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If $G$ is a group and $g_1, g_2 \in G$ let us write $g_1 \sim g_2$ if there is an automorphism $\alpha \in \operatorname{Aut}(G)$ such that $g_1^\alpha = g_2$.

Let $F = F_2$ be the free group on two generators. Let $(\gamma_n(F))_{n=1}^\infty$ denote the lower central series of $F$. Let $w_1, w_2 \in F$. Suppose that, for every integer $n$, $$(w_1 \bmod \gamma_n(F)) \sim (w_2 \bmod \gamma_n(F)).$$ Does it follow that $w_1 \sim w_2$?

This question is interesting to me because a positive answer sounds too good to be true, and yet equally I would be fascinated to see an offending pair of words $w_1, w_2$. Not sure if there is research out there about this.

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This is extremely false due to the following basic fact about nilpotent groups:

Theorem: Let $N$ be a finitely generated nilpotent group and let $f\colon N \rightarrow N$ be a homomorphism. Then $f$ is an isomorphism if and only if the induced map $f_{\ast}\colon N^{\text{ab}} \rightarrow N^{\text{ab}}$ on the abelianization is an isomorphism.

This can be proved by induction on the degree of nilpotence of $N$. For instance, I wrote out a proof in Theorem 8.1 of my paper here.

Now let $F$ be free on $x$ and $y$. Consider some $w \in F$ that maps to the same element of $F^{\text{ab}} = \mathbb{Z}^2$ as $x$. Define an endomorphism $f\colon F \rightarrow F$ taking $x$ to $w$ and $y$ to $y$. Then $f$ is almost certainly not an automorphism of $F$, but by the above theorem it induces an isomorphism $F/\gamma_n(F) \rightarrow F/\gamma_n(F)$ for all $n$. It follows that $x$ and $w$ are equivalent modulo $\gamma_n(F)$ for all $n$, but they are probably not equivalent in $F$ (for instance, it is easy to arrange such a $w$ that does not lie in part of a free basis for $F$).

Note: I had originally written something stronger than the above in haste.

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  • $\begingroup$ I can't open your paper. But this fact follows from two old facts (1) f.g. nilpotent implies Hopfian (2) if $H$ is a subgroup of an arbitrary nilpotent group $G$, $H=G$ iff $H[G,G]=G$. From (2) it follows that if $f:G\to H$ is a homomorphism between arbitrary nilpotent groups, $f$ is surjective iff the induced homomorphism of abelianizations is surjective. Applied to endomorphisms of f.g. nilpotent groups, and using Hopfianity, we get the result. $\endgroup$
    – YCor
    Commented Dec 19, 2023 at 21:22
  • $\begingroup$ @YCor: I was definitely not claiming that this was due to me (which is why I said "For instance, I wrote out a proof..."). The only reason I gave a reference to my old paper (which also emphasized that we were not the first to notice it) was that I happened to know a precise location where exactly this statement was proven. $\endgroup$
    – Andy Putman
    Commented Dec 19, 2023 at 21:31
  • $\begingroup$ (ps: was there a reason you couldn't open it? I thought the arXiv was good at generating pdf files that were easily read on most computers.) $\endgroup$
    – Andy Putman
    Commented Dec 19, 2023 at 21:32
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    $\begingroup$ If I understand correctly, a result of Nielsen states that the set of primitive elements $w$ such that $w$ and $x$ map to the same same element of $\mathbb Z^2$ is just the conjugacy class of $x$. (This is special to the rank-two free group.) It follows that $w = x^2 y x^{-1} y^{-1}$ is not primitive. Here "primitive" means "equivalent to $x$ by an automorphism", or equivalently "part of a free basis". $\endgroup$
    – Sean Eberhard
    Commented Dec 20, 2023 at 12:18
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    $\begingroup$ @SeanEberhard: that's correct. You can also check directly that $w$ is primitive using Whitehead's algorithm: the Whitehead graph of $w$ is two triangles glued along one side; since this doesn't have a cut vertex, Whitehead's lemma implies that $w$ is not primitive. $\endgroup$
    – HJRW
    Commented Dec 20, 2023 at 13:16

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