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.


$\newcommand{\index}[1]{\lvert #1 \rvert}$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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.