Cyclic groups, generators, quotient groups

Question

Consider the additive group $G=Z/9Z$ x $Z/18Z$. I know that $G$ is isomorphic with $Z_9$ x $Z_{18}$ .

1. Order of $(4,3)$ and $(3,5)$.

Order of $4$ in $Z_9 =9$, order of $3$ in $Z_{18}=6$ so the order of $(4,3)$ is the smallest common multiple $=18$.
Order of $(3,5)=18$.

2. Is it true that $(4,3) \in \langle(3,5)\rangle$ and $(3,5) \in \langle (4,3) \rangle$?

In $Z_9, 3$ can't generate $4$ and in $Z_{18}, 3$ can't generate $5$ so none of them is true.

3. Is $\{(4,3),(3,5)\}$ a system of generators for $G$?

I think it is, since 4 generates all the elements of $Z_9$ and 5 generates all the elements of $Z_{18}$.

4. Is $G$ cyclic?

Since any $Z_n$ is cyclic $G$ is also cyclic and can be generated by $(5,5)$.

5. Is $G/ \langle(4,3) \rangle$ cyclic ?

I don't know how to approach this…

Are these correct and are my arguments enough?

Answer 1

1. Your argument is correct.
2. Your argument is correct. But note that the converse would not be correct: even if $a\in\langle c\rangle$ in $\Bbb Z_9$ and $b\in\langle d\rangle$ in $\Bbb Z_{18}$, it's not necessarily true that $(a,b) \in \langle (c,d)\rangle$ in $G$ (for example, $(1,5) \notin \langle (1,1)\rangle$).
3. This reasoning is faulty. The same argument would apply to $\{(4,4),(5,5)\}$, but those two elements do not generate all of $G$: they generate the subgroup $$ \langle (1,1) \rangle = \{ (0,0), (1,1), \dots, (8,8), (0,9), (1,10), \dots, (8,17) \}. $$
4. This reasoning is also faulty. If we try to articulate the general assertion you're using, it would be "if two groups are cyclic then so is their direct product"—but in fact this never happens if the orders of the two groups have a factor in common.