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"Arturo Magidin" <magidin@xxxxxxxxxxxxxxxxx> wrote in message
news:eec06h$2ssg$1@xxxxxxxxxxxxxxxxxxxxx
> In article <eebvnd$ta$1@xxxxxxxxxxxxxxxx>,
> mina_world <mina_world@xxxxxxxxxxx> wrote:
>
> >yes, i see.
> >maybe, i need the lemma.
> >
> >lemma) Let G be a group containing normal subgroups H and K
> >such that H /\ K ={e} and H \/ K =G.
> >then G is isomorphic to H x K.
> >(H \/ K is the smallest subgroup of G containing HN.)
>
> You mean, containing HK.
>
> You hardly need that much. Just show that if H and K are normal, and
> H/\K={e}, then HK is isomorphic to H x K.
oh...no...Start with headache to me.
yes, i try...sir~
first, we showed the hk = kh for k in K and h in H.
(we used the h.k.h^-1.k^-1 = e.)
Let g : H x K -> HK be defined by g(h,k) = hk.
then, g((h,k)(h',k')) = g(hh',kk') = hh'kk' = hkh'k'
= g(h,k)g(h',k').
so, g is a homomorphism.
of course, onto.
and one to one, because,
if g(h,k)=e, then hk = e, so h = k^-1.
since h = k^-1 in H /\ K = {e}.
so, h = e = k.
so, ker g = (e,e) in H x K.
thank you very much for your teaching.
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