Let G be a finite Abelian group which contains a proper subgroup H which in contained in every proper subgroup of G. Prove G is cyclic. order of G Does it have anything to do with since G is finite abelian group it is a direct product of its slow subgroup? 
 A: I don't know if the Structure Theorem for Finitely Generated Abelian Groups is "on the table", so to speak, but this answer uses it regardless.
Furthermore, I assume that you meant for $H$ to be nontrivial, instead of proper -- otherwise the statement to prove is not even true.
By that theorem, your finite group $G$ is a direct product of cyclic groups $G_1, G_2, \ldots, G_k$: we have
$$G = G_1 \times G_2 \times \cdots \times G_k.$$
Now suppose that $k \geq 2$ and we have some subgroup $H \leq G$ which is contained in every (proper) subgroup of $G$. Well, $$G_1 \times \{1\} \times \ldots \times \{1\}$$ is a subgroup of $G$, and therefore must contain $H$; but also
$$\{1\} \times \ldots \times \{1\} \times G_k$$ is a subgroup of $G$, and thus contains $H$. But the two subgroups $G_1 \times \{1\} \times \ldots \times \{1\}$ and $\{1\} \times \ldots \times \{1\} \times G_k$ intersect trivially, so $H$ must be trivial.
So we see that if a nontrivial $H \leq G$ is contained in every subgroup of $G$, then necessarily $k = 1$, since otherwise by the argument above we would have $H$ trivial, a contradiction. So $G = G_1$, which is a cyclic group.
A: Suppose $G$ isn't cyclic. Then, for $a \in G : a \ne 1$ (multiplicative notation) $\exists b \in G : b \notin \langle a \rangle$ and trivially $b \ne 1$. Then $\{1\} < \langle b \rangle < G$ (we're assuming $G$ isn't cyclic). So 
$$\exists H: H \le \langle a \rangle, H \le \langle b \rangle, 1 < H < G$$
But then $H \le \langle a \rangle \cap \langle b \rangle = \{1\}$ (prove it) $\implies H = \{1\}$
A: The lattice of subgroups of $G$ is self dual. Hence there is also a proper subgroup $H'$ of $G$ that contains every other proper subgroup.
Now take an element $a$ of $G$ not in $H'$.  The subgroup generated by $a$ is not contained in $H'$ so must be $G$, so $G$ is cyclic with generator $a$.
