Minimal normal subgroup of a finite group.


by moont14263
Tags: finite, minimal, normal, subgroup
moont14263
moont14263 is offline
#1
Nov20-12, 01:04 PM
P: 39
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.
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