Timeline for About the orders of subgroups of $SL(n,q)$
Current License: CC BY-SA 3.0
13 events
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| Jun 20, 2013 at 13:54 | comment | added | Nick Gill | The existence of $\varphi$ is just standard rep theory. Perhaps it would be more transparent if you took $k$ to be a field of characteristic $r$ where $r$ does not divide the order of $G$, your simple group. Then $G$ is trivially a $C_{pp}$-group and the representation theory of $G$ over $k$ is ``the same" as representation over $\mathbb{C}$, so you can just consult character tables for $G$ to get irreducible representations of the type I've described. | |
| Jun 20, 2013 at 12:53 | comment | added | BHZ | Thanks for your answer, but why $\varphi$ exists | |
| Jun 20, 2013 at 12:24 | history | edited | Nick Gill | CC BY-SA 3.0 |
Removed remark and added example with $G$ simple.
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| Jun 20, 2013 at 12:18 | comment | added | Nick Gill | I'll remove the remark and add an example with $G$ simple. | |
| Jun 20, 2013 at 12:15 | comment | added | Nick Gill | I've just realised that my "remark" was somewhat misguided - you are talking about the centralizer in $G$, not in $SL_n(q)$, thus my concerns about $Z(SL_n(q))$ don't really make any difference, sorry. And this allows me to assert a different counter-example to your original question, namely by taking $G$ to be a Sylow $p$-subgroup of $SL_n(q)$. | |
| Jun 20, 2013 at 10:28 | comment | added | BHZ | Dear Nick Thanks for your answers. Let me state our question in a different manner. Let $G$ be a subgroup of $PSL(n,q)$ such that $G$ is not a subgroup of $PSL(n-1,q)$, where $q=p^\alpha$. If $G$ is a simple group and also is a $C_{pp}$ group with this condition that $p$ divides the order of $G$, can we say that there exists a primitive prime divisor $r$ of $q^n-1$ or $q^{n-1}-1$ such that $r$ divides $|G|$? Now we use the same definition of $C_{pp}$ as usual. Thanks | |
| Jun 20, 2013 at 8:29 | comment | added | Nick Gill | @BHZ Before I could give an answer I'd need you to address the issues in my "remark" above. Because if the centre of $SL_n(q)$ is non-trivial, there will be NO $C_{pp}$-groups according to your definition, and so the answer will be trivially YES. If we use the definition that I proposed in the remark, then things become more tricky. | |
| Jun 20, 2013 at 6:28 | comment | added | BHZ | Dear Nick Thank you very much for your complete and useful answer. Is it still true if $p$ divides the order of $G$? | |
| Jun 20, 2013 at 6:26 | vote | accept | BHZ | ||
| Jun 19, 2013 at 9:30 | history | edited | Nick Gill | CC BY-SA 3.0 |
deleted 1 characters in body; edited body
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| Jun 19, 2013 at 9:18 | history | edited | Nick Gill | CC BY-SA 3.0 |
Added some extras.; added 24 characters in body; added 22 characters in body; added 50 characters in body; deleted 20 characters in body
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| Jun 19, 2013 at 9:12 | history | edited | Nick Gill | CC BY-SA 3.0 |
Corrected answer to deal with small $n$; deleted 41 characters in body
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| Jun 19, 2013 at 8:55 | history | answered | Nick Gill | CC BY-SA 3.0 |