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The Question:

What is the smallest $m$ (dependent upon $n,q$) such that $$\operatorname{SL}_n(q)\hookrightarrow S_m?$$ That is, such that $\operatorname{SL}_n(q)$ embeds in $S_m$.

Context:

I want to view $\operatorname{SL}_n(q)$ as a permutation group in the most efficient way possible.

Put differently, I want the smallest $m$ such that

$$\operatorname{SL}_n(q)(\cong \circ\le)S_m,$$

where $(\cong \circ\le)$ means there exists a group $K$ such that $\operatorname{SL}_n(q)\cong K$ and $K\le S_m$.

Thoughts:

Of course, by Cayley's Theorem, $m\le |\operatorname{SL}_n(q)|$.

A Place to Look:

If I recall correctly, there's stuff about optimising such an embedding, in Robinson's book, "A Course in the Theory of Groups (Second Edition)". The problem with looking it up, though, is that it's buried deep in the text and it's been three years since I was at that point of the book.

Shaun
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    I mean, $q^n$ is an easy upper bound - $GL(n,q)$ has a faithful representation in $\mathbb F_q^n.$ The real question is whether you can get smaller. – Thomas Andrews Sep 25 '24 at 12:27
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    As long as we are talking about obvious bounds, at the very least $m\leqslant q^n$. – Captain Lama Sep 25 '24 at 12:29
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    This number $\mu(G)$ has the name "minimal permutation degree" of $G$. There are several posts here about it, e.g., this one, or this one. I will try to find more for $G=SL(n,q)$. – Dietrich Burde Sep 25 '24 at 12:41
  • The more I look into this, the more it looks like a research level question. I'll give it a few more days then perhaps I'll migrate it. I don't know though. Let me know what you think (the collective "you", that is). – Shaun Sep 25 '24 at 13:20

2 Answers2

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Let $G$ be a finite group. The invariant $$ \mu(G)=\min\{m \mid G\hookrightarrow S_m \} $$ is called the minimal faithful permutation degree of $G$. Clearly we have $\mu(SL(n,q))\le q^n-1$. There is an extensive literature on this invariant. It has been computed for all finite simple groups. To give an example, for $n=2$ we have \begin{align*} \mu(SL(2,3)) & = 8,\\ \mu(SL(2,5)) & = 24, \end{align*} which equals the bound $q^n-1$ mentioned above.

I found a recent paper by Borade et.al on Minimal permutation representations for linear groups, which proves results on $\mu(SL(n,q))$. In particular, for $(q-1)_2$ being the factor $2^e$ arising in the prime decomposition of $q-1$, we have $$ \mu(SL(2,q))=(q-1)_2\cdot (q+1). $$

Dietrich Burde
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The classification of maximal subgroups of PSL (Aschbacher, Kleidman/Liebeck, Bray, Holt, Roney-Dougal ) implies that, with a few small exceptions such as $PSL_2(5)$ and $PSL_2(7)$ the maximal subgroup of $PSL_n(q)$ of smallest index is the "point" stabilizer of index $(q^n-1)/(q-1)$, and all other maximal subgroups have significantly larger index.

For your problem that gives a lower bound of $(q^n-1)/(q-1)$, which is reached, for example, if $SL\cong PSL$. Otherwise you get at worst an extra factor $|Z(SL_n(q))|$ to represent the extension, and in some cases this is indeed equal to the naive $q^n-1$.

ahulpke
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