Skip to main content
MathOverflow

Return to Answer

corrected minor grammar thing
Source Link
Bobby Grizzard
Bobby Grizzard
  • 1.5k
  • 1
  • 10
  • 21

Not anything resembling an answer, but too long for a comment. This question seems very hard to me, from a purely diophantine perspective. We have formulas for the orders of all the finite simple groups. Let's take just the example of $PSp(4,q)$, where $q$ is an odd (for convenience) prime power. I believe the order of this group is $$f(q) = \frac{1}{2}q^4(q^4-1)(q^2-1).$$ If we fix some $k$ such as 2 or 3 and ask for the order of the group to equal $n^k$, then a group in this family of order $n^k$ corresponds to an integer point on the curve $$n^k = f(q)$$... and you impose the additional constraint that $q$ is a positive prime power. It sounds like the authors of the paper linked in the MSE post you reference have found some solutions to this equation (and I'm guessing this curve has genus zero if they are conjecturing it has infinitely many), but proving you've found them all sounds challenging. And then to do this for even all groups of the form $PSp(2m,q)$, let alone all FSG's, seems very hard to me. It wouldn't surprise me if you ran across a family of curves which was quite intractable.

Not anything resembling an answer, but too long for a comment. This question seems very hard to me, from a purely diophantine perspective. We have formulas for the orders of all the finite simple groups. Let's take just the example of $PSp(4,q)$, where $q$ is an odd (for convenience) prime power. I believe the order of this group is $$f(q) = \frac{1}{2}q^4(q^4-1)(q^2-1).$$ If fix some $k$ such as 2 or 3 and ask for the order of the group to equal $n^k$, then a group in this family of order $n^k$ corresponds to an integer point on the curve $$n^k = f(q)$$... and you impose the additional constraint that $q$ is a positive prime power. It sounds like the authors of the paper linked in the MSE post you reference have found some solutions to this equation (and I'm guessing this curve has genus zero if they are conjecturing it has infinitely many), but proving you've found them all sounds challenging. And then to do this for even all groups of the form $PSp(2m,q)$, let alone all FSG's, seems very hard to me. It wouldn't surprise me if you ran across a family of curves which was quite intractable.

Not anything resembling an answer, but too long for a comment. This question seems very hard to me, from a purely diophantine perspective. We have formulas for the orders of all the finite simple groups. Let's take just the example of $PSp(4,q)$, where $q$ is an odd (for convenience) prime power. I believe the order of this group is $$f(q) = \frac{1}{2}q^4(q^4-1)(q^2-1).$$ If we fix some $k$ such as 2 or 3 and ask for the order of the group to equal $n^k$, then a group in this family of order $n^k$ corresponds to an integer point on the curve $$n^k = f(q)$$... and you impose the additional constraint that $q$ is a positive prime power. It sounds like the authors of the paper linked in the MSE post you reference have found some solutions to this equation (and I'm guessing this curve has genus zero if they are conjecturing it has infinitely many), but proving you've found them all sounds challenging. And then to do this for even all groups of the form $PSp(2m,q)$, let alone all FSG's, seems very hard to me. It wouldn't surprise me if you ran across a family of curves which was quite intractable.

Source Link
Bobby Grizzard
Bobby Grizzard
  • 1.5k
  • 1
  • 10
  • 21

Not anything resembling an answer, but too long for a comment. This question seems very hard to me, from a purely diophantine perspective. We have formulas for the orders of all the finite simple groups. Let's take just the example of $PSp(4,q)$, where $q$ is an odd (for convenience) prime power. I believe the order of this group is $$f(q) = \frac{1}{2}q^4(q^4-1)(q^2-1).$$ If fix some $k$ such as 2 or 3 and ask for the order of the group to equal $n^k$, then a group in this family of order $n^k$ corresponds to an integer point on the curve $$n^k = f(q)$$... and you impose the additional constraint that $q$ is a positive prime power. It sounds like the authors of the paper linked in the MSE post you reference have found some solutions to this equation (and I'm guessing this curve has genus zero if they are conjecturing it has infinitely many), but proving you've found them all sounds challenging. And then to do this for even all groups of the form $PSp(2m,q)$, let alone all FSG's, seems very hard to me. It wouldn't surprise me if you ran across a family of curves which was quite intractable.