# quotient group computation

$f: \mathbb Z \rightarrow \mathbb Z_{(2)} \oplus \mathbb Z$ (the sum is direct) $f(1)=(1,2)$ so the image is $\mathbb Z*(e_{1}+2e_{2})$ however computing the cokernel of this map really puzzles me (because of the torsion term) I first thought that it was simply $\mathbb Z_{(2)}$ because the rank of my image is 1 however I am not quite sure if that is right.

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There are two generators, $e_1\in \mathbb{Z}/2$ and $e_2\mathbb{Z}$, and neither is in the image. The question is, what relations do they have in the quotient? Because $e_1=-2e_2$, the subgroup of the quotient generated by $e_2$ contains $e_1$, and thus, the quotient is cyclic. What is the order of the group? Since $4e_2=-2e_1=0$, the order divides $4$, and it should be clear that the order is not $2$. Therefore the quotient is $\mathbb Z/4$