# Varieties Of Groups & Enumeration Of Size of Isomorphic Factor Groups

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!

• @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 – Jeremy Young Dec 2 '12 at 18:51
• @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 I had envisioned something just like Derek Holt wrote (invoking only the universal property), but he has now withdrawn his answer, so I may have overlooked something.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... – Arturo Magidin Dec 2 '12 at 21:22
• (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...) – Arturo Magidin Dec 2 '12 at 21:35
• 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. – Derek Holt Dec 2 '12 at 22:25
• @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 – Jeremy Young Dec 3 '12 at 8:06

## 1 Answer

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.

• @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 ! – Jeremy Young Dec 4 '12 at 8:17
• 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 $XH'=H$. – Derek Holt Dec 4 '12 at 9:41