My question is simple: How to compute the quotient group $$\frac{\langle(1,1,0),(0,0,1)\rangle}{\langle(1,1,-1),(1,1,1)\rangle}$$ of subgroups of $\mathbb Z \oplus \mathbb Z \oplus \mathbb Z$? The answer would be graet if each of the steps are explained.
This is my try: Letting $v=(1,1,0)$ and $w=(0,0,1)$, then the problem is to compute $$\frac{\langle v,w\rangle}{\langle v+w, v-w \rangle}.$$ I wish this is isomorphic to the quotient $$\frac{\mathbb Z \oplus \mathbb Z}{\langle (1,1),(1,-1) \rangle} \simeq \mathbb Z_2.$$
How to verify this argument? Or is there any other good argument? I want to know as many ways as possible. The answer may contain any undergraduate level mathematics.
Thank you.
By definition of quotient:
\begin{alignat}{1} \frac{\langle(1,1,0),(0,0,1)\rangle}{\langle(1,1,-1),(1,1,1)\rangle} &= \{(i,i,j)+\langle(1,1,-1),(1,1,1)\rangle\mid i,j\in\Bbb Z\} \\ &= \{(i,i,j)+\{(k+l,k+l,-k+l)\mid k,l\in \Bbb Z\}\mid i,j\in\Bbb Z\} \\ &= \{\{(i+k+l,i+k+l,j-k+l)\mid i,j,k,l\in\Bbb Z\}\} \\ &= \{\{(m,m,j-m+2l+i)\mid i,j,l,m\in\Bbb Z\}\} \\ &= \{(m,m,j-m)+\{(0,0,2l+i)\mid i,l\in \Bbb Z\}\mid j,m\in\Bbb Z\} \\ &= \{(m,m,j-m)+\{(0,0,r)\mid r\in \Bbb Z\}\mid j,m\in\Bbb Z\} \\ &= \{(m,m,j-m)+\langle(0,0,1)\rangle\mid j,m\in\Bbb Z\} \\ &= \frac{\langle(1,1,-1),(0,0,1)\rangle}{\langle(0,0,1)\rangle} \end{alignat}
Now, $\varphi\colon \langle(1,1,-1),(0,0,1)\rangle\to \langle(1,1,0)\rangle$ defined by $(m,m,j-m)\mapsto (m,m,0)$ is a surjective homomorphism with kernel $\operatorname{ker}\varphi=\langle(0,0,1)\rangle$, whence (First Homomorphism Theorem):
$$\frac{\langle(1,1,-1),(0,0,1)\rangle}{\langle(0,0,1)\rangle}\cong \langle(1,1,0)\rangle$$
Finally, $\langle(1,1,0)\rangle\cong \Bbb Z_2$; in fact, consider the mapping $(m,m,0)\mapsto 0$ if $m$ is even and $(m,m,0)\mapsto 1$ if $m$ is odd.
(Remark. I wasn't smart enough to use the FHT directly on the first quotient, but it's certainly possible.)