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

Question

Let $\operatorname{GL}_2(\mathbb{F}_5)$ be the group of invertible $2\times 2$ matrices over $\mathbb{F}_5$, and $S_n$ be the group of permutations of $n$ objects.

What is the least $n\in\mathbb{N}$ such that there is an embedding (injective homomorphism) of $\operatorname{GL}_2(\mathbb{F}_5)$ into $S_n$?

Such a question has been asked today during an exam; it striked me as quite difficult. There is an obvious embedding with $n=24$, and since $|\operatorname{GL}_2(\mathbb{F}_5)|=480$ and in $\operatorname{GL}_2(\mathbb{F}_5)$ there are many elements with order $20$, we have $n\geq 9$. However, "filling the gap" between $9$ and $24$ looks hard, at least to me. Can someone shed light on the topic? I would bet that representation theory and Cayley graphs may help, but I am not so much confident to state something non-trivial. I think that proving that $\operatorname{GL}_2(\mathbb{F}_5)$ is generated by three elements (is this true?) may help, too.

I would be interested also in having a proof of something sharper than $9\leq n\leq 24$.

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Update. The following Wikipedia page claims, in paragraph Exceptional isomorphisms, that $\operatorname{PGL}(2,5)$ is isomorphic to $S_5$. This seems to suggest that $\operatorname{GL}_2(\mathbb{F}_5)$ embeds in $\mathbb{Z}_4\times S_5$ that embeds in $S_9$. Am I right?

Second Update. No, I am wrong, since $\mathbb{F}_{25}^*$ embeds in $\operatorname{GL}(2,\mathbb{F}_5)$, so there is an element with order $24$, so $n\geq 11$.

Answer 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.

Answer 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 $-1$.

Answer 3

First let's consider the general problem of determining for a finite group $G$ the smallest $n\in\mathbb N_0$ such that $G$ embedds into $S_n$. This is equivalent to asking what is the smallest possible cardinality of a set on which $G$ acts faithfuly. We know from Cayley's theorem that $n=|G|$ is possible, but this is in general far from optimal.

So let $G$ act faithfuly on a finite set $X$ and let $x_1,\ldots,x_k\in X$ be representatives of the orbits of this action. Then by the orbit-stabilizer theorem it's $$|X|=\sum_{i=1}^k|G\cdot x_i|=\sum_{i=1}^k[G:G_{x_i}].$$ Every $x\in X$ is of the form $x=gx_i$ for some $g\in G$ and $i\in\{1,\ldots,k\}$ and stabilizers of elements in the same orbit are conjugated since $G_{gx}=gG_xg^{-1}$. Then the condition that the action is faithful means that $$\bigcap_{x\in X}G_x=\bigcap_{i=1}^k\bigcap_{g\in G}gG_{x_i}g^{-1}=1.$$

OTOH, let $k\in\mathbb N_0$ and $H_1,\ldots,H_k\le G$ such that $$\bigcap_{i=1}^k\bigcap_{g\in G}gH_ig^{-1}=1.$$ The the action of $G$ on the disjoint union $\bigsqcup_{i=1}^kG/H_i$ by left multiplication is faithful and $$\left|\bigsqcup_{i=1}^kG/H_i\right|=\sum_{i=1}^k[G:H_i].$$ It follows that the smallest possible value of $n$ is $$n=\min\left(\sum_{i=1}^k[G:H_i]:k\in\mathbb N_0;H_1,\ldots,H_k\le G;\bigcap_{i=1}^k\bigcap_{g\in G}gH_ig^{-1}=1\right)$$ (it's $k\in\mathbb N_0$ because for $G=1$ it's possible to take $k=0$ since the trivial group acts faithfuly on the empty set, but for non-trivial $G$ it's $k\in\mathbb N$).

Equipped with that, we can look at $G=\mathrm{GL}_2(\mathbb F_5)$. We have $Z(\mathrm{GL}_2(\mathbb F_5))=\mathbb F_5^\times\cdot I$, and we'll need the following lemma.

Lemma: Let $H\le\mathrm{GL}_2(\mathbb F_5)$ such that $H\cap\mathbb F_5^\times\cdot I=I$. Then $|H|\le20$. Proof: The condition $H\cap\mathbb F_5^\times\cdot I=I$ implies that the composition $H\to\mathrm{GL}_2(\mathbb F_5)\to\mathrm{PGL}_2(\mathbb F_5)$ is injective. It's well known that $\mathrm{PGL}_2(\mathbb F_5)\cong S_5$, thus $H$ embedds into $S_5$. If it was $|H|>20$ then $[S_5:H]\le5$, and thus $H$ is isomorphic to $S_4$, $A_5$ or $S_5$ (see green keeper’s post [url=https://artofproblemsolving.com/community/c7h2675661]here[/url]). If $H\cong A_5$ or $H\cong S_5$ then $H$ contains $(5-1)!=24$ elements of order $5$. It's easy to see that every element of $\mathrm{GL}_2(\mathbb F_5)$ of order $5$ is conjugated to $T=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and that $|C_{\mathrm{GL}_2(\mathbb F_5)}(T)|=20$ (using the fact that $T$ is non-derogatory) and so $\mathrm{GL}_2(\mathbb F_5)$ has $[\mathrm{GL}_2(\mathbb F_5):C_{\mathrm{GL}_2(\mathbb F_5)}(T)]=480/20=24$ elements. Thus $H$ must contain all of them, in particular $T,T^T\in H$. But it's $(T^{-1}T^TT^{-1})^2=-I$ or alternatively $(T^{-1}T^T)^3=-I$, so $-I\in H$, contradicting $H\cap\mathbb F_5^\times\cdot I=I$ (in fact it's $\langle T,T^T\rangle=\mathrm{SL}_2(\mathbb F_5)$ since the same is true over $\mathbb Z$ and the reduction map $\mathrm{SL}_2(\mathbb Z)\to\mathrm{SL}_2(\mathbb F_5)$ is surjective). If $H\cong S_4$ then the conditions $H\cap\mathbb F_5^\times\cdot I=I$ and $\mathbb F_5^\times\cdot I\subseteq Z(\mathrm{GL}_2(\mathbb F_5))$ imply $\langle H,\mathbb F_5^\times\cdot I \rangle\cong H\times\mathbb F_5^\times\cdot I\cong S_4\times Z/4$. Since $D_4<S_4$, if follows that $D_4\times Z/4$ embedds into $\mathrm{GL}_2(\mathbb F_5)$. But it's $v_2(| D_4\times Z/4|)=5=v_2(|\mathrm{GL}_2(\mathbb F_5)|)$, so $D_4\times Z/4$ would be a $2$-Sylow subgroup of $\mathrm{GL}_2(\mathbb F_5)$, but that's false since $\mathrm{GL}_2(\mathbb F_5)$ contains an element of order $24$ (as discussed above) and hence also of order $8$.

Now let $k\in\mathbb N$ and $H_1,\ldots,H_k\le\mathrm{GL}_2(\mathbb F_5)$ such that $$\bigcap_{i=1}^k\bigcap_{g\in\mathrm{GL}_2(\mathbb F_5)}gH_ig^{-1}=1.$$ Then there exist $g\in\mathrm{GL}_2(\mathbb F_5)$ and $j\in\{1,\ldots,k\}$ such that $-I\notin gH_jg^{-1}$. Since $-I\in Z(\mathrm{GL}_2(\mathbb F_5))$, this means $-I\notin H_j$. This implies $H_j\cap\mathbb F_5^\times\cdot I=I$ (because $\pm I$ is the unique minimal subgroup of $\mathbb F_5^\times\cdot I$), and so by the lemma we have $$\sum_{i=1}^k[G:H_i]\ge[G:H_j]\ge\frac{480}{20}=24.$$

OTOH, it's trivial to see that the value $n=24$ is attainable.The canonical action of $\mathrm{GL}_2(\mathbb F_5)$ on the vector space $\mathbb F_5^2$ is faithful and $(0,0)^T$ is obviously fixed by it, so $\mathrm{GL}_2(\mathbb F_5)$ acts faithfully on $\mathbb F_5^2\setminus\{(0,0)^T\}$ and $|\mathbb F_5^2\setminus\{(0,0)^T\}|=5^2-1=24$.