I have a group theory question from Rotman's homological algebra textbook. The question asks:
Let $$0\rightarrow B \rightarrow A \rightarrow \mathbb{Z}_p\rightarrow 0$$ be an exact sequence of finite abelian p-groups for $p$ prime (Here, $\mathbb{Z}_p$ is the cyclic group of order $p$). If $B$ is cyclic, prove that either $A$ is cyclic or the sequence splits.
So far, it's easy to see that if $i:B\rightarrow A$ and $q:A\rightarrow \mathbb{Z}_p$ are the maps and if $x$ generates $B$, $i(x)$ generates $\ker q$, since if $a\in \ker q=\hbox{im } i$, $a=i(b)=i(x^m)=i(x)^m$ for some integer $m$. Further, if $|B|=|x|=p^l$ for some $l$, then:
$$|A/\ker q|=|\mathbb{Z}_p|=p$$ $$\Rightarrow p=|A|/|\ker q|=|A|/|\hbox{im }i|=|A|/|B| \Rightarrow |A|=p^{l+1}$$
I'm not even sure if this last fact is useful, but it's about as far as I've gotten. Any hints would be greatly appreciated!
You can think of $B$ as a subgroup of $A$. Then $B$ is cyclic of order $p^k$ with generator $b$ say. Now, $A/B$ has order $p$ and has a generator $\overline a$ with $a\in A$. Then $pa\in B$, so that $pa=rb$ where $0\le b<p^k$. If $r$ is coprime to $p$ then $rb$ is a generator of $B$ and then $A=\langle a,b\rangle=\langle a\rangle$ is cyclic. Otherwise $pa=psb$ for some $s$ and then $p(a-sb)=0$. Let $a'=a-sb$. Then $\overline a\mapsto a'$ is a splitting map from $A/B\cong\Bbb Z_p$ to $A$.