One of the problems that has come up during my research concerns $K\_4$-simple groups (simple groups with $4$ prime divisors). The only (potentially) infinite family of groups satisfying this condition is $PSL(2,q)$, and I was interested in exactly what values of $q$ satisfied this condition. After writing a quick Python program, I got this output (where each ordered pair denotes $(q, |PSL(2,q)|)$:


$$[(11, 660), (13, 1092), (16, 4080), (19, 3420), (23, 6072), (25, 7800), (27, 9828), (31, 14880), (32, 32736), (37, 25308), (47, 51888), (49, 58800), (53, 74412), (73, 194472), (81, 265680), (97, 456288), (107, 612468), (127, 1024128), (128, 2097024), (163, 2165292), (193, 3594432), (243, 7174332), (257, 8487168), (383, 28090752), (487, 57750408), (577, 96049728), (863, 321367392), (1153, 766403712), (2187, 5230175508), (2593, 8717209632), (2917, 12410213148), (4373, 41812719372), (8192, 549755805696), (8747, 334616519988), (131072, 2251799813554176), ...$$


There seems to be a huge gap in $q$ after $8747$, going immediately to $2^{17} = 131072$ and then skipping another couple hundred thounsand or so. The condition on $q$ is that it is a prime power satisfying $q(q^2-1) = 2^{\alpha\_1}3^{\alpha\_2}p^{\alpha\_3}r^{\alpha\_4}$ for primes $r > p > 3$. Yet, I found it quite surprising that these gaps exist. Does anyone know why? Are there arbitrarily large gaps in values of $q$?


This is more number-theoretic than group-theoretic, but I am definitely interested in that aspect as well. Thanks in advance!



Sparsity of q in groups PSL(2,q) that are K_4-simple
Let me expand my earlier comment to a partial answer of "why sparsity?". It is impossible for $|{\rm PSL}(2,p^{m})|$ to have four or fewer different prime factors when
$p$ is an odd prime which is neither Fermat nor Mersenne and $m > 1$ is an integer. In fact, if $m >1$ is not a power of $2$, we will see that
whenever $p >3$ is prime, then $|{\rm PSL}(2,p^{m})|$ has at least $5$ different prime divisors.


(These two facts "explain" why the non-squarefree $q$ appearing in your Python output are all either powers of $2$ or powers of Fermat or Mersenne primes, and why we only have $q$ of the form $p^{2n+1}$ ( for positive $n$) appearing for $p = 2$ or $p =3).$


If $m$ is itself even, then $p^{2m}-1$ is divisible by $p^{4}-1$. It is easy to check that $\frac{p-1}{2}, \frac{p+1}{2}$ and $\frac{p^{2}+1}{2}$ are pairwise coprime. Also $\frac{p^{2}+1}{2}$ is odd, while one of $\frac{p \pm 1}{2}$ is even, and the other is odd. Hence if the odd prime $p$ is neither a Fermat prime nor a Mersenne prime, then $p^{4}-1$ is even, and has at least three different odd prime factors, so has $4$ or more different prime factors (none of which is $p$). Thus
$|{\rm PSL}(2,p^{2n})|$ has $5$ or more different prime factors whenever $p$ is an odd prime which is neither Fermat nor Mersenne, and $n$ is any positive integer.


If $m$ is not a power of $2$, then $m$ is divisible by some odd prime $r$. We note below that that $p^{2r}-1$ has four or more prime factors whenever $r$ is an odd prime and $p$ is a prime greater than $3$.


For of the four integers $\frac{p-1}{2}, \frac{p+1}{2}, \frac{p^{r}-1}{p-1}$ and $\frac{p^{r}+1}{p+1},$ exactly one is even. Since $p >3$, the product $\frac{p+1}{2}\frac{p-1}{2}$ has at least two different prime factors, since $p-1$ and $p+1$ can't both be powers of $2$ for any prime $p$ greater than $3$.


Note also that Zsygmondy's Lemma tells us that there are primes $s$ and $t$ such that $p+s\mathbb{Z}$ has multiplicative order $r$ in the units of $(\mathbb{Z}/s\mathbb{Z})^{\times}$ and $p+t\mathbb{Z}$ has multiplicative order $2r$ in the units of $(\mathbb{Z}/t\mathbb{Z})^{\times}$. Then $s$ and $t$ are different (and both odd) and neither of them divides $p^{2}-1.$ Hence $p^{2r}-1$
has at least $4$ different prime divisors, and $|{\rm PSL}(2,p^{m})|$ has at least five prime divisors.


