$\newcommand{\lcm}{\text{lcm}}\newcommand{\Aut}{\text{Aut}}$Let $G=C_9\rtimes_\phi C_6$, where $C_9=\langle x:x^9=1\rangle$ and $C_6=\langle y:y^6=1\rangle$, and $\phi:C_6\mapsto \Aut(C_9)$ is defined by $\phi(y^i)=f^i$ for $i\in \mathbb{Z}$ and $f\in \Aut(C_9)$ is defined by $f(x)=x^2$.
I am trying to figure out the order of $(x,y)$, where $x$ and $y$ are the generators of $C_9$ and $C_6$, respectively. Well, I know that $\lvert x \rvert =9$ and $\lvert y \rvert=6$, so isn't $\lvert (x,y)\rvert =\lcm(9,6)=18$? Why would this be wrong? If someone could explain, I would really appreciate it.
Also, how would one go about showing that $\lvert Z(G)\rvert=1$? Thanks!
$x$ and $y$ do not commute, so the lcm formula does not necessarily hold. Just see $S_{3}$ for an example, taking $x = (123)$ and $y=(12)$.
We have $x^{y} = y^{-1} x y = x^{2}$. Write $x^{1+y} = x \cdot x^{y}$ and similarly for more complicated exponents that are polynomial in $y$. Then \begin{align} (y x)^6 &= y^{6} y^{-5} x y^{5} y^{-4} x y^{4} \cdots y^{-1} x y x \\&= x^{y^5 + y^{4} + \dots + y^2+y+1} \\&= x^{2^5 + 2^4 + \dots + 2^2 + 2 + 1} \\&= x^{64 - 1} = x^{63} \\&= (x^9)^7 = e. \end{align} This shows that $y x$ (and then $x y$) has order at most $6$, and then it has order $6$ because the image of $y x$ in $G/C_{9}$ has order $6$.
As to the center, no non-trivial power of $x$ is centralized by $y$. And if you take any $x^i y^j$, with $y^j \ne e$, this acts on $\langle x \rangle$ like $y^j$, and it does not centralize $x$.