# Minimal normal subgroup of a finite group.

• Nov 20th 2012, 11:45 AM
moont14263
Minimal normal subgroup of a finite group.
I have Theorem 1 from a research paper.

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

I have Theorem 2 from a book.

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

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

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