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Let $G=KH$ be a Frobenius group of even order with Frobenius kernel $K$ and Frobenius complement $H$ such that $\pi(H)=\{p\}$, where $p$ is prime. Why is a Sylow $2$-subgroup of $G$ normal in $G$?

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Must $p$ be odd? Because then it is true by Thompson's famous theorem that kernels $K$ of Frobenius groups are always nilpotent. – Nicky Hekster Nov 18 '12 at 21:58
up vote 3 down vote accepted

This is not true. $S_3$ is a counterexample. We have $S_3= K \rtimes H$ where $K=\langle (1,2,3)\rangle$ is a Frobenius kernel and $H=\langle (1,2)\rangle$ is a Frobenius complement. Here $H\in \text{Syl}_2(G)$ but $H\not\unlhd G$.

However, the theorem does not fail if we additionally assume that $p$ is odd, and in fact a stronger result holds.

Proposition: If $G=K\rtimes H$ is a Frobenius group with complement $H$ and kernel $K$ , then $G$ has a characteristic Sylow $2$-subgroup if and only if the order of $H$ is odd.

Proof: If the order of $H$ is odd, every Sylow $2$-subgroup $P$ is contained in the Frobenius kernel. $K$ is nilpotent by Thompson (see his paper about this, or Character Theory of Finite Groups by Huppert or Isaacs). Thus $P$ is characteristic in $K$. Since $K=\mbox{Fit}(G)$, $K$ is characteristic in $G$, whence $P$ is characteristic in $G$ as well.

Conversely, if $P$ is a characteristic Sylow $2$-subgroup of $G$, then because $G$ is Frobenius $P$ cannot be a subgroup of $H$. Since $|H|$ divides $|K|-1$, $P$ must then be a subgroup of $K$, so we have that $|H|$ is odd.

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This is exactly what I was looking for! Thank you. – Math3435 Nov 20 '12 at 14:35

I think that if $p$ is odd then the 2-sylow subgroup of $G$ is normal in $G$. Since every 2-sylow subgroup of $G$ is a 2-sylow subgroup of $K$ and $K$ is nilpotent then every sylow subgroup of $k$ is normal in $k$. so $k$ has only one 2-sylow subgroup and so $G$ has only one.

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