# 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?

• In what text is this proof? – Shaun Nov 3 at 10:18
• Taking the quotient by $C$ amounts to killing $c$. – Shaun Nov 3 at 10:18

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.)