Subgroup of a group with length 3 relators

Question

Let $\Gamma=\left<\mathcal{S}\,|\,\mathcal{R}\right>$ be a group defined by the presentation, where each relator $r\in\mathcal{R}$ is a reduced word of length 3 consisting of three different symbols only in $\mathcal{S}$, not in $\mathcal{S}^{-1}$.

Take a proper subset $\mathcal{S'}$ of $\mathcal{S}$, and let $\Gamma'$ be the subgroup of $\Gamma$ generated by $\mathcal{S'}$. Define $\mathcal{R'}$ be a set consisting of all of the relators in $\mathcal{R}$ whose three symbols are in $\mathcal{S'}$. Let us formalize my question:

(Question) Does the natural surjection (generator to generator) $\left<\mathcal{S'}\,|\,\mathcal{R'}\right> \twoheadrightarrow \Gamma'$ have trivial kernel?

Surely, there are easy counterexamples. For $G=\left<a,b,c\,|\,abc,bac\right>$, take a subset $\{a,b\}$ of the generating set. In this case, we see $\mathcal{R'}$ is empty. To avoid these trivial cases, we need the following property:

(C1) If $c\in\mathcal{S}$ and $c\in\left<\mathcal{S'}\right>$($c$ is in the subgroup of $\Gamma$ generated by $\mathcal{S'}$), then $c\in\mathcal{S'}$.

Now, after choosing $\mathcal{S'}$ satisfying (C1), can we answer (Question) positively? I think there are counterexamples, but it seems a bit complicated to me. Could you recommend some references on these groups?

Revision: the answer to the original question above is negative, as Roland Bacher neatly demonstrated. If we refine the question, adding the condition $\Gamma$ satisfies (C2) below, would the question be still meaningful?

(C2) For any two different symbols $a,b\in\mathcal{S}$, $a,a^{-1},b,b^{-1}$ are four different elements in $\Gamma$ as group elements.

Answer 1

Here is an explicit example showing that assumptions (C1) and (C2) are not sufficient for your question. Let $$G = \langle a,b,c,d \mid abc, bdc, dac, adb \rangle.$$ A quick computer calculation shows that $|G|=24$ and in fact $G \cong {\rm SL}(2,3)$ with generator images $$ \left( \begin{array}{cc}1&2\\1&0\end{array} \right), \left( \begin{array}{cc}0&2\\1&1\end{array} \right), \left( \begin{array}{cc}2&2\\0&2\end{array} \right), \left( \begin{array}{cc}2&0\\1&2\end{array} \right).$$ So all of the generators represent distinct elements of $G$ (in fact they all have order 6) , and there are no relators involving fewer than three generators.

Answer 2

Consider $\mathcal S=\{a,b,c,d\}$ with relations $\mathcal R=\{acd,bcd\}$ implying $a=b$. For $\mathcal S'=\{a,b\}$ there is no relation involving only elements of $\mathcal S'$ in $\mathcal R$. The group $\langle \mathcal S',\mathcal R'\rangle$ is therefore free but has the relation $a=b$ in the kernel of the obvious surjection onto the subgroup generated by $a,b$ in $\langle \mathcal S,\mathcal R\rangle$.