TITLE: Finite abelian groups in which quotients of same order are isomorphic
QUESTION [3 upvotes]: Let $G$ be a finite abelian group which is isomorphic to direct sum of some elementary abelian groups and a cyclic group such that all summands have coprime orders. Are quotients of its all subgroups of same order isomorphic?
If yes, Is it sufficient also?

REPLY [2 votes]: Interesting problem.
In other words, you have a finite abelian group in which the Sylow subgroups are elementary abelian, or cyclic. Then clearly the order of a subgroup determines its isomorphism type, as a subgroup of an elementary abelian group is elementary abelian, and a subgroup of a cyclic group is cyclic. (I'm doing this for subgroups, but the case to quotients is entirely similar.)
This also hints at the converse. In fact, if a finite abelian $p$-group $H$ is not elementary abelian, nor cyclic, by the structure theorem it will be of the form
$$
H = C_{p^{a}} \times C_{p^{b}} \times \dots
$$
with $a \ge 2$ and $b \ge 1$. Then $H$ has two subgroups of order $p^{2}$ that are not isomorphic, namely a cyclic one and an elementary abelian one.
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