distribution of non-solvable group orders - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T10:32:54Z http://mathoverflow.net/feeds/question/10267 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10267/distribution-of-non-solvable-group-orders distribution of non-solvable group orders Martin Brandenburg 2009-12-31T13:29:19Z 2010-01-03T03:04:08Z <p>let $M$ be the set of natural numbers such that there is a group of this order, which is not solvable. what is the minimal distance $D$ of two numbers in $M$?</p> <p>the examples $660$ and $672$ show $D \leq 12$. the famous theorem of feit-thompson implies $D>1$.</p> http://mathoverflow.net/questions/10267/distribution-of-non-solvable-group-orders/10269#10269 Answer by Ben Webster for distribution of non-solvable group orders Ben Webster 2009-12-31T13:41:17Z 2010-01-03T03:04:08Z <p>By the Euclidean algorithm, the answer is the gcd of all orders of all non-abelian finite simple groups. I believe that this is 4 (looking at the groups listed in Wikipedia, one can see that it is at most 4 since once can get down to 12 on the tables of low order groups, and the Suzuki groups have order not divisible by 3). My recollection is that a finite simple group actually cannot have cyclic 2-Sylow, and thus must have order divisible by 4.</p> http://mathoverflow.net/questions/10267/distribution-of-non-solvable-group-orders/10280#10280 Answer by Michael Lugo for distribution of non-solvable group orders Michael Lugo 2009-12-31T15:23:08Z 2009-12-31T15:23:08Z <p>The <a href="http://www.research.att.com/~njas/sequences/A056866" rel="nofollow">encyclopedia of integer sequences</a> gives the following criteria for a number being a non-solvable number:</p> <p>A positive integer n is a non-solvable number if and only if it is a multiple of any of the following numbers: </p> <p>a) $2^p (2^{2p}-1)$, p any prime. </p> <p>b) $3^p (3^{2p}-1)/2$, p odd prime. </p> <p>c) $p(p^2-1)/2$, p prime greater than 3 and congruent to 2 or 3 mod 5</p> <p>d) $2^4 3^3 13$</p> <p>e) $2^{2p}(2^{2p}+1)(2^p-1)$, p odd prime.</p> <p>It's not hard to check that all these orders are divisible by 4, so there will never be two non-solvable numbers differing by less than 4.</p> <p>In fact, they're all divisible by 12 except those generated by (e), which are all divisible by 20.</p> <p>So, for example, all numbers of the form 29120n are nonsolvable, since $29120 = 2^6 (2^6+1) (2^3-1)$. And all numbers of the form 25308n are nonsolvable, since $25308 = 37(37^2-1)/2$. We have the prime factorizations $25308 = 2^2 3^2 19^1 37^1$ and $29120 = 2^6 5^1 7^1 13^1$. </p> <p>So we just need to find multiples of 29120 and 25308 which differ by 4. From the Euclidean algorithm, $29120 \cdot 2483 = 72304960$ and $25308 \cdot 2857 = 72304956$.</p> <p>I haven't searched exhaustively, so it's possible that there's a smaller pair of non-solvable numbers that differ by 4; I chose 25308 and 29120 by just looking at the prime factorizations of the numbers generated by (a) through (e) until I found two that had gcd 4.</p>