Timeline for Embedding $\mathrm{SL}_n(3)$ into $\mathrm{SL}_n(p)$
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 29 at 4:42 | comment | added | user44312 | Very concise and clear answer. Thanks very much! | |
| Feb 29 at 4:34 | comment | added | testaccount | 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. | |
| Feb 29 at 3:17 | comment | added | user44312 | 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. | |
| Feb 29 at 3:04 | vote | accept | user44312 | ||
| Feb 29 at 3:04 | |||||
| Feb 29 at 3:04 | vote | accept | user44312 | ||
| Feb 29 at 3:04 | |||||
| Feb 29 at 3:04 | vote | accept | user44312 | ||
| Feb 29 at 3:04 | |||||
| Feb 28 at 22:25 | comment | added | Dave Benson | Slower and cleaner. Thanks, Geoff. | |
| Feb 28 at 16:18 | history | edited | LSpice | CC BY-SA 4.0 |
Link to comment
|
| Feb 28 at 16:11 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Explained exceptional case.
|
| Feb 28 at 14:50 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Noted exceptional case $n = 3$ and $p \equiv 1$ (mod $3$).
|
| Feb 28 at 14:24 | history | answered | Geoff Robinson | CC BY-SA 4.0 |