What is the least $n$ such that it is possible to embed $\operatorname{GL}_2(\mathbb{F}_5)$ into $S_n$?

Solution 1:

The answer is $24$. The natural action on ${\mathbb F}_5 \setminus \{0\}$ shows that ${\rm GL}_2(5) < S_{24}$.

To show that this is the smallest possible we prove the stronger result that $24$ is the smallest $n$ with $G:={\rm SL}_2(5) \le S_n$. The centre $Z = \{ \pm I_2 \}$ of $G$ has order $2$ and, since $G/Z \cong {\rm PSL}_2(5) \cong A_5$ is simple, $Z$ is the only nontrivial proper normal subgroup of $G$. So any non-faithful permutation action of $G$ has $Z$ 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 $H < G$ with $H \cap Z = 1$. Hence we are looking for the largest subgroup $H$ of $G$ with $H \cap Z = 1$.

Since $ -I_2$ is the only element of order $2$ in $G$, all subgroups of $G$ of even order contain $Z$. 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 $120/5 = 24$.

In general, for a finite group $G$ with a complicated structure, the problem of finding the least $n$ with $G \le S_n$ seems to be very difficult, and I have not come across any computer algorithms that solve it efficiently. The difficulty comes from the fact that the the smallest $n$ does not generally come from a transitive action, so you have to look at all possibilities of combining transitive actions to get trivial intersectino of kernels. In this particular case, we are lucky in that we can reduce the problem to ${\rm SL}_2(5)$, where we are guaranteed that the minimal action is transitive, so then it just becomes a search for the largest core-free subgroup, which can be done computationally if the group is not too huge.

Solution 2:

Here are some thoughts. Every group action of a group $G$ is a disjoint union of transitive group actions $G/G_i$ for various subgroups $G_i$. A group action is faithful iff for every $g \in G$ not equal to the identity there is some $i$ such that $g$ does not act by the identity on $G/G_i$. Hence there must be some coset $h G_i$ such that $gh G_i \neq h G_i$, or equivalently such that $hgh^{-1} \not \in G_i$. So the condition is that the conjugacy class of $g$ is not entirely contained in $G_i$. Finally, for a group action to be small we want the $G_i$ to be large.

This condition is hardest to satisfy when $g$ is central; in that case, the condition is that there must exist some $G_i$ such that $g \not \in G_i$. But it seems hard to me to find a large subgroup of $\text{GL}_2(\mathbb{F}_5)$ that doesn't contain a nontrivial central element. The action of $\text{GL}_2(\mathbb{F}_5)$ on $\mathbb{F}_5^2 \setminus \{ (0, 0) \}$ comes from the subgroup of matrices of the form

$$\left[ \begin{array}{cc} 1 & \ast \\ 0 & \ast \\ \end{array} \right]$$

which has order $20$, and that's the largest subgroup I can think of that doesn't have a nontrivial central element.

If we allow ourselves to ignore the center we can do much better. $\text{GL}_2(\mathbb{F}_5)$ acts on $\mathbb{P}^1(\mathbb{F}_5)$, which has $6$ elements, with kernel precisely the center. We can capture half of the center by using the determinant $\det : \text{GL}_2(\mathbb{F}_5) \to \mathbb{F}_5^{\times}$, which gives a group action with $4$ elements. The disjoint union of these two group actions has $10$ elements and has kernel just $-1$.