem 3.2, and we omit it
Theorem 3.3 Let

*G *be a finite group.

*G *is solvable if and only if for everymaximal subgroup

*M *of

*G *in

*F *there exists a normal completion

*C *in

*I*(

*M *) such
that

*C/k*(

*C*)is solvable.

As we have known [4], if

*G *is

*S*4-free, then

*G *is super-solvable if and only if for
each maximal subgroup

*M *of

*G*, there exists a maximal completion

*C *in

*I*(

*M *) suchthat

*G *=

*CM *and

*C/k*(

*C*) is cyclic. The following theorem extends this result.

Theorem 3.4 Suppose that

*G *is

*S*4-free.

*G *is super-solvable if and only if for each
maximal subgroup

*M *of

*G *in

*F *, there exists a maximal completion

*C *in

*I*(

*M *)
such that

*G *=

*CM *and

*C/k*(

*C*) is cyclic.

Proof. Let

*G *be a super-solvable group. Then every chief factor of

*G *is a cyclicgroup of prime order.

*∀M ∈ F *, it is clear that the set

*S *=

*{T ✁ G|T ≤ M } *is
not empty. Choose an

*H *to be the minimal element in

*S*. Clearly,

*H ∈ I*(

*M *) and

*H/k*(

*H*)is a chief factor of

*G*, hence

*H/k*(

*H*) is cyclic.

Let

*G *be a group satisfying the hypothesis of the Theorem. If

*F*
then

*G *= Φ (

*G*) and

*G *is super-solvable [9]. We now assume that

*F*
empty and then

*G *is solvable. In the remainder of the proof we will drop themaximality imposed on the completion

*C *in

*I*(

*M *) in the hypothesis. For eachmaximal subgroup

*M *in

*F *, there exists a completion

*C *in

*I*(

*M *) such that

*G *=

*CM *and

*C/k*(

*C*) is cyclic. From [5, Lemma 2], we can get a normal completion

*A*in

*I*(

*M *) such that

*A/k*(

*A*) is either cyclic or elementary abelian of order 22.

First suppose that there exists an

*M *in

*F *which has a normal completion

*A*
such that

*A/k*(

*A*) is elementary abelian of order 22. Let

*G *=

*G/coreG*(

*M*) and

*C*,

*M *,

*A *be the images of

*C*,

*M *and

*A *in

*G *respectively. Then

*G *=

*C · M *=

*A · M *. Itis easy to verify that

*k*(

*A*) =

*A*
*coreGM*, so

*A/k*(

*A*)

*∼*
=

*A coreGM/coreGM *=

*A*.

Since

*core M *= 1,

*k*(

*A*) = 1,

*A *is a minimal normal subgroup of

*G*.

*A *is an
elementary abelian of order 22 and

*M*
*A *= 1. Considering the permutation
representation of

*G *on 4 cosets of

*M *,

*G *is isomorphic to a subgroup of

*S*4. Again

*S*4 and

*A*4 are the only non-super-solvable subgroups of

*S*4,

*A*4 doesn’t satisfy thehypothesis of the theorem, and

*G *is

*S*4-free, so

*G *is super-solvable.

Now assume that for each maximal subgroup

*M *in

*F *,

*M *has a normal com-
pletion

*A *so that

*A/k*(

*A*) is cyclic. Let

*N *be a minimal normal subgroup of

*G*.

Obviously, that

*G *is

*S*4-free is quotient-closed. By [4, Lemma 3] and [7, Lemma3], we can assume that the hypothesis holds for

*G/N *. Using induction, we obtainthat

*G/N *is super-solvable. Similar to Theorem 3.1, we can suppose that

*N *is theunique minimal normal subgroup of

*G*. If

*N ≤ *Φ (

*G*), then

*G *is super-solvable.

If

*N ≤ *Φ (

*G*), there exists a maximal subgroup

*M *in

*F*
*coreG*(

*M*) = 1. Obviously

*N *is a normal completion in

*I*(

*M*). By hypothesis,there exists a normal completion

*A *so that

*A/k*(

*A*) is cyclic. By Lemma 2.2,

*A/k*(

*A*)

*∼*
=

*N/k*(

*N *) =

*N *. Thus

*N *is cyclic and

*G *is super-solvable.

Remark Let

*G *be a solvable group. To obtain the conclusion in Theorem 3.4,the condition of maximality imposed on the completion

*C *is nonsignificant. So wehave the following result: If

*G *is

*S*4-free and solvable,

*G *is super-solvable if andonly if for each maximal subgroup

*M *of

*G *in

*F *, there exists a completion

*C *in

*I*(

*M *) so that

*G *=

*CM *and

*C/k*(

*C*) is cyclic.

Theorem 3.5 Let

*G *be a group and

*M *be an arbitrary maximal subgroup of

*G*in

*FG*. Then

*G *is nilpotent if and only if for each normal completion

*C *of

*M*,

*|C/k*(

*C*)

*| *=

*|G *:

*M |.*
Proof.

*⇐*) Let

*G *be a group satisfying the hypothesis of the theorem. If

*FG *is
On the index complex of a maximal subgroup
empty then

*G/N *= Φ1(

*G/N*). Using [9, Lemma 2.3],

*G/N *is nilpotent. If

*G *issimple, then for every

*M *in

*FG*,

*G *is the only normal completion in

*I*(

*M*) with

*k*(

*G*) = 1. By hypothesis

*|G/k*(

*G*)

*| *=

*G *=

*|G *:

*M |*,

*M *= 1, hence

*G *is a cyclicgroup of prime order. So assume that

*G *is not simple. Let

*N *be a minimal normalsubgroup of

*G*. Without loss of generality, suppose that

*FG/N *is not empty. Forany maximal subgroup

*M/N *in

*FG/N *, suppose that

*C/N *is an arbitrary normalcompletion in

*I*(

*M/N *). From [7, Lemma 3] we have

*M *in

*FG*. Obviously

*C *is anormal completion in

*I*(

*M *) and

*|C/k*(

*C*)

*| *=

*|G *:

*M |*. Using Lemma 2.1,

*|C/N k*(

*C/N *)

*| *=

*|C/N k*(

*C*)

*/N | *=

*|C/k*(

*C*)

*| *=

*|G *:

*M | *=

*|G/N M/N |.*
Thus

*G/N *satisfies the hypothesis of the theorem. Applying induction one can see

*G/N *is nilpotent. Similar to the proof in Theorem 3.1, we may assume

*N *is theunique minimal subgroup of

*G*.

If

*N ≤ *Φ1(

*G*), by [5, Lemma 2.3]

*G *is nilpotent. If

*N ≤ *Φ1(

*G*), there exists
an

*M *in

*FG *so that

*G *=

*NM*. Clearly,

*N *is a normal completion in

*I*(

*M*).Byhypothesis

*|N/k*(

*N *)

*| *=

*|N | *=

*|G *:

*M |*. For any

*L *in

*FG *with

*coreG*(

*L*) = 1,obviously

*N ≤ L *and

*G *=

*N L*.

*N *is also a normal completion in

*I*(

*M *), so

*|N/k*(

*N *)

*| *=

*|N | *=

*|G *:

*L|*. By Lemma 2.4

*G *has a nontrivial solvable subgroup

*K*,so

*N ≤ K *and

*N *is solvable. Since

*G/N *is nilpotent,

*G *is solvable. Thus

*N *is anelementary abelian

*p*-group. If

*G *is not a

*p*-group, we assume that

*|G| *has a primefactor

*q *different from

*p*. If the subgroup

*Q *=

*a|a ∈ G *and

*|a| *=

*q ≤ M *, thiscontradicts with the fact that

*coreGM *= 1. So there exists an of order

*q *element

*a *in

*G − M *. This implies that

*G *=

*M, a *. However,

*|N | *=

*|G *:

*M | *is a powerof

*p*. This leads to another contradiction. So

*G *must be a

*p*-group and then is anilpotent group.

*⇒*) The converse holds obviously.

1. W.E.Deskins. On maximal subgroups.

*Proc. symp. in pure Math.*, 1959, 1, 100-104.

2. W.E.Deskins. A note on the index complex of a maximal subgroup.

*Arch. Math.*, 1990, 54,
3. T.K.Dutta. Some characterisations of

*π*-solvable groups using index-complex.

*Indian J. Pure*
*Appl. Math.*, 2002, 33(4), 555-564.

4. Guo Xiu-yun. On Deskins’s conjecture.

*J. of Pure and Appl. Algebra. *1998, 124, 167-171.

5. Zhao Yao-qing. On the index complex of a maximal subgroup and the supersovability of a
finite group.

*Commun. in Algebra. *1996 24(5), 1785-1791.

6. Xu Ming-yao. An Intruduction of Finite Groups. Beijing: Science Press, 1987. (in Chinese)
7. Wang Pin-chao and Yang Zhao-xing. On the normal index of maximal subgroup in finite
group.

*J. Enginer. Math.*, 11(1), 1994, 41-48.

8. R.Baer. Classes of finite groups and their properties.

*Illions J. Math.*, 1957, 1, 115-167.

9. Wang Yan-ming. C-normality of groups and its properties.

*J. Algebra. *1996, 180, 954-965.

*Beijing Institute of Civil Eng. and Arch.*
*Beijing, 100081, P.R.ChinaE-mail:jiangjl@yahoo.com.cn*
Source: http://www.spm.uem.br/bspm/pdf/vol23-1-2/art5.pdf

Oxytocin, der kleine Tausendsassa unter den Hormonen Darf‘s ein bisschen mehr sein?Alles eine Frage der Chemie: Schmet-terlinge im Bauch, Glücksgefühle ohne Ende – da haben Verstimmungen und Depressionen keine Chance. In Zeiten großen Glücks schüttet der Körper ver-mehrt das Hormon Oxytocin aus. tids beschreibt die amerikanische Psy-chologin Professorin Diane Witt von wie folgt: „E

Clinical features Onset of the MH in humans is extremely variable; in initial symptoms and in the time of onset of syndrome. There have been instances where fulminant MH has occurred in patients who have previously tolerated potent triggers without difficulty. Reason is unknown. Timing Increased creatine kinase, myoglobinuria Triggers Succinylcholine Safe drugs All intravenous