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I am working on extension of finite groups. Let $G$ be a finite group with trivial center such that it has a normal subgroup $N$ with odd order and $G/N=L_{2}(p)$ ($p$ is prime). I am looking for an example of group $G$. Please let me know if you know an example of this type of group. Thanks.

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What is $L_2(p)$? –  Mariano Suárez-Alvarez Mar 23 '13 at 7:40
    
@MarianoSuárez-Alvarez: $L_{2}(p)$ is projective linear group. –  Yakov Mar 23 '13 at 7:41
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Please add that information in the question itself, so that people do no need to read all comments to find it. –  Mariano Suárez-Alvarez Mar 23 '13 at 8:00

2 Answers 2

You must have some extra condition, such as $N \le G'$, otherwise the direct product of a centreless group of odd order by $\operatorname{PSL}(2, p)$ will do.

If it's a particular example you're looking for, consider $H = \operatorname{PSL}(2,2) \cong S_{3} = \langle s, t \rangle$, where $s$ has order $3$ and $t$ has order $2$.

Consider the action on $G$ on a vector space $N = \Bbb{Z}_{7}^{2}$, where $$ s \mapsto \begin{bmatrix}2&0\\0&4 \end{bmatrix}, \qquad t \mapsto \begin{bmatrix}0&1\\1&0 \end{bmatrix}. $$ (Note that $2$ is a primitive $3$-rd root of unity in $\Bbb{Z}_{7}$.)

Then $G = N \rtimes H$ will do. In fact $H$ clearly acts irreducibly on $N$, so the centre of $G$ is trivial. Here $N \le G'$.

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How about $\text{PSL}(2,5)\cong A_5$? Consider the group $\mathbb{Z}_3^5\rtimes A_5$.

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You mean that $A_5$ acts by permuting the base elements of $\Bbb{Z}_{3}^{5}$? Then the sum of the base elements is in the centre. However, if you factor out the group generated by that sum, you should get a centreless group as desired. –  Andreas Caranti Mar 23 '13 at 8:13
    
@AndreasCaranti, thank you for your reminder! –  Easy Mar 23 '13 at 9:53
    
You're welcome! –  Andreas Caranti Mar 23 '13 at 9:54

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