Is it true that a supesolvable group has a sylow tower? How can I construct the tower kwowing than there's a principal serier with factors of prime order? Can anyone help me with an hint or an idea? Thanks to everyone for the help!

up vote 4 down vote accepted

I think you can obtain this from the fact that a finite supersoluble group $G$ has a normal subgroup of order a power of the largest prime $p$ dividing its order. Then you argue by induction.

To prove this fact, start with a minimal normal subgroup $N$ of $G$ of order a prime $q$ - the existence of such an $N$ comes from the definition of a supersoluble group.

If $q = p$, you are done. If $q < p$, proceed by induction on the order of $G$, so that there is a normal subgroup $M/N$ of $G/N$, of order $p^{a} > 1$, for some $a$.

Clearly $M = N P$, where $P$ is a subgroup of $G$ of order $p^{a}$. Since the automorphism group of $N$ has order $q - 1 < p$, we have that $P$ centralizes $N$. Therefore $M = N \times P$, hence $P$ is characteristic in $M$, and thus normal in $G$.

Here is an outline proof. Let $G$ be a finite supersolvable group and let $p$ be the largest prime dividing $|G|$. We claim that $G$ has a normal cyclic subgroup $N$ of order $p$, and a normal Sylow $p$-subgroup. If we can prove that then the existence of the Sylow tower is easily proved by induction.

Let $N$ be a normal subgroup of $G$ of prime order $q$. If $q=p$ we are done. If $q \ne p$, then by induction $G/N$ has a normal cyclic subgroup $M/N$ of order $p$ and, since $p>q$, we have $M = P \times N$ with $P \unlhd G$ normal cyclic of order $p$.

  • 1
    This proof must be in The Book, then ;-) – Andreas Caranti Jun 19 '16 at 10:52
  • @Derek Holt thank you very much! – Sibilla Jun 19 '16 at 12:22
  • @Andreas thank you very much! – Sibilla Jun 19 '16 at 12:27

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