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Let $G$ be a group of order 2520 and let $K$ be the maximal normal soluble subgroup of $G$. If we know that $G/K\cong A_5$, $K=C_2\times (C_7 :C_3)$, $C_G(K)=SL(2,5)$ and $G= C_G(K) K$, what would be the structue of $G$?( By $A_5$ I mean the alternating group of degree 5)

Tina
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    $G$ has a normal subgroup $N$ of order 21, which has a complement $H := C_G(K)$ of order 120. $H$ is a perfect central extension of $A_5$, whose center acts trivially on $N$ (by conjugation), so $A_5$ has to act on $N = C_7\rtimes C_3$. It fixes the characteristic subgroup $C_7$ of $N$, so it is not hard to see that $A_5$ acts trivially. In total you get that $G$ is the direct product of $N$ and $H$. – j.p. Jul 27 '13 at 12:23
  • Great, thank you very much. – Tina Jul 28 '13 at 06:12

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