What can I say about the quotient group?

Question

Let $G$ be a group of order $24$, and let $H$ be a normal subgroup of order $6$. So the quotient group $ {G\over H} $ is Abelian group?. What can I say about the quotient group beside her order?

Answer 1

The quotient group has order $24/6=4$. An abelian group of order $4$ is either $C_4$ or $C_2\times C_2$.

Answer 2

$G$ is a group of order $24$, and $H$ is a normal subgroup of order $6$. As such, $G/H$ is of order $4$. The fundamental theorem of finitely generated abelian groups tells you that this group is isomorphic to $\mathbf{Z}_4$ or $\mathbf{Z}_2 \times \mathbf{Z}_2$. (Note that each individual part is of prime power order, if you had an abelian group of order $12 = 2^2 \cdot 3$, then $\mathbf{Z}_4 \times \mathbf{Z}_3$ would be an alternative for isomorphism, while $\mathbf{Z}_6 \times \mathbf{Z}_2$ would not.)