This question is related to the following three questions:

  1. Two subgroups $H_1, H_2$ of a group $G$ are conjugate iff $G/H_1$ and $G/H_2$ are isomorphic
  2. If $H_1, H_2\leq G$ are such that $H_1\cong H_2$ then $G/H_1\cong G/H_2$?
  3. Isomorphic quotients by isomorphic normal subgroups

Let $G$ be a group, and $H_1$, $H_2$ be two normal subgroups of $G$, and $\varphi : H_1 \to H_2$ be a group isomorphism. Consider the following proposition: $$(P): \text{The groups $G/H_1$ and $G/H_2$ are isomorphic.}$$

In If $H_1, H_2\leq G$ are such that $H_1\cong H_2$ then $G/H_1\cong G/H_2$?, we can see that $(P)$ does not necessarily hold, even if $G$ is assumed to be abelian and finite. On the other hand, it is quite easy to show that $(P)$ holds in the following cases :

  • If $\frac{|G|}{|H_1|} = \frac{|G|}{|H_2|}$ and if this number is prime.
  • More particularly, $(P)$ holds for all subgroups $H_1$ and $H_2$ of $G$ if $|G|$ is the product of two prime numbers.

Question: Let $G \simeq \mathbb{Z}/q_1\mathbb{Z} \times\dots\times \mathbb{Z}/q_r\mathbb{Z}$ be some finite abelian group, where $q_1,\dots,q_r$ are prime powers. Do we know a necessary and sufficient condition on $(q_1,\dots,q_r)$ so that $(P)$ holds for all subgroups $H_1$ and $H_2$? What if we restrict $H_1$ and $H_2$ to be subgroups of $G$ of cardinality $d$, where $d$ is a factor of $\prod_{i=1}^r q_i$?


The answer to you main question is: $G$ has property (P) if and only if all the $q_i$'s are equal. This is not too hard to see. If all the $q_i$'s are equal, then all subgroups that are isomorphic are actually conjugate under the automorphism group of $G$. Conversely, if not all of them are equal, you can find two subgroups of order $p$ with non-isomorphic quotients (in the obvious way, by taking them from factors of different orders).

For your second question, I think the answer is the same for every $d$, assuming it's a non-trivial divisor ($1<d<|G|$). I don't have all the details worked out but, basically, you can more or less ignore $p$ and encode your group $G$ by the list of $q_i$'s.

For example, $G$ could be the group $(1,2,3)$ and $(1,1,0)$ and $(0,1,1)$ would represent two isomorphic subgroups, with quotients $(0,1,3)$ and $(1,1,2)$, so non-isomorphic.

Now, it suffices to show that, unless you have the constant sequence, for every non-trivial sum $s$, you can always find two dominated sequences of sum $s$ which are permutations of each other, such that the difference with the original sequence are not permutations of each other.

This is essentially a combinatorics problem and I think it's true, but there is a little work to be done.

EDIT: As was pointed out in the comments, I was assuming that $G$ is a $p$-group, for some reason. But note that the general problem is easily reduced to that case: (abelian) $G$ has this property if and only if all its Sylow $p$-subgroups have this property.

  • 1
    $\begingroup$ I’m not sure I agree with your “if and only if”. You seem to have assumed all the $q_i$ are powers of the same prime. I do not see that in the original question, and of course it is easy to verify that if all the $q_i$ are powers of pairwise distinct primes, then $G$ is cyclic and the result follows by vacuity (since there is one and only one subgroup of each order). $\endgroup$ – Arturo Magidin May 11 '20 at 1:20
  • $\begingroup$ Yes, indeed, I had assumed that, I'm not sure why. It should be easy to find the right fix, I'll get to it. $\endgroup$ – verret May 11 '20 at 1:25
  • $\begingroup$ I expect what you want is your condition on each prime part; i.e., each prime part is homocyclic. $\endgroup$ – Arturo Magidin May 11 '20 at 1:27
  • $\begingroup$ Yes, that's right.Probably I had implicitly done the reduction in my head and forgot to go back to the original problem. $\endgroup$ – verret May 11 '20 at 1:29

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