Finite abelian groups whose subgroups of the same order are all isomorphic I am trying to find describe a finite abelian group whose subgroups of the same order are isomorphic. That is to say, if $H$ and $K$ are two subgroups of $G$ with $|H|=|K|$, then $H\cong K$.
My progress so far:
Let us first consider the case when $G$ is an abelian $p$-group of order $p^n$. By the fundamental theorem,
$$G\cong \mathbb{Z}_{p^{e_1}}\times\mathbb{Z}_{p^{e_2}}\times\cdots\times\mathbb{Z}_{p^{e_k}}$$
where $\sum_{j=1}^k e_k=n$. We claim that no exponent $e_i$ exceeds 1. Indeed, if there were two summands $\mathbb{Z}_{p^{e_1}}$ and $\mathbb{Z}_{p^{e_2}}$ such that $e_1,e_2>1$, then we can find subgroups (isomorphic to) $\mathbb{Z}_{p^2}$ and $\mathbb{Z}_{p}\times\mathbb{Z}_p$, which are of the same order yet not isomorphic.
Hence if $G$ is a $p$-group of order $p^n$, it is either cyclic or of the form $\mathbb{Z}_p^n$.
Question 1: Did I account for all possible cases? I concluded that $\mathbb{Z}_p^n$ works because the diagonal subgroup is of order $p$, which is unique.
In the general case, we know at least that if $|G|=p_1^{e_1}\cdots p_k^{e_k}$, then $G$ must be either of the form $G=\mathbb{Z}_{p_1^{e_1}}\times\cdots \times\mathbb{Z}_{p_k^{e_k}}$, or $G=\mathbb{Z}_{p_1}^{e_1}\times\cdots\times\mathbb{Z}_{p_k}^{e_k}$. I suspect that these are all possible options.
Question 2: Did I account for all possible cases in the general case? I honestly doubt it, but I cannot substantiate my ominous hunch...
 A: Here's an approach using the structure theorem. Firstly, as you've said, either each $p_i$-torsion of $G$ is a sum of copies of $\mathbb{Z}_{p_i}$ or isomorphic $\mathbb{Z}_{p_i^{e_i}}$. This is because in any other case, we would have two $p_i$-subgroups with one having cardinality strictly greater than $p_i$, from which we conclude that there are subgroups isomorphic to $\mathbb{Z}_{p_i^2}$ and $\mathbb{Z}_{p_i} \oplus \mathbb{Z}_{p_i}$. We are then left to analyse these two cases, for each $p_i$-torsion. 
Let's also note that if the condition holds individually for each $p_i$-torsion of $G$, then if $H,K$ are subgroups of $G$, once again by the structure theorem we have that 
$$
H \simeq \bigoplus_{i = 1}^k H[p_i] \quad , \quad K \simeq \bigoplus_{i = 1}^k K[p_i]
$$
and taking cardinals and using that $|H| = |K|$ and the fact that each $p_i$-torsion has cardinality equal to a power of $p_i$, by uniqueness of factorization in the integers, $|H[p_i]| = |K[p_i]|$ for all $i$. Since $H[p_i] = H \cap G[p_i]$ and $K[p_i] = K \cap G[p_i]$ are subgroups of $G[p_i]$ with the same cardinality, they would be isomorphic, from which we can conclude $H \simeq K$. 
This reduces the problem to the exact same question, but restricted to each $p_i$-torsion: the condition is trivially necessary, and by the previous observation, also sufficient. 
It is clear that $G[p_i] = \mathbb{Z}_{p_i^{e_i}}$ verifies the condition, since it has exactly one subgroup of cardinality $p_i^s$ with $0 \leq s \leq e_i$.   
The other case is $G[p_i] = (\mathbb{Z}_{p_i})^{e_i}$, which is also a possible outcome since any non-trivial group will be isomorphic to copies of $\mathbb{Z}_{p_i}$. In effect, by the structure theorem, any subgroup of $G[p_i]$ is  a sum of $p_i$-groups, and since every nonzero element of $G[p_i]$ has order $p_i$ each is isomorphic to $\mathbb{Z}_{p_i}$. 
In conclusion, if $G$ is an abelian group with all subgroups of equal cardinality being isomorphic and $|G| = p_1^{e_1} \cdots p_k^{e_k}$, then
$$
G \simeq H_1 \oplus \cdots \oplus H_k
$$
with $H_i \simeq (\mathbb{Z}_{p_i})^{e_i}$ or $H_i \simeq \mathbb{Z}_{p_i^{e_i}}$ for each $i \in [k]$.
