I would like to know if there is a way to generate infinitely many examples of numbers $n$ such that there is only one non-solvable group of order $n$

Question

It is known that, up to isomorphism, $A_5$ is the only non-solvable group of order 60. I would like to know if there is a way to generate infinitely many examples of numbers $n$ such that there is only one non-solvable group of order $n$. My candidate was $n=60p$, where $p$ is a prime such that $(60, p)=1$. However, for $p=11$, I saw that there are 2 non-solvable groups of order 660.

Answer 1

For a group of order $n$ being non-solvable, there needs to be a (nonabelian) simple composition factor of order $s|n$. But there always will be a direct product of the simple group of order $s$ and a group of order $d=n/s$. Thus, to have only one nonsolvable group, we must show that the only such group of order $n$ can be this direct product. Thus

• There must be only one group of order $d=n/s$. $d$ prime satisfies, but there are some other orders, such as $15$, they must be squarefree.
• The direct product must be the only extension possible between the group of order $d$ and the simple group $S$ of order $s$. This is guaranteed if $d$ is coprime to $s$ and $(d,\mbox{Aut(S)})=1$.
• We want no other cases happen to work. If $d$ is prime and $d>s$, the $d$-Sylow subgroup must be normal, and the extension splits, i.e. $C_d\rtimes S$, but $S$ cannot act on $C_d$ and thus the group is a direct product.

These are not all cases, one can obviously combine $S\times C_p$ and $S\times C_q$ to a unique $S\times (C_p\times C_q)$ if $(p,q-1)=(q,p-1)=1$.

Answer 2

With the data available at OEIS one can determine all positive integers $n\le 500$, where $g(n)=1$, i.e., the number of non-solvable groups up to isomorphism is equal to $1$. In fact, we have, for the nonzero values up to $500$ $$ g(60)=1,\; g(120)=3,\; g(168)=1,\; g(180)=1,\; g(240)=8, $$

$$ g(300)=48, \;g(336)=3,\; g(360)=6,\; g(420)=1,\; g(480)=26. $$

A general criterion for such numbers $n$ seems not to be known. The MO post cited above has some more information. 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.

Note that the question, to find the numbers $n$, such that there is exactly one solvable group, is known. For every $n$ with $\gcd(n,\phi(n))=1$ there is only one group, necessarily cyclic and hence solvable.