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Let G be a finite group,H is an arbitrary proper subgroup of G,H is solvable,but G is not solvable.then what can we know about group G?

user94782
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  • I think the answer is likely to be "not much". Any element in a group (solvable or not) generates a cyclic subgroup, which is solvable. You need more structure to have any chance of saying anything interesting. One thing you can say is that $H$ can't be too big (the index of $H$ in $G$ can't be too small), because otherwise it would be a normal subgroup. But that is about $H$ not $G$. – Mark Bennet Sep 17 '13 at 14:42

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All we know is that $G$ is not solvable.

Any $G$, even if it's nonsolvable, will have many $H\subset G$ that are solvable. For example, the trivial subgroup $e$ and the cyclic subgroups $\langle g\rangle$ for any $g\in G$ are all abelian, hence solvable.

Edit: As DonAntonio pointed out, I misread the question. Here are some thoughts on the question that was actually asked:

Assume that $G$ is not simple, and of finite length.

Then $G$ has a normal subgroup $H\subset G$, and, by the condition in the question, $H$ is solvable, and so all of $H$'s composition factors are abelian. Also, $G/H$ still satisfies the condition of the question: if it were solvable, then by combining composition series, $G$ would also be, and any proper subgroup is the image of a proper subgroup of $G$, and so will still be solvable. Finally, $G/H$ has a shorter length than $G$.

Therefore, if we repeat this process enough, $G/H$ will be simple, and we have this result:

Any $G$ with this property is an extension of a solvable group by a nonabelian simple group with this property.

I suspect that the converse holds as long as $G/H$ is not a subgroup of $G$.

William Ballinger
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  • Yes, but the OP seems to have stated that $;G;$ is a non-solvable group all of which proper subgroups are solvable... – DonAntonio Sep 17 '13 at 15:46
  • @DonAntonio Oh, yes. That makes the problem considerably more difficult. – William Ballinger Sep 18 '13 at 00:43
  • The converse needs a few tweaks: $G/H$ needs to be minimal simple, and I don't see a clean way to give the list of things that cannot be subgroups (just $G/H$ is not sufficient, there could be other $G/K$ with $K < H$). – Jack Schmidt Sep 19 '13 at 14:45