|
|
<mareg@xxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:eeb3qv$7s7$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> In article <eeb1d6$ka8$1@xxxxxxxxxxxxxxxx>,
> "mina_world" <mina_world@xxxxxxxxxxx> writes:
> >hello sir~
> >
> >let G be a group of order 2p where p is an odd prime.
> >
> >show that if G has a normal subgroup of order 2,
> >then G is cyclic.
> >
> >----------------------------------------------
> >um...i think....
> >
> >by Cauch's theorem,
> >G has an element "x" of order p.
> >so, let |<x>| = p.
> >(of course, <x> is normal because |<x>|=|G|/2.)
> >
> >since G has a normal subgroup of order 2,
> >let N = {e, y} be normal subgroup of G.
> >(of course, y^2 = e and y^-1 = y.)
> >
> >i will show that |xy| = 2p.
>
> I think your solution below is correct (I have not checked every detail),
> but it seems unnecessarily long. It is preferable to use a more general
> technique.
>
> Lemma. If normal subgroups M and N of a group G satisfy M intersect N =
{e},
> then mn = nm for all m in M and n in N.
>
> Proof By normality, the commutator m^-1 n^-1 m n lies in both M and N,
hence
> m^-1 n^-1 m n = e, so mn = nm.
>
> Now in your problem, N and <x> have coprime orders, so N intersect <x> =
{e}.
> Hence xy = yx, from which it follows easily that |xy| = 2p
> (because (xy)^2 = x^2y^2 = x^2 != e, and (xy)^p = x^py^p = y^p = y != e).
>
um...i think again with your advice.
since, |N.<x>| = (|N||<x>|)/|N intersection <x>|,
so, |N.<x>| = 2p.
so, N.<x> = G.
by lemma, G is abelian.
so, by fundamental theorem of finitely generated abelian groups,
G is isomorphic to either Z_2 x Z_p or Z_2p.
but Z_2 x Z_p is isomorphic to Z_2p.
thus, G ~ Z_2p. so, cyclic.
thank you very much.
|
|