Re: Subgroups of every order in p-groups product

newsgr.mail@xxxxxxxxx wrote:

On 16 Mag, 04:25, Timothy Murphy <gayle...@xxxxxxxxxx> wrote:
newsgr.m...@xxxxxxxxx wrote:
Let H be an elementary abelian subgroup of order 9, H is normal in G.
K is an abelian 2-subgroup of G, its order is 2^a.
Consider the group HK. Important: it is unknown whether an order 3
subgroup of H is normal in HK or not (if it were normal then my
problem would be trivial).
Suppose that K contains a subgroup C of order 2^{a-1} and that this
one is normal in HK.
I want to prove that the above hypotheses are sufficient to claim that
HK has subgroups of every order (whenever this order is compatible
with Lagrange's theorem). Please note that this could not be true, I
am just guessing that it is right. Let's try.

C is normal in HK and H /\ C = 1, so we can consider the direct
product H x C. This group has order 9*2^{a-1}. Since C is abelian, it
has normal subgroups of every order and we can find isomorphic copies
of these  subgroups in the product H x C. These copies are also normal
in H x C. Hence, there are subgroups of every order less or equal
than  9*2^{a-1}.

Is it right?

I'm no group-theorist,
but can't you just consider the group HK/C, of order 18?

The question then is, does such a group contain a subgroup of order 6?


Sure it does, because groups of order 2p^k (p prime) are
supersolvable. But I am not sure about what you were trying to
indicate to me.

If you are the OP, the question you asked was if HK contained subgroups
of all possible orders.
The only order missing was 3*2^a,
and I was simply pointing out that by passing to HK/C,
the question reduced to the same question for a group of order 18.
I left it to you to deal with this case.


.

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