Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In a misguided attempt at answering a question on divisibility of simple group orders, I looked at $\newcommand{\PSL}{\operatorname{PSL}}v_p(|\PSL(p,q)|)$ which went smooth enough for $p=2$ and $p=3$, but which just confused me for $p=5$.

Apparently if $n=2k+1$ is odd, then there are eight residue classes mod $5^{n+1}$ such that $v_5(|\PSL(5,q)|) = n$. The residue classes seem to have some structure, but I'm having trouble seeing what it is.

Here is the early data:

  • k | residues of $q$ mod $5^{2k+2}$ such that $v_5(|\PSL(5,q)|) = 2k+1$
  • 0 | 2, 3, 8, 12, 13, 17, 22, 23
  • 1 | 57, 68, 193, 307, 318, 432, 557, 568
  • 2 | 2057, 4193, 5182, 7318, 8307, 10443, 11432, 13568
  • 3 | 32318, 45807, 123932, 188568, 202057, 266693, 344818, 358307
  • 4 | 280182, 1672943, 2233307, 4186432, 5579193, 7532318, 8092682, 9485443

For number theorists, we need to solve: $$v_5(q^2-1) + v_5(q^3-1) + v_5(q^4-1) + v_5(q^5-1) - \min(1,v_5(q-1)) = n$$ for $q \mod 5^{n+2}$ but apparently this is too many cases for me.

I'd be interested in a formula for even one of the eight classes, as I think the others will follow . If this is trivial, I'd also be interested in hearing why the even case is so different (and erratic).

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.