Re: generators be bound

From: Tom St Denis (tomstdenis_at_iahu.ca)
Date: 10/24/03 

Date: Fri, 24 Oct 2003 11:21:50 GMT

"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]

What really happens is all generators are in fact primitive [to some
sub-group]. 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. There will be an element [g=3] which
generates an order 6 group but there are also elements which generate
smaller sub-groups [e.g. g=2 genrates a group of order 3].

The point is w.r.t the 3 element sub group {2, 4, 1} of Z/7Z g=2 is
primitive since it generates the entire group. w.r.t. Z/7Z multiplicative
sub-group {3,2,6,4,5,1} g=2 is not primitive.

Not to abuse notation though. I agree that without further details
"generator" should therefore be w.r.t. the multiplicative group of maximal
order [as others stated]. My point though is that it isn't invalid to say
"g=4 is a generator of prime order modulo a safe prime". It really does
generate such a sub-group.

Tom

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