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Let $G$ be a finite group having the property that for any prime $p$ dividing $|G|$, it has a subgroup H with $[G:H]=p$. What can be said about these groups? I believe I can prove that they must be solvable. But are they supersolvable?

Nicky Hekster
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    Surely, if $G$ is any finite group, then you can get a group in this class by taking the direct product of $G$ with a cyclic group of order divisible by all primes dividing $G$. – Derek Holt Oct 04 '11 at 08:01
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    $S_4$ is in this class of groups, but is not supersolvable. – Kevin Ventullo Oct 04 '11 at 08:32
  • Derek, thanks, good point! So they aren't even necessarily solvable. Class is hence very large. – Nicky Hekster Oct 04 '11 at 08:46
  • duplicate question: http://math.stackexchange.com/questions/66451/finite-groups-with-a-subgroup-of-every-possible-index – Julian Kuelshammer Oct 04 '11 at 11:12
  • Julian, thanks! Was not aware of this entry. Sure, I know about CLT (Converse Lagrange Theorem) groups, but this is a wider notion. Every CLT group has the property above of course. So my question is not really a duplicate one. The reference to the paper of Jing is interesting. The hyperlink is dead however. – Nicky Hekster Oct 04 '11 at 12:41
  • @Julian: It's not really a duplicate; CLT groups have this property, but groups with this property need not satisfy the Converse Lagrange Theorem. Derek Holt's observation shows that you can say very little about the structure of these groups. – Arturo Magidin Oct 04 '11 at 13:12
  • I didn't find a hyperlink, but here is the reference: Jing, Naihuan: The order of groups satisfying a converse to Lagrange's theorem. Mathematika 47, 2000, 197-204, 2002. And for the book also mentioned it is: Bray, Deskins, Humphreys, Puttaswamaiah, Venzke, Walls: Between nilpotent and solvable. – Julian Kuelshammer Oct 04 '11 at 13:18
  • Such an H must be a maximal subgroup. If you have control over only one maximal subgroup, I don't think you can say very much about G. You can however say a fair amount about G/Core(G,H), G mod the largest normal subgroup of G contained in H. It just comes down to permutation groups of prime degree though, so it may be old news to you. – Jack Schmidt Oct 04 '11 at 14:45
  • There is a kind of converse of Derek's construction: For any group $G$ in your class $G\times H$ will be in your class for any group $H$ with $\pi(H)\subseteq \pi(G)$ (where $\pi()$ denotes the primes dividing the order of the group). – Someone Oct 04 '11 at 15:55
  • @Jack, thanks for your comments. Yes for the smallest prime $ p$ dividing the order of $G$, you automatically have a normal subgroup of index $p$, and that is the way how I thought to start an induction argument to prove solvability. Apparently Derek showed that that would not work. – Nicky Hekster Oct 06 '11 at 09:26
  • @Nicky: yes, if you require H also to satisfy the property, the result is a supersolvable group. Otherwise, you just get a homomorphism from G into a direct product of symmetric groups of prime degree with a fairly arbitrary kernel (whose prime divisors are fixed). For instance, S4 is written as the subdirect product of S3 and S2 with a kernel of order 4. The kernel can be quite large, as in Derek Holt's examples where it would be all of "G". – Jack Schmidt Oct 06 '11 at 15:25
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    @Derek: I don't see any reason why you couldn't expand your comment into an answer. – davidlowryduda Aug 21 '12 at 22:31

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