In A. Mann's paper: Enumerating finite groups and their defining relations (1998, J. group theory), that can be found Here ,

Mann's says (see pages 62-63):

" Let $F$ be a free group of rank d, and let $F_i$ be the lower p-central series of $F$ . Let $ H= F/F_{c+1} $ for some $c$ , and let $ G= H/N$ for some subgroup $N $ of $H_c=F_c/F_{c+1} $ satisfying $|H_c/N|=|N| $ or $ p|N| $ . Since H is the free group of rank d in the variety of groups with lower p-central series of length $c$ , two such factor groups $H/N,H/M$ are isomorphic if and only if $M$ and $N$ are conjugate under $Aut(H) $ ".

Can someone help me understand the bold part ? Why is it true ? I understand that if we have an automorphism $\phi $ of $H$ , such that $\phi( N )= M $ , then the corresponding factor groups are isomorphic. But why in this case, if we have $H/M \equiv H/N$ i it implies that there is an automorphism of $H$ sending $N$ to $M$ ? Does someone have any good reference for this kind of theorem? Can someone help me understand it? Does similar results hold for general free groups? (i.e- $F/N \equiv F/M $ iff $M$ , $N $ are conjugates under $Aut(F)$ ) Are there any similar results for other varieties (such as the variety of groups with p-derived series of length $c$ ? or even derived/central series of length $c$ ?

Hope someone will be able to help me understand this

thanks everyone!

  • $\begingroup$ @Arturo Magidin: Can you give me a reference for this? If it is true for any variety, I will be glad if you will be able to explain me the reason for this... In particular, if I take $ F/N , F/M $ , where $F $ is finitely generated free, does it mean that there exists an automorphism of $F$ sending $M$ to $N $ ? Thanks $\endgroup$ – Jeremy Young Dec 2 '12 at 18:51
  • $\begingroup$ @Jeremy: Hmmm... I had envisioned something just like Derek Holt wrote (invoking only the universal property), but he has now withdrawn his answer, so I may have overlooked something. The idea is to try to leverage the maps $F\to F/M$ and $F\to F/N\to F/M$ into a map $F\to F$ that maps $N$ into $M$, and then vice-versa and use uniqueness to deduce the desired result. Let me think about it and see if I can figure out the details... $\endgroup$ – Arturo Magidin Dec 2 '12 at 21:22
  • $\begingroup$ (In fact, I was in the middle of trying to write an answer when Derek Holt posted his, and it seemed to match what I was going for, so I stopped. I'm a bit worried that he now says the details were wrong...) $\endgroup$ – Arturo Magidin Dec 2 '12 at 21:35
  • $\begingroup$ It is true that any isomorphism $\psi:H/N \to H/M$ must lift to an endomorphism $\phi:H \to H$, but $\phi$ is not necessarily uniquely determined by $\psi$, and I don't know how to prove that it is an isomorphism in general. In the particular case in Mann's paper, since $M \le F_c/F_{c+1}$, the image of $\phi$ must project onto $F/F_c$, and so $\phi$ must be an epimorphism and hence, by finiteness, an isomorphism. But I don't see how to prove it for free groups. $\endgroup$ – Derek Holt Dec 2 '12 at 22:25
  • $\begingroup$ @Derek Holt: Thanks a lot, but I still can't understand the fundemental idea behind what you are saying. Why does every isomorphism between $H/N, H/M $ must lift to an endomorphism $\phi:H \to H $ ? I guess that this is trivial and that the only nontrivial thing is to prove this lift is actaully an automorphism? Thanks again $\endgroup$ – Jeremy Young Dec 3 '12 at 8:06

This property is not true in general for free groups. I have finally remembered a counterexample! There are 19 normal subgroups $N$ of the free group $F_2$ of rank 2 with $F_2/N \cong A_5$. It was proved in

B.H. Neumann, and H. Neumann. "Zwei Klassen charakteristischer Untergruppen und ihrer Faktorgruppen". Math. Nachrichten 4 (1951), 106-125

that there are two orbits of the action of ${\rm Aut}(F_2)$ on these subgroups, with lengths 9 and 10.

As I said in my comment, in the situation in Mann's paper, we are considering normal subgroups $N$ of a free group $H$ of $p$-nilpotency class $c$, with $N \le H_c$. For two such subgroups $M$ and $N$, an isomorphism $\psi:H/M \to H/N$ lifts, by freeness, to an endomorphism $\phi:H \to H$, which must satisfy ${\rm Im}(\phi)N = H$. Now any subgroup of a nilpotent group $H$ that projects onto $H/H'$ must be the whole of $H$, so $\phi$ must be an epimorphism and hence an isomorphism.

| cite | improve this answer | |
  • $\begingroup$ @Derek: did you mean that $\phi$ must satisfy $Im(\phi)N= M$ ? I couldn't understand what you mean by "any subgroup of a nilpotent group $H$ that projects onto $H/H' $ " . What do you mean by "projects onto $H/H'$ "? thanks a lot again ! $\endgroup$ – Jeremy Young Dec 4 '12 at 8:17
  • $\begingroup$ No I meant what I wrote: ${\rm Im}(\phi) N = H$. This follows from the fact that $\phi$ is a lift of $\psi$, and ${\rm Im}(\psi) = H/N$. I want to conclude from that that in fact ${\rm Im}(\phi)=H$. By "$X$ projects onto $H/H'$", I meant $XH'=H$. $\endgroup$ – Derek Holt Dec 4 '12 at 9:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.