Type of Conjugacy Class of $SL(2,5)$ Including $\begin{pmatrix} 0 & 3\\ 3 & 3\end{pmatrix}$ – Linear Algebra, Group Theory

group-theorylinear algebramatrices

I'm trying out a bunch of calculations to better understand this (where $\mathcal{F}=\Bbb F_q$ is finite field, $q$ is a power of a prime):

We can describe the matrix representatives of conjugacy classes in $\mathcal{S}=SL(2,\mathcal{F})$ by four families of types ([6]):

  1. $\begin{pmatrix} r & 0 \\ 0 & r\end{pmatrix}$, where $r\in\mathcal{F}$ and $r^2=1$.
  2. $\begin{pmatrix} r & 0 \\ 0 & s\end{pmatrix}$, where $r,s\in\mathcal{F}$ and $rs=1$.
  3. $\begin{pmatrix} s & u \\ 0 & s\end{pmatrix}$, where $s\in\mathcal{F}, s^2=1$ and $u$ is either $1$ or a non-square element of $\mathcal{F}$, i.e. $u\in \mathcal{F}\setminus\{ x^2: x\in \mathcal{F}\}$.
  4. $\begin{pmatrix} 0 & 1 \\ -1 & w\end{pmatrix}$, where $w=r+r^q$ and $1=r^{1+q}$ for some $r\in \mathcal{E}\setminus\mathcal{F}$, where $\mathcal{E}$ is a quadratic extension of $\mathcal{F}$.

That is, any conjugacy class $A^\mathcal{S}$ of $\mathcal S$ must contain one of the above matrices.

This is from Harris et al.'s, "On Conjugacy Classes of $SL_2(q)$."

The Question:

Which type of conjugacy class of $SL(2,5)$ (i.e, 1., 2., 3., or 4.) includes

$$\begin{pmatrix}
0 & 3\\
3 & 3
\end{pmatrix}?$$

The matrix is, of course, taken modulo five.

Context:

I used GAP to compute the conjugacy classes, like so:

gap> ConjugacyClasses(SL(2,5));
[ [ [ Z(5)^0, 0*Z(5) ], [ 0*Z(5), Z(5)^0 ] ]^G,
[ [ 0*Z(5), Z(5)^2 ], [ Z(5)^0, Z(5)^3 ] ]^G,
[ [ 0*Z(5), Z(5)^3 ], [ Z(5)^3, Z(5)^3 ] ]^G,
[ [ Z(5)^2, 0*Z(5) ], [ 0*Z(5), Z(5)^2 ] ]^G,
[ [ 0*Z(5), Z(5)^2 ], [ Z(5)^0, Z(5) ] ]^G,
[ [ 0*Z(5), Z(5)^3 ], [ Z(5)^3, Z(5) ] ]^G,
[ [ 0*Z(5), Z(5)^2 ], [ Z(5)^0, Z(5)^2 ] ]^G,
[ [ 0*Z(5), Z(5)^2 ], [ Z(5)^0, Z(5)^0 ] ]^G,
[ [ Z(5)^3, 0*Z(5) ], [ 0*Z(5), Z(5) ] ]^G ]
gap> x:=[[0*Z(5), Z(5)^3],[Z(5)^3, Z(5)^3]];
[ [ 0*Z(5), Z(5)^3 ], [ Z(5)^3, Z(5)^3 ] ]
gap> Display(x);
 . 3
 3 3
gap>

I think I have all the others classified; respectively:

$$\begin{array}{|c|c|}
\text{Matrix} & \text{1, 2, 3, or 4 as above}\\
\hline
I & 1.\\
\begin{pmatrix}
0 & -1\\
1 & 3
\end{pmatrix} & 4.\\
\begin{pmatrix}
0 & 3\\
3 & 3
\end{pmatrix} & ?\\
\begin{pmatrix}
-1 & 0\\
0 & -1
\end{pmatrix} & 1.\\
\begin{pmatrix}
0 & -1\\
1 & 2
\end{pmatrix} & 4.\\
\begin{pmatrix}
0 & 3\\
3 & 2
\end{pmatrix} & 3?\\
\begin{pmatrix}
0 & -1\\
1 & -1
\end{pmatrix} & 4.\\
\begin{pmatrix}
0 & -1\\
1 & 1
\end{pmatrix} & 4.\\
\begin{pmatrix}
3 & 0\\
0 & 2
\end{pmatrix} & 2.
\end{array}$$

As you can see, I can't use the process of elimination (as every type is accounted for above). That's kind of cheating anyway.

I doubt it's 3., since $3^2\pmod 5\equiv -1\not\equiv 1$.

I get the feeling I'm missing something obvious.

There'll be at least one way to use GAP for this but I'm interested in computing it by hand, so as to enable understanding.

Edit: I just realised I miscalculated for

$$\begin{pmatrix}
0 & 3\\
3 & 2
\end{pmatrix}.$$

Hence "3?" in the table.

Best Answer

The characteristic polynomial is $x^2-3x-9 = x^2+2x+1=(x+1)^2$.

Since it is not central, it must have type 3 with $s= -1$

It is not too hard to triangularize it by hand, but a quick computer calculation shows that we can take $u=2$ or 3 .

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