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Consider a finitely presented group $G$ with presentation $P$ given by $\left\langle g_1,\ldots,g_n|\, r_1,\ldots,r_m\right\rangle$, equipped with a homomorphism $\rho\colon\, G\to H$ to a finitely generated group $H$ (generators in the presentation are ordered). Perform a sequence of Tietze transformations starting at $P$ and ending at presentation $P^\prime$ given by $\left\langle g^\prime_1,\ldots,g^\prime_n|\, r^\prime_1,\ldots,r^\prime_m\right\rangle$ which differs from $P$ by relabeling of generators (so $g_i\mapsto g_i^\prime$ for all $i$ is an isomorphism of presentations). Is it true that $\rho(g_i)=\rho(g_i^\prime)$ for all $i$ up to inner automorphism of $H$?

At first I thought this was trivial, but I'm worried that it might actually be false- if so, is there some way to measure such a failure of simple-connectedness of "presentations of $G$ over $H$".

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Daniel: (i) Have you an example of such a sequence of Tietze transformation that relabel generators only? (ii) Have you thought of using Fox derivatives or something of that form? – Tim Porter Jul 19 '12 at 8:19
So, let $G$ be a free group $\langle g_1,g_2\rangle$ and consider the alternative presentation $\langle g_2,g_1\rangle$. Now consider the obvious map $\rho:G\to \mathbb{Z}/2\times\mathbb{Z}/2$. – HJRW Jul 19 '12 at 8:25
In view of HW's example, you have to allow all automorphisms of $H$, not just the inner ones. However, the statement is false even in this form: Take $G=F_n$. You would be asking if the action of $Out(F_n)$ on $Hom(G,H)/Aut(H)$ is trivial. This is clearly false too, say, for $G=SL(2,p)$, $p\ge 5$. (Same argument as in the case of the $SL(2,C)$-character varieties.) – Misha Jul 19 '12 at 12:29
... Oh dear. So it is false. Misha: What is the argument for $SL(2,C)$ character varieties? Does the statement have a chance of being true for fundamental groups of (cusped) hyperbolic 3-manifolds? – Daniel Moskovich Jul 19 '12 at 13:32
@Dan: Think first about Dehn twists acting on the character variety (and why mapping class group acts nontrivially on the Teichmuller space). The easiest case to consider is when $F_3=F_2* Z$, then apply $\phi\in Aut(G)$, where $\phi|F_2=id$ and $\phi|Z$ is conjugation by $g_1\in F_2$. Same works if $G=F_2$, use $(g_1,g_2)\to (g_1, g_1g_2)$. – Misha Jul 20 '12 at 5:51
up vote 0 down vote accepted

The question is effectively answered by HW and by Misha in the comments. Namely, a sequence of Tietze transformations starting and ending at "the same" presentation defines an automorphism of $G$. But the action of $\mathrm{Out}(G)$ on $\mathrm{Hom}(G,H)/\mathrm{Aut}(H)$ is non-trivial in general.

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