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Let $G$ = $(\mathbb Z_3 × \mathbb Z_4, +)$ with the operation defined somewhat like vector addition: for $a, a'$$\mathbb Z_3$ and $b, b'$$\mathbb Z_4$,

$(a, b) + (a', b') = (a +_3 a', b +_4 b')$. For example, $(2, 2) + (1, 3) = (0, 1)$.

Show that $G$ is cyclic.

Note: convince yourself that G is indeed a group and that $|G| = 12$.

I know that for $x$ in $G$, the smallest subgroup of $G$ that contains $x$ is $\langle x \rangle = \{\ldots,x^{-3},x^{-2},x^{-1},1,x,x^2,x^3,\ldots\} $

And so we say $G$ is cyclic if $\langle x \rangle$ = $G$ for some $x$.

And \begin{align}\mathbb Z_3 × \mathbb Z_4 &= \{(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3),(2,0),(2,1),(2,2),(2,3)\}\\ &= \mathbb Z_{12}\end{align}

But how can I apply this definition in my problem? I mean in my case, do I have to show that $\langle(a +_3 a', b +_4 b')\rangle$ = $G$? And if so, how do I do that? Any help please?

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Try to show that your group is generated by $(1,1) \in G$, i.e. $G = \langle (1,1) \rangle$. The group $\langle (1,1) \rangle$ consists of all multiples (we have additive groups here) of $(1,1)$. Thus, if you want to work very elementary, you can just compute them:

$2(1,1) = (2,2),\; 3(1,1) = (0,3),\; 4(1,1) = (1,0)$, ...

Just pay attention that you are using the addition mod $3$ in the first argument and mod $4$ in the second arguement. You will manage to get all elements from your group $G$ in that way, which shows that $G \subset \langle (1,1) \rangle$. As we also have $\langle (1,1) \rangle \subset G$, we have an equality, such that $G$ is cyclic.

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  • $\begingroup$ yes, thank you, I get that. but how can I show that $(1,0)$ has order $3$ and $(0,1)$ has order $4$? And hence $(1,1)$ would have order $12$ $\endgroup$ – JOJO Sep 12 '19 at 11:28
  • $\begingroup$ Just compute the mulitples of $(1,0)$ and $(0,1)$. Then you will see that the smallest positive number $m$, such that $m(1,0) = (0,0)$ will be $3$ and respectively $4$ for $(0,1)$. That just relies on the orders of $1$ in $\mathbb{Z}/3\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$. $\endgroup$ – TMO Sep 12 '19 at 11:42
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Hint: What is the order of $([1]_3, [1]_4)$ ?

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    $\begingroup$ Please don't downvote perfectly appropriate answer. $\endgroup$ – Ennar Sep 12 '19 at 10:48
  • $\begingroup$ I know that it has to be of order $12$, because $(1,0)$ has order $3$ and $(0,1)$ has order $4$. But how can I show that $(1,0)$ has order $3$ and $(0,1)$ has order $4$? $\endgroup$ – JOJO Sep 12 '19 at 11:14
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you can see that, $Z_3\times Z_4=<(1,1)>$

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  • $\begingroup$ why is it so? and how can this help me? $\endgroup$ – JOJO Sep 12 '19 at 10:51
  • $\begingroup$ use definition of cyclic subgroup generated by an element, $<a>$ $\endgroup$ – sabeelmsk Sep 12 '19 at 10:55
  • $\begingroup$ and how can I find <$(1,1)$> ? I don't know how to apply the definition having 2 components which are 1,1 $\endgroup$ – JOJO Sep 12 '19 at 10:59
  • $\begingroup$ Let $x=(1,1)$. Let $d$ be its order. Then $d\mid 12$ sicne the order of the group is 12. If $d$ was less than $12$, you can see that you would have either $x^3=0$ or $x^4=0$ (exercise, tell me if you have a problem with this). But neither is true, $x^3=(0,3), x^4=(1,0)$. Therefore, $d=12.$ $\endgroup$ – Kostas Psaromiligkos Sep 12 '19 at 11:02
  • $\begingroup$ I know that $(1,1)$ has to be of order $12$, because $(1,0)$ has order $3$ and $(0,1)$ has order $4$. But how can I show that $(1,0)$ has order $3$ and $(0,1)$ has order $4$? $\endgroup$ – JOJO Sep 12 '19 at 11:16

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