# Abelian Group In The Middle Of A Short Exact Sequence

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 $$A$$?

• $A$ need not be cyclic. Consider, for example, the case that $p=2$ and $a=b=1$. Feb 2, 2019 at 3:25
• So I guess all we can conclude is that $A$ is isomorphic to a product of cyclic groups whose orders are powers of $p$. Feb 2, 2019 at 5:33
• 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. Feb 2, 2019 at 15:17
• @Shaun I edited my post so that it contains the text. Feb 2, 2019 at 20:14
• 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. Feb 2, 2019 at 20:45

The only such groups will be $$Z/(p^{a+b})$$ and $$Z/(p^{a})\times Z/(p^{b})$$. To obtain this, one can use the two comments of @DerekHolt, and showing that considering $$a\leq b$$, $$i$$ has to be either $$0$$ or $$a$$, in the first comment of @DerekHolt. The problem for other $$i$$'s can be understood from the counterexample $$p=2,a=2,b=3,i=1$$.
• This is false. Consider the subgroup of $C_8\times C_2$ generated by $(2,1)$ and the quotient by this subgroup.