# Re: Probabilistic polynomial time algorithm for solving the DLOG problem

On 11-05-26 9:22 PM, Jeffrey Goldberg wrote:
On 11-05-26 7:37 PM, Maaartin G wrote:

Using the hint that Pubkeybreaker gave that "Subgroups with size
5,7,8,9,10,11 are impossible" along with your example of even numbers
then a first guess is that [turned out to be wrong]

I'll play with this this evening. Thanks!

I got stuck on the proof. The actual pattern is instantly clear.

For a Group {0, 1, ..., m-1} with a * b = a + b mod m, then for every n
such that n divides m there is a subgroup S_n containing all the
multiples of n mod m

So in the case of m = 12 there are six subgroups including {0} and also
the not quite proper subgroup where n = 1, {0, 1, ..., 11}.

That bit is obvious. And most of the proof is simple, but there are a
couple of place where I got stuck.

First a lemma. If x divides y and cx > y, then cx mod y is a multiple of
x. Proof. As x divides y the "mod y" operation is the same as
subtracting an multiple of x from cx. The result of that subtraction is
a multiple of x.

Part 1. First to prove that all sets meeting my definition are subgroups.

Because n divides m, there will be a multiple of n that equals m. m mod
m is zero, so 0 will always be in there.

Now for closure. Consider two elements, an and bn. an + bn is a multiple
of n. There are three cases

Case 1: an + bn < m; This is trivially in the set.
Case 2: an + bn = m; 0 is in the set (as shown earlier).
Case 3: an + bn > m; by the lemma above this is a multiple of n and by
definition of "mod" it is less then m. This also in the set.

What I haven't been able to prove (though the intuitions are there) is
that only those sets will be subgroups. Maybe I'll have to look up
Legrange's work after all. But I was hoping that I could prove this all
on my own. Maybe after I've slept a bit.

Cheers,

-j

--
Jeffrey Goldberg http://goldmark.org/jeff/
I rarely read HTML or poorly quoting posts
.

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