Is the sequence of $\mathbb Z$-modules below exact?

Question

I came across the following example in my algebra textbook:

Considere the sequence of $\mathbb Z$-modules: $$0\longrightarrow \mathbb Z_2\stackrel{f}{\longrightarrow}\mathbb Z_4\stackrel{g}{\longrightarrow}\mathbb Z_2\longrightarrow 0,$$

where $f$ is the isomorphism of $\mathbb Z_2$ and the unique proper subgroup $2\mathbb Z_4$ of $\mathbb Z_4$ and $g$ is the multiplication by $2$. This is an exact short sequence.

I guess there is something strange here for if the sequence were exact $\textrm{im}(g)=\mathbb Z_2=\{\overline{0}, \overline{1}\}$ but if there were $\overline{a}\in \mathbb Z_4$ such that $2\overline{a}=\overline{1}$ in $\mathbb Z_2$ then $2|2a-1$ what certainly can't happen.

Am I missing something?

Reference: Algèbre et Modules (I. Assem, pg. 31).

Answer 1

The only way to get an exact sequence there is to have $g$ = reduction mod $2$.

"Multiplication by 2" is the description of $f$, not $g$.

Answer 2

It seems to me that the second $\Bbb Z_2$ is actually $2\Bbb Z_4$, that is, it is:

$\{[0]_4,[2]_4\}$ (this is isomorphic to $\Bbb Z_2$).

It should be clear then, that $gf = 0$, as one would expect in an exact sequence.