5
$\begingroup$

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!

$\endgroup$
1
  • 1
    $\begingroup$ Use the classification of f.g. abelian groups. $\endgroup$
    – anomaly
    Commented Dec 27, 2018 at 1:51

1 Answer 1

3
$\begingroup$

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.]

For example:

  • $G = \mathbb Z / p^2 \oplus \mathbb Z / p^2$, 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 $(p, 1)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .