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From: joe.doubtful on 29 Mar 2010 08:34
Let N,H two subgroups of a finite group G, N is normal in G. N is a p-
group for p prime.
When someone reads that a finite group G is a semidirect product of N
and H, and (n,h)=1 where n=|N| and h=|H|, is it automatic (by Sylow's
theorem) to say that the group G has just one normal subgroup of order
n ?
From: Arturo Magidin on 29 Mar 2010 10:59
On Mar 29, 7:34 am, "joe.doubtful" <joe.doubt...(a)yahoo.it> wrote:
> Let N,H two subgroups of a finite group G, N is normal in G. N is a p-
> group for p prime.
> When someone reads that a finite group G is a semidirect product of N
> and H, and (n,h)=1 where n=|N| and h=|H|, is it automatic (by Sylow's
> theorem) to say that the group G has just one normal subgroup of order
> n ?

Yes: any subgroup of order n would be a Sylow p-subgroup, since n=p^k
and the order of G is p^k*m with m and p relatively prime. Since Sylow
p-subgroups are conjugate, and N is one of them and is normal, there
is one and only one Sylow p-subgroup of G, namely N.

--
Arturo Magidin
From: Derek Holt on 29 Mar 2010 16:16
On 29 Mar, 15:59, Arturo Magidin <magi...(a)member.ams.org> wrote:
> On Mar 29, 7:34 am, "joe.doubtful" <joe.doubt...(a)yahoo.it> wrote:
>
> > Let N,H two subgroups of a finite group G, N is normal in G. N is a p-
> > group for p prime.
> > When someone reads that a finite group G is a semidirect product of N
> > and H, and (n,h)=1 where n=|N| and h=|H|, is it automatic (by Sylow's
> > theorem) to say that the group G has just one normal subgroup of order
> > n ?
>
> Yes: any subgroup of order n would be a Sylow p-subgroup, since n=p^k
> and the order of G is p^k*m with m and p relatively prime. Since Sylow
> p-subgroups are conjugate, and N is one of them and is normal, there
> is one and only one Sylow p-subgroup of G, namely N.
>

You don't even need N to be a p-group. The condition (n,h)=1 is
enough to ensure that N is the only subgroup of order n, because if M
were another such subgroup, then NM would be a subgroup of order |N||
M|/|N^M|, which would have to equal n, so N=M.

Derek Holt.

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