Why is $\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8 /\left<(1,2,4)\right>$ isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_8$? I understand that this is isomorphic to either $\mathbb{Z}_4 \times \mathbb{Z}_8$ or $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_8$, but I can't think of any way to show that it is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_8$.
 A: $(0,0,1)$ goes to an element of order $8$. $(0,1,0)$ goes to an element of order $4$ which is not a multiple of $(0,1,1)$. Every element of the quotient is the image of a unique element of the form $(0,b,c)$.
Essentially, you want a map $$\mathbb Z_4\times\mathbb Z_4\times\mathbb Z_8\to \mathbb Z_4\times\mathbb Z_8$$ which is onto and has the kernel $\langle(1,2,4)\rangle$.
So send $(a,b,c)\mapsto (b-2a,c-4a)$. (Why does $4a$ make sense sending $\mathbb Z_4\to\mathbb Z_8$?)
A: You can also play directly with the definitions:
\begin{alignat}{1}
&\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8 /\langle(1,2,4)\rangle= \\
&= \{(a,b,c)+\left<(1,2,4)\right>\mid a,b,c\in \Bbb Z \} \\
&= \{(a,b,c)+\{(l,2l,4l)\mid l\in\Bbb Z\}\mid a,b,c\in \Bbb Z \} \\
&= \{(a+l,b+2l,c+4l)\mid a,b,c,l\in\Bbb Z \} \\
&\stackrel{m:=a+l}{=} \{(m,b+2m-2a,c+4m-4a)\mid a,b,c,m\in\Bbb Z \} \\
&= \{\{(m,2m,4m)\mid m\in \Bbb Z\}+(0,b-2a,c-4a)\mid a,b,c\in \Bbb Z \} \\
&= \{(0,b-2a,c-4a)+\langle(1,2,4)\rangle\mid a,b,c\in \Bbb Z \} \\
\end{alignat}
Therefore, there is a complete set of representatives, isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_8$. So $\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8 /\langle(1,2,4)\rangle$ is, too.
A: In $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$, let $v=(1,2,4)$, $e_1=(1,0,0)$, $e_2=(0,1,0)$, $e_3=(0,0,1)$.
Then $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} = \langle v,e_2,e_3 \rangle$ because $e_1= v-2e_2-4e_3$.
Thus $\langle v,4e_1,4e_2,8e_3 \rangle = \langle v,4e_2,8e_3 \rangle$ and so
$$
\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8 /\langle(1,2,4)\rangle
=
\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} / \langle v,4e_1,4e_2,8e_3 \rangle = 
\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} / \langle v,4e_2,8e_3 \rangle
\cong
\mathbb{Z}_4 \times \mathbb{Z}_8
$$
