Re: Metacyclic groups and cyclic Sylow-p-subgroups

From: Robin Chapman (rjc_at_ivorynospamtower.freeserve.co.uk)
Date: 10/07/04 

Date: Thu, 07 Oct 2004 14:06:32 +0100

Toni Lassila wrote:

> Definition:
>
> Group G is said to be metacyclic if G' is cyclic and G/G' is cyclic.
>
>
> Now all that's missing is to show that a group is metacyclic if all
> Sylow-p-subgroups are cyclic and the result follows by chaining these
> statements. How does one show that?

This is quite tricky. Marshall Hall takes nearly a page in his
book to prove this, and relies on a difficult previous result
(if m is a factor of the order of a finite group G, then the
number of solutions of x^m = 1 in G is a multiple of m).

Here's an outline of his proof.

Write n = |G| = p_1^{a_1} ... p_k^{a_k}
where the primes p_1 < ... < p_k.
Then if m = p_j^b p_{j+1}^{a_{j+1}} ... p_k^{a_k},
x^m = 1 has exactly m solutions (this is not trivial!).
Consequently the Sylow p_k-subgroup is normal, and inductively H
is soluble. If we have G'/G'' and G''/G''' cyclic, then G'' = G'''
(this isn't too hard). By solubility G'' is trivial. Then G/G'
and G'/G'' are abelian and have cyclic Sylow subgroups, so are cyclic. 

-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9"
Francis Wheen, _How Mumbo-Jumbo Conquered the World_

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