I have to find elements of order $18$ from $Z_{12}×Z_{9}$. I did the following thing: The order of the elements in $Z_{12}$ must be ${1,2,3,4,6,12}$ and the order of elements in $Z_9$ must be $1,3,9$. I have to find elements $(a,b) ∈ Z_{12}×Z_{9}$ so that the least common multiple of the orders of $a$ and $b$ in $Z_{12}$ and $Z_{9}$ respectively is $18$, but there are no such order numbers in $Z_{12}$ and $Z_9$ that the least common multiple is $18$.
Does that mean that they don't exist, or am I missing something ?
Edit: I didn't see order $9$ elements in $Z_{9}$
You have determined the lists $\ell:=\{1,2,3,4,6,12\}$ and $\ell':=\{1,3,9\}$ of the divisors of $12$ and $9$. If $a\in{\mathbb Z}_{12}$ has order $d\in\ell$ and $b\in{\mathbb Z}_9$ has order $d'\in\ell'$ then $(a,b)$ has order $n:={\rm lcm}(d,d')$ in ${\mathbb Z}_{12}\times{\mathbb Z}_9$. Since we want $n=18$ we have to determine the pairs $(d,d')\in\ell\times\ell'$ with this property. There are two such pairs, namely $(2,9)$ and $(6,9)$.
There is exactly one element of order $d=2$ in ${\mathbb Z}_{12}$, namely $6$. On the other hand all elements $j\in{\mathbb Z}_9$ with ${\rm gcd}(j,9)=1$ have order $d'=9$. These are the six elements $\pm1$, $\pm2$, $\pm4$. It follows that $(d,d')=(2,9)$ leads to the following six elements of order $18$ in ${\mathbb Z}_{12}\times{\mathbb Z}_9$: $$(6,\pm1),\quad (6,\pm2),\quad (6,\pm 4)\ .$$
We now have to determine the elements of order $d=6$ in ${\mathbb Z}_{12}$. An element $j\in{\mathbb Z}_{12}$ has order $6$ iff $6j=0$ mod $12$, but $r j\ne0$ mod $12$ for $1\leq r\leq 5$. It follows that $j$ has to be even, but not divisible by $4$ or $6$. Therefore the only elements of order $6$ in ${\mathbb Z}_{12}$ are $\pm2$. The conclusion is that $(d,d')=(6,9)$ leads to the following $12$ elements of order $18$ in ${\mathbb Z}_{12}\times{\mathbb Z}_9$: $$(\pm2,\pm1),\quad (\pm2,\pm2),\quad (\pm2,\pm 4)\ ,$$ whereby the signs can be chosen independently.