I'm trying to show that any group of order 24 is solvable. I know $G$ is solvable if and only if both it's normal subgroup $N$ and $G/N$ is solvable. I have proved previously that a group of order 24 has a normal subgroup of order 4 or 8. But I'm stuck.. how do I proceed here? How do I show that $N$ is solvable and $G/N$ is also solvable?
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1Then you only need to prove groups of orders $3$, $4$, $6$ and $8$ are soluble. – Angina Seng May 10 '18 at 14:14
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@LordSharktheUnknown can you explain a little further? – Gregoire Rocheteau May 10 '18 at 14:18
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1The groups $N$ and $G/N$ will have orders in this list. – Angina Seng May 10 '18 at 14:19
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And for the fact (which you have proved), that there is a normal subgroup of order $4$ or $8$, see here. – Dietrich Burde May 10 '18 at 14:23