I'm trying to get the hang of factor group computations so I made up a problem for myself to try to solve. For practice, I tried computing $(\mathbb{Z}_4\times \mathbb{Z}_8)/\langle(2,4)\rangle$ using the Fundamental Theorem of Finitely Generated Abelian Groups. Here's my solution:
First, we see that $\langle(2,4)\rangle=\{(2,4),(0,0)\}$, so $|\langle(2,4)\rangle|=2.$ Therefore, we have that
$$|(\mathbb{Z}_4\times \mathbb{Z}_8)/\langle(2,4)\rangle|=\frac{|\mathbb{Z}_4\times\mathbb{Z}_8|}{|\langle(2,4)\rangle|}=\frac{32}{2}=16, $$
so according to the Fundamental Theorem, our factor group $(\mathbb{Z}_4\times \mathbb{Z}_8)/\langle(2,4)\rangle$ is isomorphic to $\mathbb{Z}_{16},\mathbb{Z}_2\times\mathbb{Z}_8,$ or $\mathbb{Z}_4\times\mathbb{Z}_4$. Consider the coset $(0,1)+\langle(2,4)\rangle$. This coset has order 8 since $8(0,1)=(0,0).$ Since $1$ generates $\mathbb{Z}_8,$ $(0,1)+\langle(2,4)\rangle$ is the element with the largest possible order in $(\mathbb{Z}_4\times \mathbb{Z}_8)/\langle(2,4)\rangle$. Since $\mathbb{Z}_2\times\mathbb{Z}_8$ is the only group out of the possibilites presented whose elements have a maximum order of $8$, we conclude $(\mathbb{Z}_4\times \mathbb{Z}_8)/\langle(2,4)\rangle\cong\mathbb{Z}_2\times\mathbb{Z}_8.$
Did I get the right answer, and is my thought process correct here?
Best Answer
Your proof can be made more direct. It is easy to check that for each $(a,b) \in \mathbb{Z}/4×\mathbb{Z}/8$ the equation $8(a,b)=(8a,8b)=e$ holds. This implies that every element has order $8<16$ even in $\mathbb{Z}/4×\mathbb{Z}/8$. So conclude that every elenent in your quotient group has order at most $8<16$ and thus as the quotient group has $16$ elements, it cannot be cyclic.