Find the order of factor group The professor provided a simple example, such as the order of $\Bbb Z _6 / \langle 3 \rangle$: $|\Bbb Z _6| = 6$ and $| \langle 3 \rangle | = 2$, so $|
\Bbb Z _6 / \langle 3 \rangle | = 3$.
I thought "well, I got!" but I started practicing the topic "factor group" and I found something like this:

find the order of factor group $(\Bbb Z _{12} \times \Bbb Z _4) / (\langle 2 \rangle \times \langle 2 \rangle)$.

I suppose that we multiply $12 \times 4$ because of $\Bbb Z _{12} \times \Bbb Z _4$ right? Do we do the same thing with the other side?
He provided this formula: $|\Bbb Z _n / \langle a \rangle| = n / |\langle a \rangle| = \dfrac n {\frac n {\gcd(n,a)}}$.
So, does $\Bbb Z _{12} \times \Bbb Z _4$ become $\Bbb Z _n$, and $\langle 2 \rangle \times \langle 2 \rangle$ become $\langle a \rangle$?
 A: The order of $2$ in $\Bbb Z _{12}$ is $6$ (because $2 + 2 + 2 + 2 + 2 + 2 = 0$), so $| \langle 2 \rangle | = 6$. Similarly, $| \langle 2 \rangle | = 2$ in $\Bbb Z _4$. Therefore, the order of $\langle 2 \rangle \times \langle 2 \rangle$ in $\Bbb Z _{12} \times \Bbb Z _4$ is $6 \cdot 2 = 12$.
The order of $\Bbb Z _{12} \times \Bbb Z _4$ is $12 \times 4$ (as you were suspecting), so by Lagrange's theorem the order of the factor group $(\Bbb Z _{12} \times \Bbb Z _4) / \langle 2 \rangle \times \langle 2 \rangle)$ is the quotient of the respective orders, that is $\dfrac {48} {12} = 4$.
In this extremely simple example the quotient group is isomorphic to $\Bbb Z _{12} / \langle 2 \rangle \times \Bbb Z _4 / \langle 2 \rangle = \Bbb Z_2 \times \Bbb Z_2$, but keep in mind that in general it is not true that $(A \times B) / (I \times J) = A/I \times B/J$.
A: We have:
$$(\mathbf Z_{12}\times\mathbf Z_4)/(2\mathbf Z_{12}\times2\mathbf Z_4)\simeq\mathbf Z_{12}/2\mathbf Z_{12}\times\mathbf Z_4/2\mathbf Z_4\simeq\mathbf Z_2\times\mathbf Z_2,$$
by the third isomorphism theorem, hence the order of the quotient is $4$.
