A group with bounded element orders and its minimal and maximal subgroups. Let $n>1$ be an integer. Is there an abelian group $G$ with all elements of order less than $n$ for which exactly one of these conditions is correct:
1) every non-trivial subgroup of $G$ contains a minimal (non-trivial) subgroup of $G$.
2) every proper subgroup of $G$ is contained in a maximal (proper) subgroup of $G$.
?
How if abelianness is omitted?
 A: Every abelian group of bounded exponent satisfies (1) and (2).
As I pointed out in my comment, every torsion group satisfies (1), because every nontrivial subgroup contains a cylic group of prime order, which is clearly minimal.
So let $G$ be an abelian group of exponent $n$. Let $H$ be a proper subgroup. Then there exists $g \in G \setminus H$, and by a standard application of Zorn's Lemma, we can find a subgroup $N$ of $G$ with $H \le N$ and $N$ maximal subject to $g \not\in N$.
So, in the quotient group $G/N$, $xN$ is contained in every nontrivial subgroup. A finitely generated abelian group with that property must be cyclic, so $G/N$ is locally cyclic, and hence all finitely generated subgroups of $G/N$ are cyclic of order at most $n$. But that implies immediately that $G/N$ itself is cyclic of order at most $n$, so it has a maximal subgroup and hence (2) holds.
I don't know whether (2) holds for all groups of exponent $n$ if you drop the abelian condition - I would guess not, but I don't knoe an example.
