I am trying to compute an ordinary character of $PSL(2,p)$ and it seems like a have flaw in my argument that I can not find it. Can you please help me out. I am concerned with the case where $p\equiv \pm1 \pmod 8$

G:= $PSL(2,p)$
lGl= $\frac{(p-1)p(p+1)}{2}$
$S_4$ exists in G as a maximal subgroup. All elements of G of order 2,3,4 are conjugate to each other in G. There exists $\frac{p(p+1)}{2}$ elements of order 2, p(p+1) elements of order 3 and 4 each.
I am trying to compute the character of permutation representation for $G/S_4$.

Let $g_i$ be the representatives of the cosets for $G/S_4$. Then for our character $\chi$(g)=#$g_i$ such that g $\in {S_4}^{g_i}$. Let $a$,$b$,$c$ $\in$ G such that $a^2=b^3=c^4=1$ and e $\in$ G so that e does not have order 2 or 3 or 4. Then obviously $\chi$(1)=$\frac{lGl}{24}$ and $\chi$(e)=0. I claim that $\chi$(a)=p-1 and $\chi$(b)=$\chi$(c)=$\frac{p-1}{2}$ because, lets say for $a$, for an element of $S_4$ there are p-1 elements that conjugate it to $a$ and there are 9 elements of order 2 in $S_4$. Hence in total 9(p-1) conjugates of $S_4$ include $a$ however we are concerned about the number of such representatives for the cosets $G/S_4$. Which is 2(p-1) because there are 2 inner conjugacy classes with elements of order 2 in $S_4$. The same logic follows for b and c and the character looks like
$$ \begin{array}{c|lcr} ch & \text{1} & \text{a} & \text{b} & \text{c} & \text{e} \\ \hline \chi & \frac{(p-1)p(p+1)}{48} & 2(p-1) & \frac{p-1}{2} & \frac{p-1}{2} &0 \\ \end{array} $$
where $e$ is a generic element of order not 2 or 3 or 4. I think the solution is incorrect because scalar product of the character with itself is supposed to be an integer. But I find $<G/S_4, G/S_4>=\frac{lGl}{24^2} + 5(p-1)$ which is not an integer for p=23

  • $\begingroup$ Are you sure about the number of elements of order $3$ when $3|(p+1)$? $\endgroup$ – ancientmathematician Jul 12 '18 at 6:42
  • $\begingroup$ Conjugacy classes of $sl(2,p)$ is determined by the trace expect when the trace is $\pm$ 2. An element of $psl(2,p)$ has order 3 if its preimage in $sl(2,p)$ has trace $\pm 1$. There are p(p+1) matrices with trace 1 in $sl(2,p)$ and the same amount for trace -1 as well. When you quotient it with the center, I think both conjugacy classes collapse into one conjugacy class with p(p+1) elements. This was my line of thinking. $\endgroup$ – Baran Zadeoglu Jul 12 '18 at 7:45
  • $\begingroup$ Take for example $p=11$; an element of order $3$ is centralised precisely by a cyclic subgroup of order $6$ and so there are $660/6=110$ ($=p(p-1)$) conjugates of the element of order $3$. $\endgroup$ – ancientmathematician Jul 12 '18 at 8:08
  • $\begingroup$ you are probably right. 660 is not divisible by 132. But then where is the flaw in my argument. Also how do we know that an element of order 3 is centralized by a cyclic subgroup of order 6? $\endgroup$ – Baran Zadeoglu Jul 12 '18 at 8:32
  • $\begingroup$ Oh wait. 660 is divisible by 132 and I rechecked the cycle lengths in $sl(2,p)$ and they are correct as well. There are supposedly p(p+1) elements with trace 1 and p(p+1) elements with trace -1. $\endgroup$ – Baran Zadeoglu Jul 12 '18 at 8:46

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