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My doctor gave me the following theorem:

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

My question is the following:

Suppose that $G$ is a finite group with $2$ divides the order of $G$ and $G$ has a $\{2,p\}$-Hall subgroup for every odd prime $p \in \pi (G)$. Does this imply that either $G$ is abelian or $G$ has a normal subgroup $N$ such that $\{1\} \neq N \neq G$ where $1$ is the identity of $G$?

Thanks in advance.

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    $\begingroup$ Mi doctor usually gives me only medicines and bills to pay... $\endgroup$ – DonAntonio Nov 23 '12 at 16:44
  • $\begingroup$ The answer is yes, assuming your doctor told you the truth. In fact I think that any finite group which has a Hall $\{2,p\}$-subgroup for every odd prime $p$ is solvable in that case. But to return to the original question: according to your doctor, the group $G$ is not non-Abelian simple, so it is either Abelian, or else it has a proper non-trivial normal subgroup. $\endgroup$ – Geoff Robinson Nov 23 '12 at 17:06
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This question has been answered in a comment:

The answer is yes, assuming your doctor told you the truth. In fact I think that any finite group which has a Hall {2,p}-subgroup for every odd prime p is solvable in that case. But to return to the original question: according to your doctor, the group G is not non-Abelian simple, so it is either Abelian, or else it has a proper non-trivial normal subgroup. – Geoff Robinson Nov 23 '12 at 17:06

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