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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 $9\leq n\leq 24$.


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 $n\geq 11$.

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  • $\begingroup$ I wonder if we can feed GAP with this sort of question. $\endgroup$ – Martin Brandenburg Sep 10 '14 at 22:23
  • $\begingroup$ $\mathrm{GL}_2(\mathbb F_5)$ contains a subgroup isomorphic to $\mathbb{F}_{25}^{\times}$, so it should have an element of order 24, right? $\endgroup$ – Thomas Andrews Sep 10 '14 at 22:23
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    $\begingroup$ @Thomas: right. That bumps up the lower bound to $3 + 8 = 11$, I think. $\endgroup$ – Qiaochu Yuan Sep 10 '14 at 22:30
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    $\begingroup$ @Jack: no, that doesn't follow. You need to know that the short exact sequence $\mathbb{Z}_4 \to \text{GL}_2(\mathbb{F}_5) \to \text{PGL}_2(\mathbb{F}_5)$ is trivial to conclude that, and as far as I know that's not the case. $\endgroup$ – Qiaochu Yuan Sep 10 '14 at 22:31
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    $\begingroup$ Note that $24$ is the highest order of an element. If the minimal polynomial for a matrix $A$ is prime or a product of distinct linear factors, that polynomial is a divisor of $X^{24}-1$ (since $X$ is never a factor.) So the only case we care about are of the form $(X-u)^2$. In that case, it is a factor of $X^{20}-1=(X^4-1)^{5}$. $\endgroup$ – Thomas Andrews Sep 11 '14 at 0:01
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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.

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    $\begingroup$ Very nice! I was having thoughts along these lines (mostly the third paragraph, not the second) but I failed to realize that restricting my attention to $\text{SL}_2(\mathbb{F}_5)$ would make the argument easier. $\endgroup$ – Qiaochu Yuan Sep 11 '14 at 8:24
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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 $-1$.

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    $\begingroup$ The argument I had previously written down about a lower bound was nonsense. $\endgroup$ – Qiaochu Yuan Sep 11 '14 at 0:00

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