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Let $G$ be a non-trivial finite solvable group.Then $G$ has a normal subgroup of prime index.

I use the below fact to solve the problem.

"Every finite solvable group has a composition series that every factor of the series is cyclic of prime order."

I wonder if there is any other way to solve the problem.

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Hint: choose a maximal normal subgroup $M$. Then $G/M$ is simple and is solvable, hence cyclic of prime order.

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  • $\begingroup$ Thank you but maybe it doesn't have a maximal subgroup which is normal. $\endgroup$ – Yasmin Nov 5 '18 at 5:16
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    $\begingroup$ You are mixing up two things: there is a difference between normal maximal and maximal normal. If $M$ is maximal normal then there is no normal proper subgroup that contains $M$ properly. These groups always exist in finite groups (and $M$ could be trivial). Of course, not every maximal subgroup is normal (in nilpotent groups this is the case and is equivalent to nilpotency) an example being $S_3$ with subgroups of order $2$. $\endgroup$ – Nicky Hekster Nov 5 '18 at 7:42
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    $\begingroup$ I understand,thank you. $\endgroup$ – Yasmin Nov 5 '18 at 8:05

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