Let $G = \langle a,b,c,d \mid [a,[c,d]]=[b,[c,d]]=[c,[a,b]]=[d,[a,b]]=1\rangle$, where $[x,y]=xyx^{-1}y^{-1}$. I want to know whether the center of this group is trivial.
It seems that any element $g$ in its center must satisfy the strong condition: if we cancel any two of the generators in $g$, then $g$ becomes a trivial word. (For example, if $g=abcdb^{-1}a^{-1}$ and we cancel $c,d$ then $g$ becomes $abb^{-1}a^{-1}=1$.) However, this condition still cannot rule out the words such as $aba^{-1}cac^{-1}b^{-1}ca^{-1}c^{-1}$, and I have no idea how to go on.
My approach may seem weird, so any possible approach to this question is welcome.
Thank you all.
Yes, the center of this group is trivial. This can be seen by viewing it as a free product with amalgamation, and using the corresponding action on the Bass-Serre tree. (I guess you could also apply the normal form theorem for free products with amalgamation, but that is less fun.)
So, $G$ decomposes as $(F(a, b)\times\langle z_1\rangle)\ast_{z_1=[c, d], z_2=[a, b]}(F(c, d)\times\langle z_2\rangle)$, and let $\Gamma$ be the Bass-Serre tree of this decomposition, so $G$ acts on $\Gamma$ such that $G_1:=F(a, b)\times\langle z_1\rangle$ and $G_2:=F(c, d)\times\langle z_2\rangle$ fix adjacent vertices $v_1, v_2$, and the edge between them is fixed by the subgroup $H:=\langle z_1, z_2\rangle\cong\mathbb{Z}^2$.
Consider $x\in Z(G)$. Letting $i=1, 2$, as $H\lneq G_i$ there exists $g_i\in G_i$ such that the only part of $\Gamma$ fixed by $g_i$ is $v_i$. Now, $$ (v_i\cdot x)\cdot g_i=v_i\cdot xg_i=v_i\cdot g_ix=v_i\cdot x $$ and so $g_i$ fixes $v_i\cdot x$. As $g_i$ only fixes the vertex $v_i$ we have that $v_i\cdot x=v_i$, so $x$ fixes $v_i$ for $i=1, 2$. Hence, $x$ fixes the edge between $v_1$ and $v_2$, and so $x\in H$. Therefore, $x\in (H\cap Z(G_1))\cap (H\cap Z(G_2))=\langle z_1\rangle\cap\langle z_2\rangle$, and so $x$ is the identity. Hence, $Z(G)$ is trivial as claimed.