Re: Centerless groups of size p+1, with prime p of the form 4k+3
mareg_at_mimosa.csv.warwick.ac.uk
Date: 07/20/04
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Date: Tue, 20 Jul 2004 09:23:48 +0000 (UTC)
In article <gerry-CDA156.16341120072004@sunb.ocs.mq.edu.au>,
Gerry Myerson <gerry@maths.mq.edi.ai.i2u4email> writes:
>In article <12245242.0407192213.350b004@posting.google.com>,
> alireza_abdollahi@yahoo.com (Alireza Abdollahi) wrote:
>
>> Let T be the set of all prime numbers p as the form 4k+3 (k integer)
>> for which there exists "no" finite group G with the properties that
>> |G|=p+1 and |Z(G)|=1, where Z(G) is the center of G.
>>
>> Is it true that T is infinite?
>
>I don't know.
>
>What do the groups of order 4q look like, when q is a prime?
>Of course there are two commutative ones, and the dihedral
>group, and the product of a dihedral and a cyclic, and each
>of those groups has |Z(G)| > 1, but are there any others?
Yes if q = 1 (mod 4), there is a group with trivial centre:
< x, y | x^q = y^4 = 1, y^-1 x y = x^t >, where t^2 = -1 (mod q)
>It is believed that there are infintely many primes p = 4k + 3
>such that p + 1 = 4q with q prime, but no one has a proof.
If p + 1 = 4q with q prime, then there exists a centreless group of
order p+1 if and only if q = 1 (mod 4). So, to prove the conjecture
with this approach, you would need to prove that there are infinitely
many primes q with q = 3 (mod 4) and 4q - 1 prime.
More generally, you could prove it by showing that there are infinitely
many primes p such that p+1 = 2^n q, where q is a prime not congruent to
1 mod 2^n and q not dividing 2^r - 1 for any r <= n.
And as someone else pointed out, you could also prove it by showing that there
are infinitely many Mersenne primes.
Derek Holt.
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