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.

Thanks in advance.