Why is $\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8 /\left<(1,2,4)\right>$ isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_8$?

Question

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 $\mathbb{Z}_4 \times \mathbb{Z}_8$.

Answer 1

$(0,0,1)$ 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 $\mathbb Z_4\to\mathbb Z_8$?)

Answer 2

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.

Answer 3

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 $$ \mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8 /\langle(1,2,4)\rangle = \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} / \langle v,4e_1,4e_2,8e_3 \rangle = \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} / \langle v,4e_2,8e_3 \rangle \cong \mathbb{Z}_4 \times \mathbb{Z}_8 $$