# Show that $G$ = $(\mathbb Z_3 × \mathbb Z_4, +)$ is cyclic

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?

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.

• 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 $12$ – JOJO Sep 12 '19 at 11:28
• 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 $\mathbb{Z}/4\mathbb{Z}$. – TMO Sep 12 '19 at 11:42

Hint: What is the order of $$([1]_3, [1]_4)$$ ?

• Please don't downvote perfectly appropriate answer. – Ennar Sep 12 '19 at 10:48
• 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 $4$? – JOJO Sep 12 '19 at 11:14

you can see that, $$Z_3\times Z_4=<(1,1)>$$

• why is it so? and how can this help me? – JOJO Sep 12 '19 at 10:51
• use definition of cyclic subgroup generated by an element, $<a>$ – sabeelmsk Sep 12 '19 at 10:55
• and how can I find <$(1,1)$> ? I don't know how to apply the definition having 2 components which are 1,1 – JOJO Sep 12 '19 at 10:59
• 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, $d=12.$ – Kostas Psaromiligkos Sep 12 '19 at 11:02
• 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 $4$? – JOJO Sep 12 '19 at 11:16