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Suppose that $G$ is a finite Abelian group that has exactly one subgroup for each divisor of |$G|$. Show that $G$ is cyclic.
What I have so far:
By the Fundamental Theorem of Finite Abelian Groups, we may write $G$ as $G=Z_{n_1}\oplus\dots\oplus Z_{n_k}$ for a set of $k$ integers $n_1$ through $n_k$ that are prime.
If $m$ divides the order of a finite Abelian group $G$, then $G$ has a subgroup of order $m$.
I am not sure what else I need to know.
Hint: Show that $\mathbb Z_{p^a}\oplus \mathbb Z_{p^b}$ has two subgroups of the size $p$. Use this to prove the general theorem.
Alternatively, if $ m\mid n$ then there is more than one subgroup of size $m$ to $\mathbb Z_{m}\oplus \mathbb Z_{n}$.
I give you a hint : If $|G|=n<\infty$, then $G$ is cyclic if and only if $G \simeq \mathbb{Z}/n\mathbb{Z}$
The theorem you mention is not precisely stated: a finite abelian group can be written as $$ G\cong \mathbb{Z}/p_1^{n_1}\mathbb{Z}\oplus \mathbb{Z}/p_2^{n_2}\mathbb{Z}\oplus\dots\oplus \mathbb{Z}/p_k^{n_k}\mathbb{Z} $$ where $p_1,p_2,\dots,p_k$ are primes, not necessarily distinct, and $n_i>0$ $(i=1,2,\dots,k)$.
You must exclude the case that two of them are equal and then you're done, because $\mathbb{Z}/r\mathbb{Z}\oplus \mathbb{Z}/(rs)\mathbb{Z}$ if $r$ and $s$ are coprime.
Or you can decompose it as $$ G=\mathbb{Z}/m_1\mathbb{Z}\oplus \mathbb{Z}/m_2\mathbb{Z}\oplus \dots\oplus \mathbb{Z}/m_k\mathbb{Z} $$ with $m_1\mid m_2\mid\dots\mid m_k$. In this case you should exclude that $k>1$.
