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The free group on two generators $F_2=\langle x,y|\rangle$ is the fundamental group of $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$. Now, there are plenty of galois covers of this space whose galois group is not solvable. Thus the "maximal solvable cover" (i.e. the limit over all galois covers with solvable galois group) is not the universal cover, but rather a quotient thereof. In other words, the natural map: $$F_2\to\lim_{\begin{smallmatrix}\longleftarrow\cr H\unlhd F_2\cr F_2/H\text{ finite solvable}\end{smallmatrix}}F_2/H$$ is not injective.

Can someone exhibit an explicit element of the kernel? What about the shortest element (by word length in $F_2$) in the kernel? In other words, the question is: what universal word in $x,y$ always vanishes when $x,y$ are specialized to elements of some solvable group $G$? (note that since $G$ is solvable, so is the subgroup generated by $x$ and $y$).

Such an element now has the following seemingly impossible property. Consider it as a closed path $\gamma$ in $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$. Now try to lift $\gamma$ to the cover of $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$ corresponding to the linking number with $0$ (i.e. we take the universal cover of $\mathbb P^1(\mathbb C)\setminus\{0,\infty\}$ and take the inverse image of $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$). Of course $\gamma$ lifts to a closed curve, since otherwise we just exhibited an abelian quotient of $F_2$ in which $\gamma$ is not sent to zero. The cover we just considered is $\mathbb P^1(\mathbb C)\setminus\mathbb Z$, and we can try to lift $\gamma$ to some abelian cover of this, etc. Of course, $\gamma$ always lifts to a closed curve, since all these covers are solvable! I'm having a hard time visualizing what such a curve $\gamma$ would look like geometrically in $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$ (I once thought all elements of $F_2$ could be "broken" by a sequence of such covers!)

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The obstruction isn't that there are elements not contained in any subgroup with a finite solvable quotient. The problem is that each subgroup with a finite solvable quotient contains elements that are not in the kernel of $A_5$. You can keep reducing the number, but there will always be infinitely many. You might get a kernel if you take the profinite completion, though. –  Will Sawin Jan 30 '12 at 22:48
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Andy's answer is a good one and proves that the map you right down actually IS injective. The natural map which is NOT injective is the one from the profinite completion of F_2 to the inverse limit you write down. –  JSE Jan 31 '12 at 2:32
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"right" ==> "write." Why oh why can't we edit comments? –  JSE Jan 31 '12 at 19:19
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4 Answers 4

up vote 20 down vote accepted

There are no such elements -- the intersection of the derived series of a free group is trivial. In fact, even more is true -- the intersection of the lower central series of a free group is trivial. This is a theorem of Magnus, and by now there are many proofs. The classical one is in the final chapter of Magnus-Karass-Solitar's book on combinatorial group theory.

By the way, a topological proof of this fact (lifting curves to covers to resolve self-intersections, etc) is contained in my paper "On the self-intersections of curves deep in the lower central series of a surface group" with Justin Malestein.

EDIT : I see that you really want finite solvable quotients, not general solvable quotients. It is still true. Fixing a prime $p$, there is a ``mod $p$ lower central series'' of a group whose quotients are $p$-groups (so finite nilpotent if the group is finitely generated). For a free group, Zassenhaus proved in his paper

H. Zassenhaus, Ein Verfahren, jeder endlichen p-Gruppe eine Lie-Ring mit der Charakteristik p zuzuordnen, Abh. Math. Sem. Hamburg Univ. 13 (1939), 200-207.

that the intersection of the mod $p$ lower central series of a free group is trivial. This can also be deduced from the paper I mentioned with Justin Malestein, at least for the prime $2$ (one of the proofs we give actually yields regular covers whose order is a power of $2$).

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The fact that free groups are residually finite p-groups for any prime p follows quite nicely from the fact that $\begin{pmatrix} 1 & p\\ 0 &1\end{pmatrix}$ and $\begin{pmatrix} 1 & 0\\ p &1\end{pmatrix}$ generate a free group of rank 2. –  Steve D Jan 30 '12 at 23:06
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Andy - or, for the finite case, you can combine Magnus' Theorem with the easy fact that nilpotent groups are residually finite. –  HJRW Feb 2 '12 at 19:35
    
@HW : Good point! I don't know why that slipped my mind when I was writing this. I had just read Zassenhaus's paper (which proves a lot more than what I said) when this question arrived, so it was very fresh in my mind. –  Andy Putman Feb 2 '12 at 22:34
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In other words, the question is: what universal word in x,y always vanishes when x,y are specialized to elements of some solvable group G? (note that since G is solvable, so is the subgroup generated by x and y).

This paper of Miklos Abert answers a related question: On the probability of satisfying a word in a group.

Given a finite group $G$, is there a word $\omega(x,y) \in F_2$ such that $\omega(g_1,g_2)=e$ exactly when $\langle g_1, g_2 \rangle \leq G$ is solvable?

Abert proves the answer is yes! The paper is quite nice.

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If C is a family of finite groups that is closed to taking quotients, normal subgroups, and extensions, then the free group on two generators embeds into the pro-C completion. Such C's can be the finite p-groups or the finite solvable groups, and others. See, e.g., Fried-Jarden's Field Arithmetic, Prop. 17.5.11, for a proof.

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Dror Speiser suggested to me that maybe a different question lies behind this one. Namely, instead of $F_2$, perhaps, it was meant to consider the profinite completion $\widehat{F_2}^{profinite}$ of $F_2$ (which is the etale fundamental group of $\mathbb{P}^1\smallsetminus\{0,1,\infty\}$). In this case the natural map $\widehat{F_2}^{profinite}\to \widehat{F_2}^{prosolvable}$ is not injective, for, essentially, the reason that was mentioned in the question, i.e., since there are covers of $\mathbb{P}^1\smallsetminus\{0,1,\infty\}$ with non-solvable Galois groups. Here $\widehat{F_2}^{prosolvable}$ is the prosolvable completion of $F_2$.

If this was indeed what was mentioned, then the kernel is well understood, namely it is the free pro-$\mathcal{C}$ group of countable rank, where $\mathcal{C}$ is the family of all finite groups whose decomposition factors are non-cyclic. I can't seem to find the reference to the paper that proves this at this moment, but if it interests someone I can look harder.

Regarding the minimal length of a word in the kernel, it will be infinite, exactly for the reason Andy Putman explained.

(An example for a nontrivial profinite word in the kernel is something of the form $[\cdots [[[[[x,y],x^2],y^2],x^3],y^3],\cdots]$, where the powers are growing so that it will converge. I didn't check that this element really works.)

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This question is interesting. However, in your example I don't see how (with which choice of exponents) you can arrange this word to be converging to a nontrivial element. [by the way, any limit point of a sequence as you construct is in the kernel of the map to the pronilpotent, not prosolvable, completion]. –  YCor May 6 '12 at 22:18
    
You are right, it is in the kernel on the pronilpotent completion. To check the the kernel $K$ onto the maximal prosolvable quotient is non trivial, just take $U$ open normal such that $\widehat{F_2}^{profinite}/U \cong A_5$. This will assure you that $KU=\widehat{F_2}^{profinite}$, hence by iso-2, $K/K\cap U=A_5$. To find an explicit element in $K$ shouldn't be too difficult, I think. –  Lior Bary-Soroker May 7 '12 at 9:48
    
I agree this kernel is nontrivial. But writing down an explicit nontrivial element in this kernel is certainly doable, but doesn't seem too immediate. [I really mean explicit, not only finding a sequence for which some limit points are nontrivial elements in the kernel.] –  YCor May 7 '12 at 10:46
    
One can reduce the question to non-finitely generated free profinite groups, since the commutator subgroup of the free profinite group on 2 generators is free of countable rank. –  Lior Bary-Soroker May 7 '12 at 10:57
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