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According to a result of Higman and Sims (which I learned about in this paper of Poonen's) the typical p-group is 3-step nilpotent of a particular form. In particular the typical group is a 3-step nilpotent 2-group of a particular form. By typical here I mean that eventually the number of these groups dominate. Is anything known about what the typical non-solvable group looks like? Probably some sort of modification of PSL_2(F_p)? |
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It seems like, if solvable groups dominate everything else, then groups with a single $A_5$ factor and all other factors cyclic should likewise dominate every other type of nonsolvable group. |
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Even if you are asking about a "typical" group of order $\leq n$, the answer is nobody knows, so this $A_5$ idea for a much more delicate question is little more than a pure speculation. If I understand the state of art correctly, it is completely open whether solvable groups are a majority. The data, of course, supports a stronger conjecture: that asymptotically almost all groups are $2$-groups. What is known now, is a number of good asymptotic estimates, notably a Pyber's paper in the Annals (1993), getting a tight asymptotic for the log of the number of groups of order $\le n$ (he gets an upper bound matching the Higman-Sims lower bound). For more on this, read up this terrific recent monograph. |
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It seems plausible that the typical nonsolvable finite group has a $PSL_2(F_p)$ composition factor, yes. I guess you could try to prove this by looking at the classification of finite simple groups and showing that the PSL_2's dominate everything else. I don't have a sense for how much work it would be to carry this out (assuming, of course, that it is true): it seems sort of like classical questions in analytic number theory but with Jordan-Holder factors replacing the primes. |
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I don't have reasonable access to internet at the moment, but I will edit this and add references when I can. There is an old paper called "Almost Every group is solvable" where one considers a finite group and its jordan holder decomposition. Ignoring all the factors which are cyclic groups, one multiplies the size of the remaining factors and divides by the size of the group. This gives a number which is <=1, and is equal to 1 only for nonabelian simple groups. They show in that paper that the "average" over all groups of this statistic is 0. In other words, most simple composition factors are cyclic abelian groups. I do not know enough about PSL_2(F_p) to say whether this fits the bill (in other words, as p increases, what is the chart of this statistic). |
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