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Let $F$ be the free group on the generators $a_1,\ldots,a_n$. Define homomorphisms $\phi_i:F\to F$ by $\phi_i(a_j)=a_j^{1-\delta_{ij}}$, where $\delta_{ij}$ is the Kronecker delta; basically, $\phi_i$ kills the $i$-th generator and fixes the others. Denote $N=\bigcap_{i=1}^n\ker\phi_i$.

I want to identify the subgroup $N$ or at least find its index in $F$. Clearly we have $F'\subseteq N$, where $F'$ is the commutator subgroup. Is it in fact the case that $N=F'$?

Edited: it has been pointed out in the comments that I was very hasty in the previous paragraph. The inclusion I stated holds only if $n\leq2$. However, $N$ is not trivial as it contains the element $[\ldots[a_1,a_2],a_3]\ldots]$

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    What's $F'$ here? As written, I'm pretty sure $N$ is the trivial group if $n > 2$.2011-08-24
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    [Commutator subgroup](http://en.wikipedia.org/wiki/Commutator_subgroup#Definition), maybe?2011-08-24
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    @Dylan: I suspected it might be, but then the statement $F' \subset N$ only holds if $n \leq 2$ (for example, the element $aba^{-1}b^{-1}$ is not killed by killing $c$).2011-08-24
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    When $n=1$ you get $N=F$, when $n=2$, you get $N=[F,F]$. But for $n=3$, it's already somewhat complicated: the subgroup generated by any two of the three free generators intersects $N$ trivially, since no element of $\langle x_i,x_j\rangle$ vanishes under $\phi_k$, where $\{i,j,k\}=\{1,2,3\}$.2011-08-24
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    More generally, each of the $n$ subgroups generated by all but one of the generators in $F_n$ intersect $N$ trivially.2011-08-24
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    Yes, $N$ will contain any basic commutator that involves all the variables, for example. But for $n\gt 2$, none of the lower central or derived series of $F_n$ will eventually be in $N$; as I said, the free group generated by, say, $x_1,\ldots,x_{n-1}$ intersects trivially with $N$, and $F/N$ contains $n$ free groups of rank $n-1$ that together generate the whole group.2011-08-24
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    Whoops, apparently I was too hasty as well, thinking that $N$ would be trivial for big $n$. So it seems that Arturo has proven that $N$ has infinite index in $F_n$, and of course $N$ is free as it's a subgroup of a free group. I guess I'm curious as to whether it's finitely generated (my guess would be it's not).2011-08-24
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    I *think* that $N$ is generated by all basic commutators on $x_1^{\pm 1},\ldots,x_n^{\pm 1}$ that involve each variable at least once. It is easy to see that $N$ is contained in the commutator subgroup (look at the induced homomorphisms in the abelianization), and that it contains all such basic commutators, but I have not tried to prove that it is *equal* to this subgroup.2011-08-24
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    @MartianInvader: Yes, $N$ certainly has infinite index in $F_n$, since even $\mathrm{ker}(\phi_i)$ has infinite index in $F_n$ (if $n\gt 1$).2011-08-24
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    @Arturo To be nitpicky, you probably mean $normally$ generated by those commutators. After all, you can conjugate such a commutator by, say, $x_1$, and get an element of $N$ that's not a straight product of commutators.2011-08-24
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    @MartianInvader: Sort of; note that $x^y = x[x,y]$. If $x$ is a basic commutator and $y$ is a generator, then $[x,y] = [x\leftarrow y]\times c_1^{a_1}c_2^{a_2}c_3^{a_3}\cdots$, where $[x\leftarrow y]$ is a the basic commutator that you get by "shuffling" $y$ "into" $x$ until you get a basic commutator, and the $c_i$ are basic commutators of weight at least $wt(x)+wt(y)$ in which all the letters of $x$ and $y$ appear.2011-08-25
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    In fact, it's easy to verify that $N$ is contained in the $n$th term of the lower central series of $F$ (by mapping to the relatively free nilpotent group of class $n-1$ and rank $n$, much like mapping to the abelianization shows that the $n=2$ is contained in the commutator subgroup). Moreover, if $H$ is the subgroup generated by the basic commutators in which every letter occurs, then $N\gamma_k = H\gamma_k$ for all $k$, where $\gamma_k$ is the $k$th term of the lower central series. However, this does *not* necessarily guarantee $N=H$.2011-08-25
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    With respect to the finitely generated question: Your group is isomorphic to the free group on infinitely many generators. Let $F_n$ be the free group on $n$ generators. If $H$ is a subgroup of index $j$ then $H$ is the free group on $j(n-1)+1$ generators. If $j$ is infinite then $H$ is the free group on (countably) infinitely many generators. (Thm 2.10 of Magnus, Karrass and Solitar "combinatorial group theory")2011-08-25
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    @Swlabr: I remember proving your result (Schreier's index formula) via topological methods. I don't have a reference at hand, so let me ask this: how would I arrive at a freely generating set for $N$?2011-08-25
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    I'm not entirely sure. Could you not use Reidermeister-Schreier to re-write your subgroup, and then un-re-write it?...(I mean, your subgroup is normal, and the collapsing tree you'll take is dead easy because you're in a free group so there are no loops in the Cayley graph). But this is just a whim - I'm not sure really...2011-08-26

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