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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 $(3,5)=18$.

  1. 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.

  1. 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 $Z_{18}$.

  1. Is $G$ cyclic?

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

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

I don't know how to approach this...

Are these correct and are my arguments enough?

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    $\begingroup$ No, it is not correct. $G$ is not cyclic since $gcd(9,18)\neq 1$, see Chinese Remainder Theorem. $\endgroup$ – Dietrich Burde May 30 '19 at 15:07
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    $\begingroup$ In (1) you mean "smallest common multiple" not "smallest common factor". (The numbers are right. The words are wrong.) $\endgroup$ – Ethan Bolker May 30 '19 at 15:15
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  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.
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    $\begingroup$ Your last assertion appears to be off a little bit, as, as I'm sure you know, when $(m,n)=1$, we have $\Bbb Z_m×\Bbb Z_n=\Bbb Z_{mn}$. $\endgroup$ – Chris Custer May 30 '19 at 16:10
  • $\begingroup$ Any hints on how to prove that $\{ (4,3),(3,5)\}$ is a system of generators? $\endgroup$ – Florin1335 May 30 '19 at 16:28
  • $\begingroup$ @ChrisCuster good catch! edited. $\endgroup$ – Greg Martin May 30 '19 at 20:52
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My answer deals with only 4 and 5:

$G$ is not cyclic. If it were, then the number of elements of order $3$ would be exactly $\phi(3)=2$. But here the number of elements of order $3$ is not two. For example, the three elements $(0,6), (3,0),$ and $(3,6)$ have order $3$.

For fifth part, note that this quotient group has order $9$, so it is isomorphic to either $\Bbb Z_9$ or $\Bbb Z_3 \times \Bbb Z_3$. Now the element $(1,1) \langle (4,3) \rangle$ has order $9$. So this quotient group is cyclic!

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    $\begingroup$ Note that what you really need is that $\langle(4,3)\rangle$ does not contain any element of the form $(a,a)$ or $(a,a+9)$ (see my answer). $\endgroup$ – Greg Martin May 30 '19 at 16:01
  • $\begingroup$ Could you explain why, if G was cyclic, the number of order 3 elements is 2? Also, another question, is $Z_n$ x $Z_m$ cyclic if and only if $(m,n)=1$? $\endgroup$ – Florin1335 May 30 '19 at 16:25
  • $\begingroup$ See theorem 4.4 of this file. (google.com/url?sa=t&source=web&rct=j&url=https://…) $\endgroup$ – Chinnapparaj R May 30 '19 at 16:45

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