Let $\mathbb{Z}_4$ and $\mathbb{Z}_{12}$ be the finite cyclic groups with $4$ and $12$ elements, respectively, and let $\langle(2,2)\rangle$ be the finite cyclic subgroup of $\mathbb{Z}_4\times \mathbb{Z}_{12}$ generated by $(2,2)$. How many elements does the quotient group $\dfrac{\mathbb{Z}_4\times \mathbb{Z}_{12}}{\langle(2,2)\rangle}$ have?

I know how to do this problem if $4$ and $12$ were relatively prime, but in this case, the only idea I have is to make all the $24$ possibilities and count them, so how can I do this problem easier?


If $G$ is a finite group and $H$ is a normal subgroup of $G$, then the quotient group $G/H$ has order $[G:H]=\frac{|G|}{|H|}$.

In this case $|G|=4\cdot 12=48$. Since $H$ is cyclic, its order is equal to the order of $(2,2)$ in $G$, which is the LCM of the orders of $2$ in $\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}/12\mathbb{Z}$, namely $6$.

Therefore the quotient group has order $\frac{48}{6}=8$.


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