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Given the group $G=\mathbb{Z/6\mathbb{Z}}\times\mathbb{Z}/\mathbb{4Z}$, I know $G$ has $24$ elements and $(2,1),(2,3),(4,1),(4,3)$ have order $12$ and $(0,1),(0,3),(3,1),(3,3)$ have order $4$. Now, if I choose one element $a$ of order $12$ and one element $b$ of order $4$, how can I determine the elements of $$H:=\langle a \rangle \cap \langle b \rangle$$

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Hint:

Let $c\in H$. Then $c\in \left <a\right >$ and also $c\in \left <b\right >$. What is the order of $c$?

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