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Is there any finite group $G$ with order $p(p^2-1)$ and normal subgroup $N$ such that $G/N\cong PSL( 2, p)$?

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Assuming the orders work out correctly, you can always take $G$ to be $PSL(2,p)\times C$ with $C$ a cyclic group, no? –  Mariano Suárez-Alvarez Dec 27 '12 at 6:30
    
@Mariano Suarez-Alvarez: Thank you. The normal subgroup $N$ is central. Thus $G$ has elements of order $2r$ for every $r$. Therefore, you are right. –  user2132 Dec 27 '12 at 7:09

1 Answer 1

up vote 2 down vote accepted

There is a presentation for $PSL(2,p)$ as follows: $$PSL(2,p)=\langle x,y\mid x^2=1, y^p=1, (xy)^3=1, (xy^4xy^\frac{p+1}{2})^2=1\rangle$$ so if we assume $N=\langle x,y\mid x^2=1, (xy)^3=1,(xy^4xy^\frac{p+1}{2})^2=y^{-p} \rangle$ then we can show that $N$ is a Schure Extension for $PSL(2,p)$. I think this is what you want.

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Thank you. Your answer also is so nice. –  user2132 Dec 27 '12 at 7:13
    
@user2132: Honestly, I saw something like your problem among my academic works and notes, so I tried to bring you what is near to your answer. Thank you. –  B. S. Dec 27 '12 at 7:17
    
But Mariano Suarez-Alvarez's answer is much simpler. –  Derek Holt Dec 27 '12 at 10:06
    
Nicely done, of course! + –  amWhy Mar 2 '13 at 0:47

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