## Minimal normal subgroup of a finite group.

I have Theorem 1 from a research paper.

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

I have Theorem 2 from a book.

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

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

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