Minimal normal subgroup of a finite group.

moont14263

I have Theorem 1 from a research paper.

Theorem 1. Suppose that [itex]G[/itex]
is a finite non-abelian simple group. Then there exists an odd prime [itex]q \in \pi(G)[/itex]
such that [itex]G[/itex]
has no [itex]\{2, q\}[/itex]
-Hall subgroup.

I have Theorem 2 from a book.

Theorem 2. If [itex]H[/itex]
is a minimal normal subgroup of [itex]G[/itex]
, then either [itex]H[/itex]
is an elementary abelian [itex]p[/itex]
-group for some prime [itex]p[/itex]
or [itex]H[/itex]
is the direct product of isomorphic nonabelian simple groups.

I like to know if the contradicition in this argument is true.

Let [itex]N[/itex]
be a minimal normal subgroup of a finite group [itex]G[/itex]
with [itex]2[/itex]
divides the order of [itex]N[/itex]
and [itex]N[/itex]
is not a [itex]2[/itex]
-group. Let [itex]P[/itex]
be a Sylow [itex]2[/itex]
-subgroup of [itex]N[/itex]
. Suppose that for each prime [itex]q \neq 2[/itex]
dividing the order of [itex]N[/itex]
, there exists a Sylow [itex]q[/itex]
-subgroup [itex]Q[/itex] of [itex]N[/itex]
such that [itex]PQ[/itex]
is a subgroup of [itex]N[/itex]
. Cleary, [itex]N=T_{1}×T_{2}×...×T_{r}[/itex]
for some postive intger [itex]r[/itex]
, where each [itex]T_{i}[/itex]
is a nonabelian simple group. Let [itex]T[/itex]
be one of the [itex]T_{i}[/itex]
. It is clear that if some prime divides the order of [itex]N[/itex]
then this prime must divide the order of [itex]T[/itex]
as [itex]|N|=|T|^{r}[/itex]
. Since [itex]T[/itex]
is normal in [itex]N[/itex]
, then [itex]T \cap PQ=(T \cap P)(T \cap P)[/itex]
(I Know that this statement is true so do not bother checking it). So, [itex]T[/itex]
is a finite nonableian simple group with Hall [itex]\{2,q\}[/itex]
-subgroup for each odd prime dividing the order of [itex]T[/itex]
which contradicts Theorem 1.

Thanks in advance.