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Let $G$ be a finite solvable group whose order is divisible by at least three distinct primes. If every Hall $p'$-subgroup of $G$ is nilpotent, show that $G$ is nilpotent.

I feel like the best approach would be to show that every Sylow subgroup is normal. We know that there exists Slow p-subgroups for the three distinct primes, but can't like it to the Hall $p'$-subgroups.

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$\newcommand{\index}[1]{\lvert #1 \rvert}$Let $r$ be an arbitrary prime dividing the order of $G$. Let $p, q$ be two other primes dividing the order of $G$, so that $p, q, r$ are distinct.

Let $X$ be a Hall $p'$-subgroup, and $Y$ be a Hall $q'$-subgroup. The intersection $X \cap Y$ contains a Sylow $r$-subgroup $R$, and by assumption this is normal in both $X$ and $Y$. Moreover $G = \langle X, Y \rangle$, so that $R$ is normal in $G$.


This result is useful. It shows among others that we have $\index{G:X \cap Y} = \index{G:X} \cdot \index{G:Y}$ and $G = X Y = \langle X, Y \rangle$ here.

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  • $\begingroup$ Thank you for your answer, however I don't understand why the intersection $X\cap Y$ contains a Sylow $r$-subgroup. We know know $r$ divides the order of $X$ and $Y$, why does this imply that it divides the order of $X\cap Y$? $\endgroup$ – user234542 Apr 27 '15 at 2:18
  • $\begingroup$ It follows from the formula for the index of the intersection. $\endgroup$ – Andreas Caranti Apr 27 '15 at 6:04

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