Timeline for Embedding $\mathrm{PGL}(n,q^h)$ in $\mathrm{PGL}(nh,q)$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 11, 2015 at 8:17 | vote | accept | Tom De Medts | ||
| Apr 29, 2015 at 9:02 | comment | added | Derek Holt | Yes, maybe it is incorrect to say that the embedding is induced. It turns out that the image of ${\rm GL}(3,27)$ in ${\rm PGL}(9,3)$ splits as a direct product $C_{13} \times {\rm PGL}(3,27)$, so it does occur as a subgroup, but only because of the splitting. | |
| Apr 29, 2015 at 8:09 | comment | added | peliukas | It might be that $\mathrm{SL}_n$ is very closely related to $\mathrm{PGL}_n$ in those cases and the embedding is obtained via $\mathrm{SL}(n, q^h)\rightarrow \mathrm{GL}(nh, q)$. | |
| Apr 29, 2015 at 8:07 | comment | added | Tom De Medts | I'm confused. If we multiply a matrix in $\mathrm{GL}(3,27)$ by a scalar in $\mathbb{F}_{27} \setminus \mathbb{F}_3$, then the corresponding matrix in $\mathrm{GL}(9,3)$ is not obtained by multiplying by a scalar, so I don't see how it can induce a well-defined map $\mathrm{PGL}(3,27) \to \mathrm{PGL}(9,3)$. What am I overlooking? | |
| Apr 29, 2015 at 7:57 | comment | added | Derek Holt | But there are some cases where the semilinear embedding ${\rm GL}(n,q^h) \to {\rm GL}(nh,q)$ does induce ${\rm PGL}(n,q^h) \to {\rm PGL}(nh,q)$. For example ${\rm PGL}(3,27) \to {\rm PGL}(9,3)$. This depends on $\gcd(q^h-1,n)$ and $\gcd(q-1,nh)$. It shouldn't be hard to write down the precise conditions for that. I might do that later. | |
| Apr 29, 2015 at 7:26 | comment | added | Tom De Medts | Thank you, that is very helpful and confirms my guess. I'm still wondering whether there is an (easy?) argument showing that the answer is negative for all other values of $n$ and $h$. Do you have any ideas? | |
| Apr 28, 2015 at 17:38 | history | edited | Derek Holt | CC BY-SA 3.0 |
added 642 characters in body
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| Apr 28, 2015 at 15:08 | history | answered | Derek Holt | CC BY-SA 3.0 |