having trouble showing that an element belongs to a centre of a group presentation.
Let $G = \langle x,y,z\mid x^2=y^3=z^3=xyz\rangle$
I have to show that $ a = xyz$ belongs to the centre of $G$. I understand what the centre of a group is (the elements $h \in G$ s.t $hg=gh$ $\forall$ $g \in G$, but my problem is i'm fairly new to working with group presentations and I'm not totally what the elements of G would be. Although I think if i was to make a table using the relations and then compose them with eachother, would that give me the elements of G? Then could i just 'brute force' it?
Hint: for each generator, show that it commutes with $a$. To do so, use the relations to simplify the expressions.
For example, we first want to show that $xa = ax$. Notice that $$ xa = x^2 yz = xyzyz = a yz = ax. $$