1

I'm traying to prove the following theorem :

"Theorem: Let $G$ be a finite group, and $P \in \text{Syl}_p(G)$ where $p$ is the smallest prime dividing the order of $G$. Suppose that every subgroup of $P$ is normal in $G$, then $G \cong P \times A$ for some group $A$."
For the proof we use the Schur-Zassenhaus theorem and we argue that $[P, A] = 1$ due to the action of $A$ on $P/M \cong \mathbb{Z}_p$ being trivial, where $M$ is arbitrary maximal subgroup of $P$. This makes $[P, A] \leq \Phi(P)$. To complete the proof I need to conclude that $[P, A] = 1$. Could someone kindly suggest some details?"

Thank you for your assistance!

Khaled
  • 75
  • It’s not clear to me who is $A$. In which way you found it? And after that, with “suppose that every subgroup of $P$ is normal in $G$” you mean proper subgroup or is $P$ normal in $G$? – Mario Dec 13 '23 at 11:25
  • I am assuming that every $p$-subgroup of $G$ (including $P$) is normal in $G$ where $p$ is the smallest prime divisor of $|G|$. Since $P$ is normal in $G$ then $G/P$ is group of odd order, which must be solvable by Fiet-Thompson theorem. The solvability of $P$ and $G/P$ implies the solvability of $G$. By Schur-Zassenhaus Theorem $P$ has a complement say $A$ in $G$. So, $G=PA$ where $A$ is Hall $p'$-subgroup of $G$. Now we pick $M$ a maximal subgroup of $P$ and let $A$ act on the left cosets of $M$ in $P$ via conjugation. I hope that clears some thing about $A$. – Khaled Dec 13 '23 at 17:52
  • But is this theorem on a book? Since I think that there are too much hypothesis. $P$ and $A$ are normal, they commute and they have trivial intersection. If it is a your intuition, why do you think this is so? – Mario Dec 14 '23 at 07:58

1 Answers1

2

By the Schur-Zassenhaus theorem, $G=P\rtimes A$. Let $1\ne x\in P$. Then $C=\langle x\rangle\unlhd G$ by hypothesis. Hence, $A$ acts on $C$. Since $|Aut(C)|=|C|(p-1)/p$ and $p$ is the smallest prime divisor of $|G|$, $A$ must act trivially on $C$. Since $x$ was arbitrary, $A$ acts trivially on $P$. Hence, $G=P\times A$. (You don't need the Feit-Thompson theorem, since the existence part of Schur-Zassenhaus does not have a solvability hypothesis; moreover $G/P$ does not necessarily have odd order anyway).

Brauer Suzuki
  • 4,060
  • You have a very bold choice of name, @Brauer Suzuki. To be cross-examined by one about one's research was exciting; by both was fearsome. Happy days. – ancient mathematician Dec 14 '23 at 20:01
  • 1
    Thank you all for your valuable comments. Your thoughtful responses have been immensely helpful, and I appreciate the time and expertise you've shared. – Khaled Dec 14 '23 at 22:30
  • @ancientmathematician I'm not sure what you actually meant. Can you explain? Brauer was one of my academic ancestors. – Brauer Suzuki Dec 24 '23 at 09:41
  • I was put through my paces at a post-seminar party one evening by Brauer, and some months later by Suzuki (who must have been the referee of my first paper, though naturally he didn't say so). – ancient mathematician Dec 24 '23 at 13:46
  • Wow, that is impressive. I personally know very few people who have met Brauer (Paul Fong and John Thompson). – Brauer Suzuki Dec 24 '23 at 14:02
  • Well now I am in boasting mode the external examiner of my thesis was Feit. ;-) – ancient mathematician Dec 24 '23 at 15:13