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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 $M$?

the examples $660$ and $672$ show $D \leq 12$. the famous theorem of feit-thompson implies $D>1$.

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up vote 12 down vote accepted

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.

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That's right -- it's an elementary exercise that a simple group cannot have order twice an odd number: a permutation representation argument shows that a group of order 2 mod 4 has a subgroup of index 2. – Pete L. Clark Dec 31 '09 at 14:09
how do you arrive at simple groups? – Martin Brandenburg Dec 31 '09 at 14:43
A group is nonsolvable iff it has at least one nonabelian simple group as a composition factor. Moreover, if G is a finite nonabelian simple group of order a, then for all positive integers x, ax is the order of a nonsolvable group: G x Z_x. – Pete L. Clark Dec 31 '09 at 14:47
Pete is also implicitly using the odd order theorem, to know that the nonabelian simple factor cannot be odd. – David Speyer Dec 31 '09 at 16:22
To avoid confusion: when Ben writes "the Suzuki group", he means a member of the family of Suzuki groups of Lie type, not the Suzuki sporadic group Suz. – Bjorn Poonen Dec 31 '09 at 19:48

The encyclopedia of integer sequences gives the following criteria for a number being a non-solvable number:

A positive integer n is a non-solvable number if and only if it is a multiple of any of the following numbers:

a) $2^p (2^{2p}-1)$, p any prime.

b) $3^p (3^{2p}-1)/2$, p odd prime.

c) $p(p^2-1)/2$, p prime greater than 3 and congruent to 2 or 3 mod 5

d) $2^4 3^3 13$

e) $2^{2p}(2^{2p}+1)(2^p-1)$, p odd prime.

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.

In fact, they're all divisible by 12 except those generated by (e), which are all divisible by 20.

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 $29120 = 2^6 5^1 7^1 13^1$.

So we just need to find multiples of 29120 and 25308 which differ by 4. From the Euclidean algorithm, $29120 \cdot 2483 = 72304960$ and $25308 \cdot 2857 = 72304956$.

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.

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again, I wish I could accept two answers. thank you both :-) – Martin Brandenburg Dec 31 '09 at 19:10
Ben's answer is better, in the sense that he actually knows group theory. I couldn't have done this without reading his answer. – Michael Lugo Dec 31 '09 at 19:46

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