Timeline for Embedding $\mathrm{SL}_n(3)$ into $\mathrm{SL}_n(p)$
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 29 at 14:12 | comment | added | Dave Benson | Sorry, I meant it contains the $m$th derived subgroup, which is more to the point. | |
| Feb 29 at 13:11 | comment | added | user44312 | Ah, I see. Thanks very much! | |
| Feb 29 at 8:36 | comment | added | Dave Benson | That's one way to put it, but if you put it that way, you're left with a computation for $S_n$, which I avoided. My point was that if a subgroup of an $\ell$-group has index $\ell^m$ then it's in the $m$th derived subgroup, which seems simpler. As you can see, it's quite a coarse bound, but it works. | |
| Feb 29 at 3:13 | comment | added | user44312 | 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? | |
| Feb 29 at 3:04 | vote | accept | user44312 | ||
| Feb 29 at 3:04 | vote | accept | user44312 | ||
| Feb 29 at 3:04 | |||||
| Feb 28 at 16:19 | comment | added | LSpice | @YCor, re, even that smudge is now edited out. | |
| Feb 28 at 16:19 | history | edited | LSpice | CC BY-SA 4.0 |
`\DeclareMathOperator`
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| Feb 28 at 15:47 | comment | added | Dave Benson | 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. | |
| Feb 28 at 15:34 | comment | added | Dave Benson | 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 $\ell = 3$. | |
| Feb 28 at 14:39 | comment | added | YCor | I don't view this as "dirty" (except the non-roman SL :) ). Where does the $\log_3$ come from? | |
| Feb 28 at 13:43 | history | answered | Dave Benson | CC BY-SA 4.0 |