Let $p$ be a prime. How to make an exact sequence of $\mathbb Z$-modules (abelian groups):
$$0\longrightarrow \mathbb Z_p\stackrel{f}{\longrightarrow} \mathbb Z_{p^2}\stackrel{g}{\longrightarrow} \mathbb Z_{p^2}\stackrel{h}{\longrightarrow} \mathbb Z_p\longrightarrow 0.$$
Since $\mathbb Z_{p}\subset \mathbb Z_{p^2}$ I thought taking $f$ as the inclusion. And I tried definying $h$ as: $$[a]_{p^2}\longmapsto [a]_p.$$ This is cleary surjective and its kernel is given by $p\mathbb Z_{p^2}$. The natural choice for $g$ would be $$[a]_{p^2}\longmapsto [pa]_{p^2}.$$ Then $\textrm{im}(g)=p\mathbb Z_{p^2}$ but it seems its kernel would also be $p\mathbb Z_{p^2}$ and not $\mathbb Z_p$ as desired.
Can anyone help me out?
Thanks
Along the lines of what you suggested, take the sequence $$0\longrightarrow p\mathbb Z/p^2\mathbb{Z}\stackrel{}{\longrightarrow} \mathbb Z/p^2\mathbb{Z}\stackrel{g}{\longrightarrow} \mathbb Z/{p^2}\mathbb{Z}\stackrel{h}{\longrightarrow} p\mathbb Z/p^2\mathbb{Z}\longrightarrow 0$$ with $g, h$ each multiplication by $p$.