Group presentation, central subgroup In a proof that I am reading there is the following statement.
For a group $G$ with presentation
$$ G= \langle \gamma_1 ,\gamma _2,\gamma _3 ,c \mid \gamma _1^pc^{-1} =\gamma_2^qc^{-1}= \gamma _3^r c^{-1}=\gamma_1\gamma _2\gamma _3 c^{-1} =1 \rangle $$
the element $c$ generates a central subgroup $C$, with quotient $G/C$ isomorpic to the group
$$ \langle \gamma_1 ,\gamma _2,\gamma _3  \mid \gamma _1^p =\gamma_2^q= \gamma _3^r =\gamma_1\gamma _2\gamma _3  =1 \rangle  . $$
Can someone maybe give me a hint why that is true?
 A: Here is a general fact: Let $G=\langle S\rangle$. Then $g\in Z(G)$ if and only if $gx=xg$ for all $x\in S$.
That is, an element is contained in the center of a group $G$ if and only if the element commutes with every element of a generating set for $G$.
In your set-up, $G=\langle\gamma_1, \gamma_2, \gamma_3\rangle$, while the element $c$ is a power of each of generator $\gamma_i$, and so commutes with each $\gamma_i$, so is central by the above fact. For example, $\gamma_1c=\gamma_1\gamma_1^p=\gamma_1^p\gamma_1=c\gamma_1$.
Therefore, $G/\langle c\rangle$ makes sense and we obtain the presentation by adding the relator $c=1$ to get:
\begin{align*}
&\langle \gamma_1 ,\gamma _2,\gamma _3 ,c \mid \gamma _1^pc^{-1} =\gamma_2^qc^{-1}= \gamma _3^r c^{-1}=\gamma_1\gamma _2\gamma _3 c^{-1} =1, c=1 \rangle\\
&\cong
\langle \gamma_1 ,\gamma _2,\gamma _3 ,c \mid \gamma _1^p=\gamma_2^q= \gamma _3^r=\gamma_1\gamma _2\gamma _3=1, c=1 \rangle&\text{simply using $c=1$}\\
&\cong
\langle \gamma_1 ,\gamma _2,\gamma _3 \mid \gamma _1^p=\gamma_2^q= \gamma _3^r=\gamma_1\gamma _2\gamma _3=1 \rangle
\end{align*}
as required. (In the last step we removed the generator $c$ via a Tietze transformation.)
