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"Tonico" <Tonicopm@xxxxxxxxx> wrote in message
news:1158222017.861450.181060@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>
> mina_world wrote:
> > 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.)
> >have other nice solution, i hope to know that.
> ********************************************************
> Hi:
> Wow, what along answer! Simpler: if G has sbgp.of order 2, then since
> it surely has
> a normal (Sylow) sbgp. of order p (why?), then G is the (inner) direct
> product of these
> two sbgps., and thus obviously abelian.
i can't understand the sentence with "then G is the (inner) direct
product of these two sbgps."
i know that "finite generated abelian => direct product".
but your sentence is inverse of this.
so, i can't understand it.
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