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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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    I'm not entirely sure. $C$ould 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...2019-03-13

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