Let ${\rm Gl}_2 \left(\mathbb{F_5} \right)$ be the group of invertible $2 \times 2$ matrices over $\mathbb{F}_5$ and $\mathcal{S}_n$ be the group of permutations of $n$ objects. What is the least $n \in \mathbb{N}$ such that there is an embedding of ${\rm Gl}_2 \left(\mathbb{F_5} \right)$ into $\mathcal{S}_n$ ?
Solution
We shall prove a stronger result. We will prove that $24$ is the smallest $n$ with $\mathcal{G}:={\rm Gl}_2(5) $ $\le \mathcal{S}_n$. The centre $\mathcal{Z} = \{ \pm \mathbb{I}_2 \}$ of $G$ has order $2$ and, since $\mathcal{G}/\mathcal{Z} \cong {\rm PSL}_2(5) \cong \mathcal{A}_5$ is simple, $\mathcal{Z}$ is the only nontrivial proper normal subgroup of $\mathcal{G}$. So any non-faithful permutation action of $\mathcal{G}$ has $\mathcal{Z}$ in its kernel. It follows that the smallest degree faithful representation is transitive, and so it is equivalent to an action on the cosets of a subgroup $\mathcal{H} < \mathcal{G}$ with $\mathcal{H} \cap \mathcal{Z} = 1$. Hence we are looking for the largest subgroup $\mathcal{H}$ of $\mathcal{G}$ with $\mathcal{H} \cap \mathcal{Z} = 1$.
Since $ -\mathbb{I}_2$ is the only element of order $2$ in $\mathcal{G}$, all subgroups of $\mathcal{G}$ of even order contain $\mathcal{Z}$. There is no subgroup of order $15$, so the largest odd order subgroup has order $5$, and the permutation action on its cosets has degree $120/5 = 24$.
Solution
We shall prove a stronger result. We will prove that $24$ is the smallest $n$ with $\mathcal{G}:={\rm Gl}_2(5) $ $\le \mathcal{S}_n$. The centre $\mathcal{Z} = \{ \pm \mathbb{I}_2 \}$ of $G$ has order $2$ and, since $\mathcal{G}/\mathcal{Z} \cong {\rm PSL}_2(5) \cong \mathcal{A}_5$ is simple, $\mathcal{Z}$ is the only nontrivial proper normal subgroup of $\mathcal{G}$. So any non-faithful permutation action of $\mathcal{G}$ has $\mathcal{Z}$ in its kernel. It follows that the smallest degree faithful representation is transitive, and so it is equivalent to an action on the cosets of a subgroup $\mathcal{H} < \mathcal{G}$ with $\mathcal{H} \cap \mathcal{Z} = 1$. Hence we are looking for the largest subgroup $\mathcal{H}$ of $\mathcal{G}$ with $\mathcal{H} \cap \mathcal{Z} = 1$.
Since $ -\mathbb{I}_2$ is the only element of order $2$ in $\mathcal{G}$, all subgroups of $\mathcal{G}$ of even order contain $\mathcal{Z}$. There is no subgroup of order $15$, so the largest odd order subgroup has order $5$, and the permutation action on its cosets has degree $120/5 = 24$.
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