Given the group $G = \mathbb{Z}_6 \times\mathbb{Z}_6$ and the following operation: $$(a,b)\cdot (c,d)=(a+(-1)^bc,b+d)$$ find (or classify) $Z(G)$ and $G/Z(G)$.
For the first part, I found that $(a,b) \in Z(G) \Leftrightarrow(a,b)\cdot(c,d)=(c,d) \cdot (a,b) \text{ for every }(c,d) \in G \Leftrightarrow a(1+(-1)^{d+1})=c(1+(-1)^{b+1}) \Rightarrow Z(G)=\langle(0,2)\rangle \simeq \mathbb{Z}_3$.
I'm having troubles on the second part though; I know that $|G/Z(G)|=12=2^23$, and I also know that there's either a normal $2$-Sylow or a normal $3$-Sylow, but at this point I'm stuck. Some hints on how to proceed?
$\newcommand{\Z}{\mathbb{Z}}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$The two subsets $A = \Set{(x, 0) : x \in \Z_{6}}$ and $B = \Set{(0, y) : y \in \Z_{6}}$ are subgroups, and $B$ normalises $A$, as $$\begin{align} (0, 1)^{-1} (1, 0) (0, 1) &= (0, -1) (1, 0) (0, 1)\\ &= (-1, -1) (0, 1)\\ &= (-1, 0). \end{align}$$ So $G$ is a semidirect product of $A$ by $B$, with $b = (0, 1)$ inducing the inversion automorphism on $A$. It follows that $Z(G) = \Span{(3, 0), (0, 2)}$ has order $6$. A Sylow $3$-subgroup is normal, while Sylow $2$-subgroups are not, as $a = (1, 0)$ conjugates $\Set{(0, 0), (3, 0), (0, 3), (3, 3)}$ to $\Set{(0, 0), (3, 0), (1, 3), (4, 3)}$.