Let $G$ be a finite group and let $P$ be sylow p subgroup of $G$. Prove the following :

$1)$ If $M$ is a normal p-subgroup of G then $M\leq P$.

$2)$ There is a normal p-subgroup of G that contains all normal p-subgroups of G.

what i have done sofar is :

For 1, by Sylow theorem, Given a sylow p-subgroup $P$ and any p-subgroup, $Q$ there exists $g\in G$ such that $gQg^{-1}\leq P$

Inparticular, for Normal p-subgroup $Q$, we have $Q=gQg^{-1}\leq P$ i.e., $Q\leq P$ for all Normal subgroups.So, I guess $(1)$ is done.

for $(2)$, I have done only a particular case : particular case of sylow p-subgroup being unique,

In that case sylow p-subgroup is Normal and we apply $(1)$ to conclude that there exists a sylow p-subgroup (and hence a Normal p-subgroup whose existence is already known) which contains all Normal p-subgroups.

I am not very sure about the way to see when sylow p-subgroup is not unique.

I would be thankful if someone can give me a hint.

Thank You,.

  • 3
    $\begingroup$ Well, a normal $p$-subgroup is contained in all Sylow $p$-subgroups. So you might want to look at the intersection of all Sylow $p$-subgroups. $\endgroup$ – Daniel Fischer Aug 31 '13 at 14:28
  • 1
    $\begingroup$ Let $|G|=p^nm,~~(p,m)=1$, so $|P|=p^n$ and $|M|=p^kt$ in which $k\leq n,~~t\mid m$. $M$ is normal in the group so we can speak about the subgroup $MP$. We have: $$|MP|=\frac{p^n\times p^kt}{p^s}\Big|~ p^nm,~~s\leq k$$ $\endgroup$ – mrs Aug 31 '13 at 14:32

Your Answer

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