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This is probably elementary but I know that in group theory the devil is in the details so I want to check. Assume I've a finitely generated group $G$ and a normal subgroup $U$ (if it does matter, in my situation $G$ is a semidirect product of $U$ by some $H$). Let $g_0,\dots,g_n$ be the generators of $G$. Assume that I know a presentation of $U$ and that $g_0 \not \in U$ (or even that $g_0 \in H$). Then, I want to find a presentation of the image of $U$ in the quotient $G/\langle g_0\rangle$.

Formally what I'm looking for is the intersection between the normal closure of $g_0$ and $U$ but it's hard to make it explicit. On the other hand, it seems to me that the only way taking the quotient can add relations is as follows: since $U$ is normal, $g_0$ acts on it by conjugation, so for $u\in U$, let $w_u$ be $g_0ug_0^{-1}$ written as a word in the generators of $U$. Then, taking the quotient by $g_0$ forces this action to be trivial, hence add the relation $$w_u\equiv u$$

Indeed I believe that there are no other new relations but didin't manage to write down a rigorous proof.

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You wrote $G/\langle g_0 \rangle$, but that only makes sense if $\langle g_0 \rangle$ is a normal subgroup of $G$, which it is not in general. Presumably you mean $G/\langle g_0^G \rangle$, where $\langle g_0^G \rangle$ is the normal closure of $g_0$ in $G$.

I am afraid that your conjecture is not true, even when $g_0$ lies in a complement $H$ of $U$. Let $G = S_4$, let $U$ be the normal subgroup $\langle (1,2)(3,4), (1,3)(2,4) \rangle$ of order 4, and let $g_0 = (1,2)$. Then $\langle g_0^G \rangle=G$, and so the quotient $G/\langle g_0^G \rangle$ is trivial. But if you just factor out elements of $U$ the form $g_0 u g_0^{-1} u^{-1}$ from $U$, then you are left with a quotient of $U$ order 2, because the only nontrivial element of $U$ of that form is $(1,2)(3,4)$.

Your argument would be correct (I think) if $H = \langle g_0 \rangle$ is a complement of $U$ in $G$, but in general the new relator $g_0=1$ can result in further new relators in $H$, which then introduce new relators in $U$.

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You're right, thanke ! Indeed by a Reidemeister-Schreier like argument, I think a complete set of relations is given by all conjugates of $w_u\equiv u$ by elements of $H$, and it's probably enough to restricts to the cases $u$ is a generator of $U$. –  Adrien May 28 '12 at 11:49

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