I'm looking at finite simple groups of order $p+1$ where $p$ is a prime number.
But they don't seem to fall into any classification - have these all been determined? Is the number of them even finite?
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1$\begingroup$ n!/2-1 is prime for n=5, 6, 9, 31, 41, 373 ... (sequence A082671 on the OEIS). Is it known if this sequence is finite? $\endgroup$– Alon AmitCommented Dec 8, 2010 at 9:54
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3$\begingroup$ If it were known that the sequence was finite, it would surely say so in the OEIS. Given that you're asking, I'm guessing it doesn't say so. So I am guessing it's not known, based purely on the information you have given me :-) $\endgroup$– Kevin BuzzardCommented Dec 8, 2010 at 9:57
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1$\begingroup$ As suggested by Jason and Wikipedia, the list of finite simple groups is probably (though not certainly) known. But the question being asked here should first be investigated for the well-known infinite families starting with alternating groups and then groups of Lie type. The completeness of the classification is not an immediate issue. $\endgroup$– Jim HumphreysCommented Dec 8, 2010 at 13:01
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6$\begingroup$ Is there any particular reason why you are looking at finite simple groups of order $p+1$ ? $\endgroup$– Derek HoltCommented Dec 8, 2010 at 13:04
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1$\begingroup$ Funnily enough, I recently ran across another context where this same question was asked: cameroncounts.wordpress.com/2010/03/17/how-wrong-was-cauchy . Such groups provide counterexamples to a "theorem" of Cauchy. $\endgroup$– Harry AltmanCommented Dec 8, 2010 at 18:01
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4 Answers
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The philosophical point here is that if all you know about a group $G$ is its order $\lvert G \rvert$, then by far the most relevant information is the prime factorization of that order. (Back when sporadic groups were still being discovered, there are anecdotes about phoning John Thompson with the order of your hypothetical new group, and after some calculations he would tell you whether it 'checked out' or not - and of course he would just be using the knowledge of which primes divided the order and to which exponents). So questions about the prime factorization of $\lvert G \rvert - 1$ are going to be dominated by the (generally unsolved) number theoretical problems that relate the prime factorization of $n$ and $n+1$, e.g. the existence of infinitely many Sophie Germain primes, Mersenne primes, or primes $p$ for which $\frac{p^2+1}{2}$ is also prime, etc.
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$\begingroup$ Ah, I see. So any potential answer here would just be a purely number-theoretic statement; the group wouldn't really enter into it. Sounds like this question is very unfeasable then. Oh well. Thanks for the insight. $\endgroup$ Commented Dec 8, 2010 at 17:11
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As suggested by the other answers and comments, this is unknown (and a hard arithmetic question). Here's another example that might help indicate why:
The order of the finite simple group $PSL_2(F_q)$ is (for $q$ not a power of $2$) $q(q^2-1)/2$. You'd therefore like to know when $q(q^2-1)/2-1$ is a prime, for $q$ a prime power. The question is (at least superficially, I hope we can agree) similar to that of when the Mersenne number $2^n-1$ is prime.
For $q=2^n$ a power of $2$ the question boils down to asking when $2^n(2^{2n}-1)-1$ is prime.
There are similar formulas for the other simple groups of Lie type, and I'll bet money no one in the world knows whether infinitely many of the relevant numbers are prime.
answered Dec 8, 2010 at 11:39
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2$\begingroup$ I'd just add that there are infinitely many known infinite families of simple groups of Lie type, indexed by prime powers and having explicit order formulas. Each such family yields an infinite sequence of group orders; not random, but not easily analyzed for patterns. Moreover, the question asked has (as far as I can see) no connection with the finite group theory. It's not a nonsensical question but seems almost impossibly difficult to answer one way or the other. $\endgroup$ Commented Dec 8, 2010 at 12:57
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Standard heuristics (together with orders from the list of finite simple groups ) suggest that by far the most common orders of the form $p+1$ for $p$ prime will come from $A_1(q) = PSL_2(\mathbb{F}_q)$, of order $\frac{q^3-q}{2}$, as $q$ ranges over odd primes (or prime powers, if you want an additional small contribution). In particular, for large $N$, one should expect roughly $\frac{\sqrt[3]{N}}{(\log N)^\alpha}$ satisfactory numbers $p+1$ less than $N$, for some fixed positive number $\alpha$, and this sequence of numbers certainly grows without bound.
As others have remarked, the question of proving that the set of suitable primes satisfies the rough asymptotics I gave above, or even proving that the set is infinite, seems to be beyond current technology. For example, we still don't know if there are infinitely many primes of the form $n^2+1$ for $n$ an integer.
answered Dec 8, 2010 at 15:21
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5$\begingroup$ The following sporadic simple groups have order $p+1$ for $p$ a prime: $M_{11}$, $HS$, $M_{23}$, $O'N$, $Fi_{22}$, $J_4$. $\endgroup$– S. Carnahan ♦Commented Dec 8, 2010 at 15:59
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As Alon remarks, it is extremely hard to find such groups even between groups of one (most known) series $A_n$.
answered Dec 8, 2010 at 11:17