Re: Generator of a group
From: Gregory G Rose (ggr_at_qualcomm.com)
Date: 01/07/05
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Date: 6 Jan 2005 21:44:12 0800
In article <crkgol$gj0$00$1@news.tonline.com>,
MokKong Shen <mokkong.shen@tonline.de> wrote:
>Greg Rose wrote:
>> I'm not aware of any way to do it without knowing
>> the factors of p1, and I have a strong feeling
>> that there can't be one, but I'd be happy to be
>> corrected on this.
>I don't yet see why the case (p1)/2 is prime could be
>more advantageously treated as compared to the case where
>the factorization of (p1)/2 is unknown (and one resorts
Gee, MokKong, you've outdone yourself here.
If (p1)/2 is prime, you *know* its factorization.
Any generator of the multiplicative group must either be:
1. the identity (generates the order 1 subgroup)
2. 1 (generates the order 2 subgroup)
3. a generator of the group of quadratic residues
4. a generator of the whole group.
The order of a subgroup must divide the order of
the group. The order of the multiplicative group
mod p is p1. P is prime, so it's odd, so p1 is
even, so 2 divides p1. If (p1)/2 is prime, there
just ain't any other possibilities.
I know that you usually argue for argument's sake,
which is why I won't usually argue against you,
but this is simply fatuous.
Greg.
 Greg Rose 232B EC8F 44C6 C853 D68F E107 E6BF CD2F 1081 A37C Qualcomm Australia: http://www.qualcomm.com.au
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