This is Problem 2.13.10 from Herstein, Topics in Algebra:
Let $G$ be a finite abelian group such that it contains a subgroup $H_0 \neq (e)$ which lies in every subgroup $H\neq (e)$. Prove that $G$ must be cyclic. What can you say about $o(G)$?
I have a proof, but would appreciate if someone could check its validity.
Let $|H_0| = p$. For any $x\in H_0$, we know $\langle x\rangle \le H_0 \le \langle x \rangle$. It follows that there are $p-1$ generators for $H_0$, and so $p$ is prime.
Suppose some prime $q$ other than $p$ divided $|G|$. Then by Cauchy's theorem, there is an element of $G$ with order $q$, hence a subgroup of $G$ of order $q$. But $H_0$ must be contained in this subgroup, contradiction. Thus $|G|=p^n$ for some $n$.
Now, suppose there is $y\in G\setminus H_0$ with order $p$. Then $H_0 \le \langle y\rangle$; since $|H_0|=p$, we then have $\langle y \rangle = H_0$ and $y\in H_0$, contradiction. Thus $H_0$ is precisely the set of elements in $G$ with order $p$.
Since $G$ is abelian, $g\mapsto g^p$ is a homomorphism; let $I_1$ be its image. Then the kernel of this homomorphism is $H_0$, so $|I_1| = p^{n-1} > 1$. Thus $H_0 < I_1$. So for $x$ a generator of $H_0$, there exists $g$ such that $x=g^p$. The order of $g$ is then $p^2$.
Let $H_1$ be the set of elements in $G$ with order dividing $p^2$. Evidently $H_0 \le H_1$. However, we also found $g \not\in H_0$ with order $p^2$, so $|H_1|\ge p^2$. Now suppose $h\in G$ has order $p^2$. Then $h^p$ has order $p$, so $h^p \in H_0$. If $h^p \in H_0$, then $h^p = x^i = g^{ip}$ for some $1\le i \le p-1$ (since $h^p$ has order $p$, it is not $e$). Then we see $(hg^{-i})^p = e \implies hg^{-i} \in H_0$, so $h = x^j g^i$ for some $1\le j\le p$. There are exactly $p(p-1) = p^2 - p$ such $h$; accounting for the $p$ elements of $H_0$ as well, we see $|H_1| = p^2$.
Again, we note $g\mapsto g^{p^2}$ is a homomorphism, with kernel $H_1$, so its image $I_2$ has $|I_2| = p^{n-2}>1$. Thus, $H_1<I_2$ and we can find a base element with order $p^3$. Continuing this process inductively proves $|I_{n-1}| = p$. It follows there is $y\in G$ with $x = y^{p^{n-1}}$. The order of $y$ must be $p^k$ for some $1\le k\le n$. If $k\le n-1$, then $x=y^{p^{n-1}} = e$, contradiction. So $k=n$ and $y$ generates $G$.
