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?)
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3$\begingroup$ I think this question would be much clearer if you used letters to refer to the various groups and subgroups you're thinking about. $\endgroup$ – Zev Chonoles Mar 2 '13 at 21:53
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$\begingroup$ Hope my reformulating the question helps :) $\endgroup$ – Annonymous Mar 2 '13 at 22:14
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1$\begingroup$ That is helpful :) If you can think of a more descriptive title, I think that'd also be quite useful. $\endgroup$ – Zev Chonoles Mar 2 '13 at 22:30
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5$\begingroup$ 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. $\endgroup$ – Gerry Myerson Mar 2 '13 at 22:52
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$\begingroup$ Per chance we shouuld apply group cohomology to compute these groups? $\endgroup$ – awllower Mar 4 '13 at 10:09
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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 $L(G)$.
answered Mar 9 '13 at 23:45
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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.
answered Mar 9 '13 at 23:19
