This is problem 2.13.12 in Herstein's Topics in Algebra, 2nd edition.
Let $G$ be a finite abelian group. Using Problem 11 show that $G$ is isomorphic to a subgroup of a direct product of a finite number of cyclic groups.
Problem 11 asked to prove that a finite abelian group with a subgroup contained in every nontrivial subgroup is cyclic. These problems are right before the proof of the fundamental theorem of finite abelian groups.
I have written a solution but it's so hand-wavy that it's possibly even wrong, so I wanted to know if it's correct nonetheless and if it's unnecessarily complicated.
Solution. Let $|G|=p_1^{a_1}\cdots p_n^{a_n}$ be the prime factorization of $|G|$. Write $G$ as the direct product of its Sylow subgroups $P_1, P_2, \dots, P_n$. For each $P_i$, if $P_i$ is cyclic we are done. Otherwise, by problem 11, $P_i$ has at least two distinct subgroups of order $p_i$, say $H$ and $K$. Now, by a previous problem, since $H$ and $K$ are normal and $H \cap K = \{ e\}$, $\:P_i$ is isomorphic to a subgroup of $P_i/H \times P_i/K$. Let's denote $P_i/H=:P_{i,2}'$ and $P_i/K=:P_{i,2}''$. If they are both cyclic, we are done. Else $P_{i,2}'$ or $P_{i,2}''$ is not cyclic (can one be cyclic and not the other?). We can repeat this process with the quotient groups and it must eventually stop at step $j<a_i$ (the exponent of $p_i$ in the prime factorization), with both $P_{i,j}'$ and $P_{i,j}''$ cyclic groups, for each $P_i$ is finite and with every step the order of $P_{i,x}$ decreases by a factor of $p_i$, so it will definitely stop when $|P_{i,x}|=p_i$. This shows that every $P_i$ is isomorphic to a subgroup of a direct product of cyclic groups, $M_i$. Then $G$, the direct product of the $P_i$'s, will be isomorphic to a subgroup of the direct product of the $M_i$'s.