# For which abelian groups $G$ is there a short exact sequence $0 \rightarrow \mathbb{Z}/p^2 \rightarrow G \rightarrow \mathbb{Z}/p^2 \rightarrow 0$?

I am trying to find for which abelian groups $$G$$ is there a short exact sequence.

$$0 \rightarrow \mathbb{Z}/p^2 \rightarrow G \rightarrow \mathbb{Z}/p^2 \rightarrow 0$$?

I have reasoned as follows: consider the sequence with the maps as follows $$0 \xrightarrow[]{\phi} \mathbb{Z}/p^2 \xrightarrow[]{f} G \xrightarrow[]{g} \mathbb{Z}/p^2 \xrightarrow[]{\psi} 0$$

Since we want the sequence to be exact, we need that $$\ker f= im(\phi) = 0$$ (so $$f$$ is injective) and we need that $$im(g) = \ker (\psi) = \mathbb{Z}/p^2$$ (so $$g$$ is surjective). On top of that, we know that $$im( f) = \ker g$$. Then, by the first isomorphism theorem, we know that $$\mathbb{Z}/p^2 / \ker f = \mathbb{Z}/p^2 \cong im(f) = \ker g$$. So we know $$\ker g = \mathbb{Z}/p^2$$ and $$im(g) = \mathbb{Z}/p^2$$. This leads me to think that one possibility for $$G$$ is $$\mathbb{Z}/p^2 \oplus \mathbb{Z}/p^2$$. However, the answers say that other possibilities could be $$\mathbb{Z}/p^4$$ and $$\mathbb{Z}/p^3 \oplus \mathbb{Z}/p$$ and I don't see how they came with this possibility, can anyone explain this solution? Also, is my argument for why $$G$$ can be $$\mathbb{Z}/p^2 \oplus \mathbb{Z}/p^2$$ correct? Thanks for your help!

• Use the classification of f.g. abelian groups. – anomaly Dec 27 '18 at 1:51

This short exact sequence is a fancy way of saying that $$G$$ has a subgroup $$H$$ such that $$H$$ and $$G / H$$ are both isomorphic to $$\mathbb Z / p^2$$.
[Think of $$f$$ as the embedding of $$H$$ into $$G$$, and think of $$g$$ as the projection of $$G$$ onto $$G/H$$. It's clear that $$f$$ is injective, and $$g$$ is surjective, and $${\rm im}(f) = {\rm ker}(g)$$, which is what your short exact sequence says.]
• $$G = \mathbb Z / p^2 \oplus \mathbb Z / p^2$$, with $$H$$ generated by $$(1, 0)$$.
• $$G = \mathbb Z / p^4$$, with $$H$$ generated by $$p^2$$.
• $$G = \mathbb Z / p^3 \oplus \mathbb Z / p$$, with $$H$$ generated by $$(p, 1)$$.