Exercise 4.5.9 (Sylow $3$-subgroups of $\mathrm{SL}_2(\mathbb{F}_3)$)

Question

Exhibit all Sylow $3$-subgroups of $\mathrm{SL}_2(\mathbb{F}_3)$ (cf. Exercise 9, Section 2.1).

Richard Ganaye · Accepted Answer

\begin{proof} 
By Exercise 2.4.9, $|\mathrm{SL}_2(\mathbb{F}_3)| = 24 = 2^3 \cdot 3$ (see the complete list in Exercice 2.4.9).

Among these $24$ elements, 8 have order $3$, which can be paired in 4 pairs $\{A,A^2\}$, thus there are exactly $4$ Sylow $3$-subgroups:
$$\langle \begin{pmatrix} 0&1\-1 & -1\end{pmatrix} angle, \qquad \langle \begin{pmatrix} 0&-1\1 & -1\end{pmatrix} angle, \qquad \langle \begin{pmatrix} 1&0\1 & 1\end{pmatrix} angle, \qquad \langle \begin{pmatrix} 1&1\0 & 1\end{pmatrix} angle. $$
\end{proof}


With Sagemath:
\begin{verbatim}
sage: F = GF(3);
sage: G = SL(2,F);G.order()
24
sage: l = []
....: for g in G:
....:     if g.order() == 3:
....:         l.append(g)
....: l
....: 

[
[0 2]  [1 2]  [2 2]  [0 1]  [1 0]  [1 1]  [1 0]  [2 1]
[1 2], [0 1], [1 0], [2 2], [2 1], [0 1], [1 1], [2 0]
]
sage: [g^2 for g in l]

[
[2 1]  [1 1]  [0 1]  [2 2]  [1 0]  [1 2]  [1 0]  [0 2]
[2 0], [0 1], [2 2], [1 0], [1 1], [0 1], [2 1], [1 2]
]
\end{verbatim}

Exercise 4.5.9 (Sylow $3$-subgroups of $\mathrm{SL}_2(\mathbb{F}_3)$)

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Exercise 4.5.9 (Sylow $3$-subgroups of $\mathrm{SL}_2(\mathbb{F}_3)$)

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