Re: generators be bound
From: Marcel Martin (mm_at_ellipsa.no.spam.net)
Date: 10/24/03
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Date: Fri, 24 Oct 2003 17:56:17 +0200
Tom St Denis a écrit :
>
> "Mok-Kong Shen" <mok-kong.shen@t-online.de> wrote in message
> news:3F988551.610D9516@t-online.de...
> >
> >
> > Tom St Denis wrote:
> > >
> > [snip]
> > > So what should I have said to mean a generator of a sub-group of a
> given
> > > order?
> >
> > My math knowledge is poor and maybe I misunderstood you,
> > but, if you have a subgroup (of whatever order) of a cyclic
> > group, then it is also cyclic and there must be an element
> > of it that generates that subgroup and that element is by
> > definition a generator of that subgroup. Or am I missing
> > something?
>
> That's just my point. According to the folk here that's not a generator.
> Unless it generates the entire group it's just a ??? [blank]
It is a "generator of order k" in the considered group. Because it
doesn't generate the whole considered group but a subgroup of order k.
> What really happens is all generators are in fact primitive [to some
> sub-group].
Yes. But "to some subgroup" must be explicit.
> Take Z/7Z for instance. There will be sub-groups of order 2, 3
> and 6. Note that Z/7Z actually has 7 elements so the group of order 6 must
> be a sub-group of it as well.
Are you saying that "since the multiplicative group U(Z/7) has less
elements than Z/7, it is a subgroup of Z/7"?
Take care, this, this is not redundant, this is totally meaningless.
mm
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