New to group theory. Is the following correct? $a, b, c, d$ are independent elements generating the free abelian group $A = \langle a, b, c, d\rangle = \mathbb{Z}a \oplus \mathbb{Z}b \oplus \mathbb{Z}c \oplus \mathbb{Z}d$, and $B$ is the subgroup $B = \langle a, 2b-a, 2c-b, 2d-c\rangle$.
$$ A / B \cong \mathbb{Z}/2 \oplus \mathbb{Z}/4 \oplus \mathbb{Z}/8 $$
Thank you!
In $A/B$, we have $c=2d$, $b=2c=4d$, $a=2b=8d$, $a=0$.
Therefore, $A/B$ is generated by the class of $d$ and this class has order $8$. Thus, $A/B \cong \mathbb Z/8 \mathbb Z$.
An explicit isomorphism is induced by the map $A \to \mathbb Z/8 \mathbb Z$ given by $\alpha a + \beta b + \gamma c + \delta d \mapsto 4 \beta + 2\gamma + \delta \bmod 8$.