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Let $G$ be a finite group. If $G$ is $2$-nilpotent then $G$ is solvable.

I know a way to prove this result using the Feit-Thompson Theorem:

Let $H$ be a normal $2$-complement of $G$. Then $H$ is solvable by Feit-Thompson since $|H|$ is odd and $G/H$ is solvable because $G/H$ is a $2$-group. Therefore, $G$ is solvable since solvability is closed under extensions.

But I think this is too much. I'm interested in a proof that uses Burnside's Normal Complement Theorem.

nom
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  • Let $H$ be a group of odd order. Then $G=C_2\times H$ is $2$-nilpotent (and has a non-trivial Sylow $2$-subgroup, in case that was a concern for you), and is soluble iff $H$ is. – David A. Craven Nov 22 '22 at 23:29
  • For groups of order/cardinality $2^n p^a q^b$ with 2-nilpotent property, https://en.wikipedia.org/wiki/Burnside%27s_theorem gives the result that the group is solvable. – Balaji sb Nov 23 '22 at 01:09
  • The theorem would also follow if we can prove that every group of odd order has a normal solvable subgroup $N$ which is not {e}. We can then apply induction to prove that original group is solvable since $N$ is solvable and $G/N$ is solvable by induction. Interesting to see if this is true. cheers. – Balaji sb Nov 23 '22 at 01:16

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Since groups of odd order are $2$-nilpotent, the claim would imply that every group of odd order is solvable. So you cannot avoid the Feit-Thompson theorem.

spin
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