Is there any finite group $G$ with order $p(p^2-1)$ and normal subgroup $N$ such that $G/N\cong PSL( 2, p)$?
Finite group with given order
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$\begingroup$
group-theory
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0@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. – 2012-12-27
1 Answers
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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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0Nicely done, of course! + – 2013-03-02