*American Mathematical Monthly*. The paper gives a classification of the groups of order sixteen without appealing to the general theory of

*p*-groups.

Anyway, towards the end of the paper things are put into the historical context of the theory of group extensions, and we learn that:

The groups that can be obtained from the trivial group by iteratively building cyclic extensions are called solvable. Statistically speaking “most” groups are solvable. For instance, all groups of order less than sixty are solvable.Of course, groups of order less than sixty make up a negligible portion of all groups; it's certainly possible that all groups of order less than sixty are solvable, but most groups of order

*n*are

*not*solvable when

*n*is large enough. There is the Feit-Thompson theorem, which states that every group of odd order is solvable, but numerical data for

*n*up to 2015 shows that there seem to be more groups of even order than odd order. This isn't surprising; the number of groups of a given finite order depends strongly on the exponents of the primes in its factorization. Nor is this special to 2 -- in general numbers with lots of prime factors are the order of lots of groups.

It looks like the statement is true, though, in any reasonable sense; from Sloane's Encyclopedia, there are only 2240 integers

*n*below 10

^{5}such that there exists a non-solvable group of order

*n*. In fact, it looks like the sequence of such numbers has roughly constant density, of about 0.0224. That should be provable or disprovable from the note given in the Encyclopedia, which gives an easy way to find such numbers. The statement that most

*numbers*are solvable (where a number is solvable if all groups of that order are solvable) is even stronger than the statement that most groups are solvable.

But talking about a ``randomly chosen group'' seems a bit silly, anyway; the groups of a given order don't strike me as the sort of set one picks things from at random. Choosing a random

*element*of a group seems like a much more natural operation. (Feel free to correct me if I'm wrong!)

## 4 comments:

A couple ways the statement could make sense.

If we let f(n) denote the number of solvable groups of order n, and g(n) denote the total number of groups of order n, the statement could mean:

1. f(n)/g(n) tends to 1 as n tends to infinity (a group chosen randomly among the groups of a fixed large order tends to be solvable).

2. (The sum of f(i) from i=1 to n)/(The sum of g(i) from i=1 to n) tends to 1 as n tends to infinity (a group chosen randomly among the groups of at most a given fixed large order tends to be solvable).

3.

Hmmm… I'm not familiar enough with group theory to understand the definition of solvable group you've given here, but in my abstract algebra class we've briefly studied solvable groups in the context of Galois theory and the idea of polynomials being solvable by radicals. (I assume the definition I learned is equivalent.) In this sense it might make sense to talk about picking a group at random--if I pick a polynomial at random and want to know if it is solvable by radicals, I'll look at its Galois group and determine whether that is solvable.

Susan, here's the connection: consider what it looks like to take an nth root in the complex plane. You divide the plane up into n slices, and rotating the plane one nth of a turn is a symmetry of the whole situation. But rotations by nths of turns make up the cyclic group Cn! So a step in a solution by radicals that involves taking an nth root is tied to building up the symmetry group of your polynomial by a cyclic group of order n.

I am told by a local expert that "almost all" groups are solvable, in fact, nilpotent of class at most two, meaning solvable in at most two steps. So depending on how exactly this density is counted (I can think of at least four ways), I would say that a "random" group has 100% probability of being solvable. JV

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