Is the center of this group trivial? 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.
 A: 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.
