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.