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Let the group $G$ have the presentation $\langle x_1, \dots, x_n \;|\; r_1, \dots, r_m \rangle$. Let $\Gamma$ be a labelled directed graph corresponding to Van Kampen diagram over the above presentation of $G$. Let $\Gamma_1$ be the undirected graph corresponding to $\Gamma$.

Let $\mathcal{C}$ be an arbitrary set of cycles of $\Gamma_1$ such that each edge of $\Gamma_1$ belongs to at least one of the edges of the cycles of $\mathcal{C}$. Let $R_1, \dots, R_k$ be the labels of cycles in $\mathcal{C}$, (where by the label of a cycle it is meant the word read on the cycle.)

Is it true that $\langle x_1, \dots, x_n \;|\; R_1, \dots, R_k \rangle$ is a presentation for $G$?

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  • $\begingroup$ I am confused. $\Gamma$ might not involve all of the relations $r_1,\ldots,r_m$. For example $\Gamma$ could just consist of a single loop labelled $r_1$, in which case $k=1$ and $R_1=r_1$. $\endgroup$
    – Derek Holt
    Commented Jul 14, 2015 at 18:10
  • $\begingroup$ I too am confused. A van Kampen diagram is a geometric proof that one element $\gamma$ of the free group on the generators $x_1,\ldots,x_n$ maps to the trivial element in $G$. It does not tell you anything about the whole of $G$. Perhaps you meant 'presentation complex'? $\endgroup$
    – HJRW
    Commented Jul 14, 2015 at 21:13
  • $\begingroup$ @DerekHolt: What I mean by the graph $\Gamma$: it is a directed labelled graph in which we have at least one cycle corresponding to each relation $r_i$. The graph $\Gamma$ is not unique: e.g., one may consider $m$ disjoint cycles each of them corresponds to a relation $r_i$; or one may glue properly the cycles to have a graph with less number of connected components. May be referring to the Van Kampen diagrams is not necessary and confusing; I am trying to draw such a graph $\Gamma$ for an specific example but at the moment I do not know how to do it in MathJax... $\endgroup$
    – Alireza Abdollahi
    Commented Jul 15, 2015 at 7:49
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    $\begingroup$ I am still unsure whether I understand. Let $G = \langle a,b,c,d \mid ab^{-1},bc^{-1},cd^{-1} \rangle$ and conisder a van Kampen diagram for $ad^{-1}$ with three interior regions. Can we not take $\mathcal{C}$ to be the cycles labelled $R_1=ab^{-1}$ and $R_2=cd^{-1}$? $\endgroup$
    – Derek Holt
    Commented Jul 15, 2015 at 8:17
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    $\begingroup$ @DerekHolt: Yes. This answers negatively my question: You take the graph $\Gamma$ as follows $1 \overset{a}{\longrightarrow} 2$, $1 \overset{b}{\longrightarrow} 2$, $1 \overset{c}{\longrightarrow} 2$, $1 \overset{d}{\longrightarrow} 2$. A graph with two vertices $1,2$ and four directed edges between these two vertices. Are there examples in which we have no multiple edges between vertices? It depends to how one choose the graph $\Gamma$. If one consider the graph $\Gamma$ with three disjoint cylces, then the answer to my question is positive. Anyway, thanks for your try to answer my question $\endgroup$
    – Alireza Abdollahi
    Commented Jul 15, 2015 at 9:07

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