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Let $p$ be a prime number. Determine all isomorphism classes of abelian groups $A$ that can appear as the middle term of a short exact sequence: $$0 \rightarrow \mathbb{Z}/(p^a) \rightarrow A \rightarrow \mathbb{Z}/(p^b) \rightarrow 0$$

Here is my attempt at this problem:


By the exactness of the sequence, $f: \mathbb{Z}/(p^a) \hookrightarrow A$, $g: A \twoheadrightarrow \mathbb{Z}/(p^b)$, and $Im(f) = Ker(g)$, so $Im(f) \cong \mathbb{Z}/(p^a)$, and thus by the First Isomorphism Theorem, $A/Ker(g) = A/Im(f) \cong A/(\mathbb{Z}/(p^a)) \cong Im(g) = \mathbb{Z}/(p^b)$. Since $\mathbb{Z}/(p^a)$ and $\mathbb{Z}/(p^b)$ are finite groups, $A$ is a finite group and hence finitely generated. Thus by Lagrange's Theorem, $|A| = |\mathbb{Z}/(p^a)||\mathbb{Z}/(p^b)| = p^{a+b}$. By the Fundamental Theorem of Finitely Generated Abelian Groups, $\displaystyle A \cong \mathbb{Z}^r \times \prod_{i = 1}^s \mathbb{Z}/(n_i)$ for some $r,s \geq 0$, $n_i \geq 2$, and $n_{i+1}|n_i$ for all $1 \leq i \leq s - 1$. Since $A$ is finite, $r = 0$ and $\displaystyle |A| = p^{a+b} = \prod_{i = 1}^s |\mathbb{Z}/(n_i)| = \prod_{i = 1}^s n_i$. Thus, the prime decomposition of each $n_i$ cannot have any primes other than $p$, so each $n_i$ is some power $m_i$ of $p$ so that $\displaystyle \sum_{i = 1}^s m_i = a+b$.


I want to prove that $A \cong \mathbb{Z}/(p^{a+b})$ but I am stuck at the last step of this proof. What we can conclude from the Fundamental Theorem is that $A$ is isomorphic to a product of cyclic groups of $p$-power order, but how can we prove that there is only one cyclic group factor in the factorization of $A$?

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  • $\begingroup$ $A$ need not be cyclic. Consider, for example, the case that $p=2$ and $a=b=1$. $\endgroup$ – Andreas Blass Feb 2 at 3:25
  • $\begingroup$ So I guess all we can conclude is that $A$ is isomorphic to a product of cyclic groups whose orders are powers of $p$. $\endgroup$ – Frederic Chopin Feb 2 at 5:33
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    $\begingroup$ In future, please do not rely on pictures of text and instead type up or copy & paste such things here. It makes them easier to search for and, for some people, much easier to see. $\endgroup$ – Shaun Feb 2 at 15:17
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    $\begingroup$ @Shaun I edited my post so that it contains the text. $\endgroup$ – Frederic Chopin Feb 2 at 20:14
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    $\begingroup$ Since $A$ has a normal subgroup $N$ such that both $N$ and $A/N$ are cyclic, $A$ can be generated by two elements and so it can have at most two nontrivial direct factors. $\endgroup$ – Derek Holt Feb 2 at 20:45

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