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Let $G$ be a finite group. Show that $G$ is not solvable iff it contains a normal non trivial subgroup $H$ with $H=H'$.

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I have done the following :

it holds that $H'=H^{(1)}=[H,H]$, derived group.

We assume that $G$ is solvable. Then each subgroup and quotient group of $G$ is solvable. So $H$ is solvable.

A group $H$ is solvable iff the derived series ends with the trivial group.

That cannot hold with a non trivial group $H$ with $H=H'$, because all derived groups $H^{(i+1)}=[H^{(i)}, H^{(i)}]$ are equal to $H$ which is non trivial.

So we get a contradiction.

Therefore $G$ cannot be solvable.

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Is everything correct and complete? Or isn't this enough for the iff statement?

Shaun
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Mary Star
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  • It seems to me that you have proved that there cannot exist such a non-trivial subgroup $H$ in a solvable $G$, but you have not proved the converse, that is, that there exists such a subgroup in any non-solvable group. – J. Darné Sep 14 '21 at 12:44
  • Could you give me a hint for the converse? Do we do the same just from the other way around? @J.Darné – Mary Star Sep 14 '21 at 12:46
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    Consider the derived series $G \ge G' \ge G'' \ge \cdots$. If the gorup is not solvable, then this series cannot reach the trivial subgroup. So, since $G$ is finite, one of the groups $H$ in the derived series must satisfy $H'=H$. Note that this argument (and also the result) fails when $G$ is infinite. – Derek Holt Sep 14 '21 at 13:14
  • The best proof is the one indicated above by Derek Holt. Here's another one though. We'll prove that if G is not soluble then it has a non-trivial characteristic perfect subgroup. Induct on $|G|$, the base case ($G = A_5$) being trivially true since non-abelian simple groups are perfect. Now if $G = G'$ we are done so we may assume that $G' < G$. If $G'$ is soluble then $G$ is soluble (since it is soluble-by-abelian). Thus we may further assume $G' <G$. By the induction hypothesis applied to $G'$ there exists a non-trivial perfect $H$ characteristic in $G'$. I think you can take it from here. – the_fox Sep 14 '21 at 14:16
  • Well, the suggested proof by induction is essentially the same as Derek Holt's (just a wee bit more fleshed out). Anyway... – the_fox Sep 14 '21 at 14:27
  • @the_fox: I agree it is essentially the same proof. I want to remark though that for the inductive proof, you could take $|G| = 1$ as the base case, since in that case the result is vacuously true. – spin Sep 15 '21 at 13:39
  • @DerekHolt Why that H is normal? – Jam Jan 24 '23 at 18:02
  • @Jam All subgroups in the derived series are characteristic and hence normal. – Derek Holt Jan 24 '23 at 19:34
  • @DerekHolt Didnt know that ! Thanks. Any suggestion where to read about normal series in general. Seems everyone is giving a brief exposure on basic algebra texts. THough i havent checked Dummit n Foote. – Jam Jan 25 '23 at 15:15

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