Re: Metacyclic groups and cyclic Sylow-p-subgroups
From: Jyrki Lahtonen (lahtonen_at_utu.fi)
Date: 10/07/04
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Date: Thu, 07 Oct 2004 16:19:40 +0300
Toni Lassila wrote:
>
> I was trying to prove myself that all groups of squarefree order are
> solvable. I got these intermediate results but am missing the final
> part:
>
[snipped two easy parts]
> 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 somewhat tricky. The resident experts will be able to tell you
more, but I think that the argument depends essentially on our knowledge
of the automorphisms of the cyclic groups. I recall seeing a lot of related
results in a paper "Über endliche Fastkörper", by H.Zassenhaus
(Abh. Math. Sem. Hansischen Universität, 1936, pp. 187-220). The experts
here may help you to a more accessible reference. Zassenhaus seeks to
classify the so called fixed-point-free groups, so he does a lot more
work).
As an example, let's consider the case of group G of order 105=3*5*7.
>From the Sylow theorems we may immediately infer that the number
of Sylow 3-subgroups is either 1 or 7. So if P_3 is a cyclic subgroup
of order 3, we see that its normalizer N(P_3) has order 105 or 15.
Anyway we see that P_3 is normalized by an element x of order 5. But
the group P_3 has only two automorphisms, so none of order 5. Hence
x must CENTRALIZE P_3, and therefore we have a cyclic subgroup of
order 15 (the product of two commuting elements of respective orders
3 and 5 is of order 15).
Let then P_5 be a Sylow 5-subgroup. The number of these is either 1 or
21. We just saw that at least one of these (hence all) is centralized
by an element of order 3. Thus the case N(P_5)=P_5 is excluded, so
we may conclude that P_5 is a normal subgroup of G. But the automorphism
group of P_5 is cyclic of order 4. Hence it has no automorphisms of order 7,
so any element of order 7 must again centralize P_5. So any element of
order 5 in G commutes with any element of order 7. Consequently G will
have a cyclic subgroup of order 35.
The number of Sylow 7-subgroups is either 1 or 15. We saw in the previous
paragraph that any P_7 is centralized (hence normalized) by an element
of order 5. Therefore N(P_7) is larger than P_7, so we see that there
is a unique P_7 that is a normal subgroup of G. The product of two normal
subgroups is normal, so as a product of P_5 and P_7 we get a cyclic normal
subgroup N of order 35.
The group P_7=C_7 has automorphisms of order 3 (e.g. squaring in P_7 is
such an automorphism), so our group is not necessarily abelian (if it
where, it would be cyclic), but anyway the only possible nontrivial
commutators are those of elements of orders 3 and 7, and as P_7 is a
normal subgroup, such a putative non-trivial commutator must be of order 7.
So either G'=1, or G'=P_7. In the first case G is cyclic and in the latter
case G/G' is of order 15, hence cyclic. Thus G is always metacyclic.
BTW I think that this is not the easiest way to prove your goal result.
I shall leave the stage now and await the experts...
Cheers,
Jyrki
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