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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 $n > 2$. –  MartianInvader Aug 24 '11 at 17:44
    
Commutator subgroup, maybe? –  Dylan Moreland Aug 24 '11 at 17:52
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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 $\{i,j,k\}=\{1,2,3\}$. –  Arturo Magidin Aug 24 '11 at 18:19
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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. –  Arturo Magidin Aug 24 '11 at 19:55
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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") –  user1729 Aug 25 '11 at 10:32

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