# 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.t-online.com>,

Mok-Kong Shen <mok-kong.shen@t-online.de> wrote:

*>Greg Rose wrote:
*

*>> I'm not aware of any way to do it without knowing
*

*>> the factors of p-1, 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 (p-1)/2 is prime could be
*

*>more advantageously treated as compared to the case where
*

*>the factorization of (p-1)/2 is unknown (and one resorts
*

Gee, Mok-Kong, you've outdone yourself here.

If (p-1)/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 p-1. P is prime, so it's odd, so p-1 is

even, so 2 divides p-1. If (p-1)/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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