Let $G$ be the group $G=\mathbb{Z/6Z} \times \mathbb{Z/4Z}$. I am asked:
-How many elements does G have?
-Determine every element of order 12 and 4.
I know $G$ has $24$ elements, the elements of order $12$ are $(1,1),(1,3),(2,1),(2,3),(4,1),(4,3),(5,1),(5,3)$.
And the elements of order 4: $(0,1),(0,3),(3,1),(3,3)$
Now if $a=(2,1)$, $b=(0,1)$ and $H=\langle a \rangle \cap \langle b \rangle$, to which product of cyclic groups is isomorphic $G/H$?
To solve that, first I need to know how many elements does $G/H$ has, and I don't even know how many elements $H$ has. So how can I determine the elements of $H$? Once I have the number of elements in $G/H$, how can I know tho which product of cyclic groups is isomorph to?
Edit: I think that, by Lagrange's theorem, the elements of $H$ should be $1$, $2$ or $4$. Since $H$ is a subgroup of $\langle a \rangle$ and a subgroup of $\langle b \rangle$.