Re: Metacyclic groups and cyclic Sylow-p-subgroups

From: Derek Holt (mareg_at_warwick.ac.uk)
Date: 10/11/04 

Date: 11 Oct 2004 04:02:15 -0700

Toni Lassila <toni@nukespam.org> wrote in message news:<36cam0lggkjpo4gg9r3c8s1fmrrbp97dvj@no.spam>...
> I was trying to prove myself that all groups of squarefree order are
> solvable.

I think the most instructive way to prove this result is to use Burnside's
Transfer Theorem, which says that if a group G has a Sylow p-subgroups P
such that P is in the centre of the normalizer in G of P, then G has a normal
subgroup N which complements P - that is, G = PN with P intersect N trivial.

For a group os squarefree order, let p be the smallest prime dividing G, and
let P be a Sylow p-subgroup of G. Since Aut(P) has order p-1, the elements in
the normalizer in G of P must induce the trivial automorphism on P, and so P
is in the centre of its normalizer, and we can use Burnside's Transfer
Theorem to infer the existence of a normal subgroup N complementing P. You
can now prove the theorem by applying induction to N.

You may not know Burnside's Transfer Theorem, but it is not too difficult,
and is worth learning. This does not prove the stronger result that G is
metacyclic, but you were not asked to prove that.

And by the way, your definition of metacyclic below is non-standard. The
standard definition is G is metacyclic if it has a normal subgroup N such
that N and G/N are both cyclic.

Derek Holt.

> I got these intermediate results but am missing the final
> part:
>
> Lemma:
>
> Let |G| be squarefree. Then all Sylow-p-subgroups of G are cyclic.
>
> Proof:
>
> We can write |G| = p_1^(k_1) p_2^(k_2) ... p_n^(k_n) by the
> Fundamental Theorem of Arithmetic, where all p_i are different
> primes. Since |G| is squarefree, we have that k_i <= 1 for all i,
> thus |G| = p_1 p_2 ... p_j. Then, by Sylow's First Theorem, the
> Sylow-p_i-subgroups of G have prime order and are cyclic.
>
>
> Definition:
>
> Group G is said to be metacyclic if G' is cyclic and G/G' is cyclic.
>
>
> Lemma:
>
> Metacyclic groups are solvable.
>
> Proof:
>
> G' is normal in G and G/G' is abelian. Thus (G, G', {1}) is a
> solvable series for G, since G'/{1} = G' is abelian.
>
>
> 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?

Relevant Pages

  • Re: algebra with Sylow.
    ... I know how to conclude this, but not from Sylow's Theorems. ... Theorems are generally useless for p-groups. ... H_1 is a normal subgroup of H_2. ... or I can prove by normalizer. ...
    (sci.math)
  • Re: normalizers and centralizers proof check
    ... paul12 wrote: ... C_Gof a subgroup H is a normal subgroup of the normalizer N_G. ... Columbia, SC ...
    (sci.math)