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The third isomorphism theorem states that we can relate an isomorphic relation between two normal subgroups of a group $G$. My question is can we infer anything about the two groups structures itself given that the factor/quotient groups in the derived series of two solvable groups are isomorphic to one another? In other words, if we have two solvable groups $G=\langle g_1,g_2,\ldots, g_n \rangle$ and $H=\langle h_1,h_2,\ldots h_p \rangle$, given that $G^{j}/G^{j+1}$ is isomorphic to $H^{j}/H^{j+1}$, what can be said about $G$ and $H$? (I don't think they are isomorphic but anything less strong be said?)

Alexander Gruber
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Vaas
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    I think this question would be much clearer if you used letters to refer to the various groups and subgroups you're thinking about. – Zev Chonoles Mar 02 '13 at 21:53
  • Hope my reformulating the question helps :) – Vaas Mar 02 '13 at 22:14
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    That is helpful :) If you can think of a more descriptive title, I think that'd also be quite useful. – Zev Chonoles Mar 02 '13 at 22:30
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    They are certainly not (necessarily) isomorphic; they certainly do have the same order; I'm not sure much more can be said. For example, there are zillions of groups of order $2^n$, even for modest values of $n$, and they all have isomorphic factor groups. – Gerry Myerson Mar 02 '13 at 22:52
  • Per chance we shouuld apply group cohomology to compute these groups? – awllower Mar 04 '13 at 10:09
  • @GerryMyerson even i am not to sure something more can be said, although i would have liked some reasonable condition on them even if not very tight! thanks – Vaas Mar 04 '13 at 14:12
  • @awllower Are you suggesting that it involves toruses and other ideas, im not too familiar wit hcohomology, but if you have an intuition thats a positive direction it would be worth reading. thanks :) – Vaas Mar 04 '13 at 14:14
  • This is both an excellent and difficult question. I am not an expert, but I believe that much research has been done on p-groups, with connections to Lie algebras. The work of Philip Hall and Norman Blackburn would be the right place to start. – Nicky Hekster Mar 04 '13 at 23:09
  • @Annonymous Because they have the same factors, they can be viewed as extensions of some groups: that is what I meant by saying group cohomology, not so directly involved with topology, though. – awllower Mar 05 '13 at 04:06
  • @GerryMyerson: But those groups don't all have isomorphic derived series. –  Mar 05 '13 at 06:34
  • @SteveD But they are not required to have isomorphic derived series. I think you mean thy do not have isomorphic factor groups? – awllower Mar 05 '13 at 09:42
  • I think the problem is formulated too generally. It would be desirable to see a more specific question. – Boris Novikov Mar 06 '13 at 22:10

2 Answers2

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There is a related concept that is very useful. Given a residually nilpotent group $G$, one can consider the set $$ L(G) = \bigoplus_{i}^{\infty} G_{i}/G_{i+1} $$ where the $G_{i}$ are the terms of the lower central series, defined by $G_{1} = G$, and $G_{i} = [G_{i-1}, G]$ for $i > 1$. Then one can show that $L(G)$ can be given the structure of a Lie ring, where the Lie bracket comes from the group commutator. Several problems on $G$ can be dealt with more efficiently by looking at $L(G)$.

However many non-isomorphic groups $G$ may have isomorphic $L(G)$, so of course not all properties of $G$ can be read off $L(G)$. This is the case for instance for $p$-groups of maximal class $G$ of a fixed order $p^{n}$, with $n > p+1$, that share the same $L(G)$.

Andreas Caranti
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If finite groups $G, H$ are metacyclic (i.e. $G'$ and $G/G'$ are cyclic) then your demand implies that their Sylow subroups are isomorphic (since in this case Sylow subroups are cyclic); see M.Hall, The Theory of Groups, Theorem 9.4.3.

Boris Novikov
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