$\DeclareMathOperator\SL{SL}$Let $p$ be an odd prime. It is easy to show that $\SL_2(3)$ can be embedded into $\SL_2(p)$.
Now, let $n$ be an integer larger than $2$.
Question: In which circumstances, $\SL_n(3)$ can be embedded into $\SL_n(p)$?
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4$\begingroup$ I think this can be attacked by considering the Sylow $3$-subgroups of the respective groups. $\endgroup$– Geoff RobinsonCommented Feb 28 at 12:26
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5$\begingroup$ And I'll bet the answer is never, for $n>2$ and $p\ne 3$. $\endgroup$– Dave BensonCommented Feb 28 at 12:28
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4$\begingroup$ I would agree with that assessment. $\endgroup$– Geoff RobinsonCommented Feb 28 at 12:30
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$\begingroup$ Yes, I agree. It is already rare that $|\mathrm{SL}_n(3)|$ divides $|\mathrm{SL}_n(p)|$. $\endgroup$– user44312Commented Feb 28 at 12:47
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$\begingroup$ That $|\mathrm{SL}_n(3)|$ divides $|\mathrm{SL}_n(p)|$ is rare but still true in many cases. It seems automatic for $n=2$, and for each $n$ seems true for infinitely many $p$ (I haven't checked, but it might be true inside one coprime arithmetic progression). It's more and more rare as $n$ grows. For instance for $n=7$ the smallest such $p$ is $535573$. $\endgroup$– YCorCommented Feb 29 at 17:25
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2 Answers
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$\DeclareMathOperator\SL{SL}$Here is a quick and dirty argument. For $\SL(3,3)$ we can check directly that there is no representation of dimension less than $11$ over a field of characteristic $\ne 3$. So we can assume that $n\geqslant 11$. On the other hand, the Sylow $3$-subgroup of $\SL(n,3)$ has derived length at least $\log_2(n)$, whereas the Sylow $3$-subgroup of $\SL(n,p)$ for $p>3$ has derived length at most $\log_3(n)+1$. If $\log_2(n)\leqslant \log_3(n)+1$ then $n\leqslant 2^{1/(1-\log_3(2))} < 7$. This contradiction shows that there is no $n\ge 3$ for which $\SL(n,3)$ embeds in $\SL(n,p)$ with $p>3$. I hope I haven't made any stupid mistakes here.
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3$\begingroup$ I don't view this as "dirty" (except the non-roman SL :) ). Where does the $\log_3$ come from? $\endgroup$– YCorCommented Feb 28 at 14:39
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$\begingroup$ It comes from the fact that in characteristic not equal to $\ell$, after algebraically closing, all representations of a finite $\ell$-group are monomial. If a representation is faithful, a little bit of jiggling around will convince you that if the dimension is $n$ then the derived length is at most $\log_\ell(n)+1$. Now set $\ell = 3$. $\endgroup$ Commented Feb 28 at 15:34
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$\begingroup$ I guess the point is that each subgroup from which an irreducible constituent is induced has index a power of $\ell$ that's at most $n$, so the $\lfloor\log_\ell(n)\rfloor$-th derived group acts diagonally, and is therefore abelian. $\endgroup$ Commented Feb 28 at 15:47
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$\begingroup$ I think I get your point now. Suppose that $\mathrm{SL}_n(p)$ has a Sylow $\ell$-subgroup $L$ where $\ell\neq p$. Then $L$ has a faithful complex irreducible representation of dimension less than or equal to $n$. So, $L$ embeds into $N_{\mathrm{GL}_n(\mathbb{C})}(T)$ where $T$ is the diagonal subgroup of $\mathrm{GL}_n$. As $N_{\mathrm{GL}_n(\mathbb{C})}(T)/T\cong S_n$, I guess the $\log_{\ell}(n)$ comes from the derived length of Sylow $\ell$-subgroup of $S_n$. Am I right? $\endgroup$ Commented Feb 29 at 3:13
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$\DeclareMathOperator\SL{SL}$The idea I had in mind originally when I made the comment was more simple-minded. I'll deal with the case that $n$ is even for ease of exposition.
Each $3$-subgroup $S$ of $\SL(n,p)$ has an Abelian normal subgroup of index dividing the $3$-part of $n!$, so certainly dividing $3^{\frac{n-1}{2}}$.
A Sylow $3$-subgroup $T$ of $\SL(n,3)$ has order $3^{\frac{n(n-1)}{2}}$, and the largest Abelian subgroup of $T$ has order $3^{\frac{n^{2}}{4}}$, so $T$ has no Abelian subgroup of index less than $3^{\frac{n^{2}}{4} -\frac{n}{2}}$.
If $T$ embeds in $\SL(n,p),$ then we must have $\frac{n^{2}}{4} \leq \frac{n}{2} + \frac{n-1}{2} < n$, so $n < 4$ (since $n$ is even, this forces $n=2$).
The argument for $n$ odd is similar.
Later edit: In fact, I think the case $n = 3$ and $p \equiv 1$ (mod $3$) needs a separate argument, since $\SL(3,p)$ may contain extra-special subgroups of order $27$ in that case. Last edit: For completeness, I will give an argument for this case: A maximal parabolic $P$ of $\SL(3,3)$ is a semidirect product of an elementary Abelian subgroup of order $9$ with $\operatorname{GL}(2,3)$. In particular, $P$ contains a Frobenius group of order $72$ with kernel of order $9$. It follows that $P$ has no faithful complex character of degree less than $8$ (in fact, it does have two faithful irreducible characters of degree $8$). Since $P$ is solvable, it follows by the Fong–Swan theorem that $P$ has no faithful representation of degree less than $8$ over any field of characteristic $p > 3$. Hence $P$ does not embed as a subgroup of $\SL(3,p)$ when $p >3$.
answered Feb 28 at 14:24
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$\begingroup$ Slower and cleaner. Thanks, Geoff. $\endgroup$ Commented Feb 28 at 22:25
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1$\begingroup$ Thanks very much for the detailed explanation. It is very clear. I just want to add a reference here for the first statement which is the Lemma 14.17 of Isaacs's book Character theory of finite groups. $\endgroup$ Commented Feb 29 at 3:17
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2$\begingroup$ Similar argument: by complete reducibility, for $p \neq 3$ every elementary abelian $3$-subgroup of $SL(n,p)$ has order $< 3^n$. Now in $SL(n,3)$, you can find an elementary abelian $3$-subgroup of order $3^{\lfloor n^2/4 \rfloor}$ , as a subgroup $\begin{pmatrix} I & * \\ 0 & I \end{pmatrix}$ for suitable block sizes. So $SL(n,3) \leq SL(n,p)$ implies $\lfloor n^2 /4 \rfloor \leq n$, which can only happen for $n = 2$, $n = 3$. Then check $n = 2, 3$ separately. $\endgroup$ Commented Feb 29 at 4:34
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1$\begingroup$ Very concise and clear answer. Thanks very much! $\endgroup$ Commented Feb 29 at 4:42