Determining the distinct conjugacy classes by obtaining Rational Cannonical forms using minimal polynomial

Question

Consider the mapping $\phi: \mathbb{Z}_3 \rightarrow Aut(\mathbb{Z}_p \times \mathbb{Z}_p)$, where $p>3$ is a prime. The action of $s \in \mathbb{Z}_3$ on $\mathbb{Z}_p \times \mathbb{Z}_p$ can be defined using a linear transformation $T$.

Let $\{s,t\}$ generate $G = (\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes_{\phi} \mathbb{Z}_3$, where $|s|=3$ and $|t|=p$.

$T: \mathbb{Z}_p \times \mathbb{Z}_p \rightarrow \mathbb{Z}_p \times \mathbb{Z}_p$, such that $T(v)=s^{-1}vs$. Let $m(x)$ be the minimal polynomial of $T$ and $u = T(t)=s^{-1}ts$.

1. $|s|=3$ and $T^3=I$. Hence, $m(x)$ divides $x^3-1=(x-1)(x^2+x+1)$.
2. Since, $|[G,G]|=p^2$, 1 is not an eigen value of $T$ ($[G,G]$ is the commutator subgroup of $G$).
3. Since, $<s,t>=G$, $u=T(t) \notin <t>$, so the minimal polynomial of $T$ has degree 2 (and $\{t,u\}$ is a basis of $\mathbb{Z}_p \times \mathbb{Z}_p$). Therefore, it can be concluded that the minimal polynomial of $T$ is $x^2+x+1$ and the Rational Cannonical form of $T$ can be written as, $$\begin{pmatrix}0&-1\\1&-1\end{pmatrix}$$ The order $3$ subgroup of $GL(2,p)$ can be generated by $$\begin{pmatrix}0&-1\\1&-1\end{pmatrix}$$

Similarly, considering the group $G^\prime = (\mathbb{Z}_{11} \times \mathbb{Z}_{11}) \rtimes_{\phi^\prime} \mathbb{Z}_5$ (note that 5|(11-1)), let $M^\prime =\{s^\prime,t^\prime\}$, where $|s^\prime|=5, |t^\prime|=11$ be the minimal generating set and an action of $s^\prime$ on $\mathbb{Z}_{11} \times \mathbb{Z}_{11}$ be defined following the above definition.

$T^\prime: \mathbb{Z}_{11} \times \mathbb{Z}_{11} \rightarrow \mathbb{Z}_{11} \times \mathbb{Z}_{11}$, such that $T^\prime(v)={s^\prime}^{-1}v s^\prime$. Let $m^\prime(x)$ be the minimal polynomial of $T^\prime$ and $u^\prime = T(t^\prime)={s^\prime}^{-1} t^\prime s^\prime$.

Then, since $|s^\prime|=5$, ${T^\prime}^5=I$ and the minimal polynomial $m^\prime(x)$ divides $x^5-1$.

$x^5-1=(x-1)(x^4+x^3+x^2+x+1)$. How to continue from here to determine the possible rational canonical forms representing distinct conjugacy classes of order 5 in GL(2,11)?

Answer 1

Over ${\mathbb F}_{11}$ we have $$x^5-1 = (x-1)(x-3)(x-4)(x-5)(x-9).$$

The minimal polynomial can be any factor of this of degree $1$ or $2$, except for $x-1$ which would correspond to trivial action, so would not give rise to an element of order $5$ in ${\rm GL}(2,11)$.

There are $14$ such factors, four of degree $1$ and $10$ of degree $2$, and this is the number of conjugacy classes of elements of order $5$ in ${\rm GL}(2,11)$.

(But note that in the corresponding semidirect products $G$, in the four cases with minimal polynomial $(x-1)(x-a)$, we have $|[G,G]|=11$, not $11^2$ as stated in the post.)