When an Abelian group is cyclic

Question

Let G be a finite abelian group.It contains a non trivial subgroup which is contained in every non trivial subgroup.Then G must be cyclic. This is a problem of Herstein book(Pg 108,#11 2nd edition).I can't solve it Plz help me.

Answer 1

Suppose that $H$ is that unique minimal subgroup of $G$. Then for any $g\in G$:

1. $H\subset \langle g\rangle$ implying $H$ is cyclic,
2. Since $H$ is minimal $|H|=p$ is a prime,
3. $\langle g\rangle$ has order $p^n$ for some $n$,
4. $H=\langle g^{p^{n-1}}\rangle$

Now suppose that $h$ is an element of $G$ with the maximal order $ord(h)=p^m$, then $H=\langle h^{p^{m-1}}\rangle$. For any $g\in G$ with order $p^n\le p^m$, $g^{p^{n-1}}$ is a generator of $H$. So we can assume that $$g^{p^{n-1}}=h^{p^{m-1}}.$$ It follows that $$(gh^{-p^{m-n}})^{p^{n-1}}=1.$$ If $gh^{-p^{m-n}}\ne 1$, then it generates a cyclic group of order at most $p^{n-1}$. This group again contains $H$, we can then proceed by reduction to show that $g$ is a power of $h$.