What choice of $X \in \mathrm{Ab}$ and maps between the groups would make the following sequence exact? $$0 \rightarrow \mathbb{Z}/3 \rightarrow X \rightarrow\mathbb{Z}/2 \rightarrow 0$$
I'm thinking either (a) $X=\mathbb{Z}/3$ or (b) $X=\mathbb{Z}/6$, but in the first case I don't know what the inclusion would look like, and in the second case what either the inclusion or projection would look like. Is for example $x+(3) \mapsto x+(6)$ well-defined? I don't even know how to test that.
You need homomorphisms; the map that takes $x+(3)$ to $x+(6)$ is not a homomorphism. What elements of $\Bbb Z/6$ have order $3$? Those are the candidates for the image of $1+(3)$ and $2+(3)$, and there’s a very nice way to map $1+(3)$ and $2+(3)$ to them. For the projection you should try imitating the natural map from $\Bbb Z$ to $\Bbb Z/2$.
There is always the possibility to choose the direct product to make such a sequence of abelian groups exact: $0\rightarrow A \rightarrow A\times B\rightarrow B \rightarrow 0$. This is $\mathbb{Z}/6\simeq \mathbb{Z}/3\times \mathbb{Z}/2$ in this case. The inclusion is $a\mapsto (a,0)$ and the projection is $(a,b)\mapsto b$.
