Let $SL(2,3)=SL(2,\mathbb{F}_3)$. Prove that there is no element of order $8$ in $SL(2,3)$.
My attempt:
Let $A$ be a matrix in $SL(2,3)$.
Then $A=U X U^{-1} $ for some invertible $U$ where $X$ is the diagonal matrix of eigenvalues of $A$. Then $A^n=UX^nU^{-1}$ and then we take the cases for different eigenvalues of $A$ but this does not seem to work as it implies every matrix in $SL(2,3)$ has order at most $2$.
Thanks in advance
Possibly the way that involves the least working with specific matrices and the most group theory:
Lemma1: The $p$-Sylow in $SL_n(\mathbb{F}_q)$ is not normal when $q$ is a power of $p$.
Proof: It is straightforward to check that either the set of upper- or lower triangular matrices with $1$'s on the diagonal is a $p$-Sylow, and since these are distinct, a $p$-Sylow is not normal.
Lemma2: If $|G| = 2^rm$ with $m$ odd and the $2$-Sylow is cyclic, then $G$ has a normal subgroup of order $m$.
Proof: Consider the action of $G$ on itself by left translation as a map to $S_{|G|}$ and not that a generator for a $2$-Sylow corresponds to $m$ disjoint $2^r$-cycles and is hence an odd permutations. This means that the image of $G$ is not contained in $A_{|G|}$ and thus the preimage of $A_{|G|}$ is a subgroup of $G$ of index $2$. Now the claim follows by induction on $r$ (since a normal subgroup of order $m$ is a Hall subgroup and thus characteristic).
Finally, we see that there is no element of order $8$ in $SL_2(\mathbb{F}_3)$ since this would mean it had a cyclic $2$-Sylow and by Lemma2 thus a normal subgroup of order $3$, i.e. a normal $3$-Sylow, which contradicts Lemma1.
Let $A$ be a matrix and $q(x)\in\Bbb F_3[x]$ its monic minimal polynomial. Since its characteristic polynomial $p(x)=\det(A-xI)$ has degree $2$, it holds $\deg q\le 2$. Notice that $p(x)$ is monic as well.
A matrix in $SL(2,\Bbb F_3)$ has order $8$ if and only if $q$ divides $x^8-1$, but it does not divide $x^4-1$.
Since the prime factorisation in $\Bbb F_3[x]$ of $x^8-1$ is \begin{align}x^8-1&=(x^4-1)(x^4+1)=\\&=\left[(x-1)(x+1)(x^2+1)\right]\left[(x^2+x-1)(x^2-x-1)\right]=\\\end{align} it must hold $q(x)=x^2-x-1$ or $q(x)=x^2+x-1$.
Since both $p(x)$ and $q(x)$ are monic, it must hold $p(x)=q(x)$. But then $q(0)=\det A=-1$, which contradicts the fact that $A\in SL(2,\Bbb F_3)$.
Well, if $A$ is not diagonalizable over $\bar{\mathbf F}_3$ then we may assume that $A=D+N$ with $$ D=\begin{pmatrix} \lambda&0\\0&\lambda\end{pmatrix} \quad\text{and}\quad N=\begin{pmatrix} 0&1\\0&0\end{pmatrix} $$ for some $\lambda\in\mathbf F_3^\star$. Since $D$ and $N$ commute, and since $N^2=0$, one has $$ A^8=D^8+8D^{7}N $$ whose element $(1,2)$ is nonzero.
If $A$ is diagonalizable over $\bar{\mathbf F}_3$ then we may assume that $$ A=\begin{pmatrix} \lambda&0\\0&\lambda^{-1}\end{pmatrix} $$ for some $\lambda\in \mathbf F_9^\star$. Of course, if $\lambda\in\mathbf F_3^\star$ then $A^2=I$. We may assume, therefore, that $\lambda\not\in\mathbf F_3$. But then $\lambda^{-1}$ should be the conjugate of $\lambda$ over $\mathbf F_3$, i.e., $\lambda^{-1}=\lambda^3$. It follows that $\lambda^4=1$ and $A^4=I$. Hence, $\mathrm{SL}(2,\mathbf F_3)$ does not contain any element of order $8$.
Note that $G={\rm SL}(2,3)$ has order $3\cdot 8$, so an element of order eight would afford a cyclic Sylow subgroup. Since all Sylow subgroups are conjugate (so in particular isomorphic), it suffices you produce a nonabelian subgroup in $G$ of order $8$.
