# If $H_1$ and $H_2$ are isomorphic normal subgroups of $G$, when do we have an isomorphism between $G/H_1$ and $G/H_2$?

This question is related to the following three questions:

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 ($$1). I don't have all the details worked out but, basically, you can more or less ignore $$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.

• 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). – Arturo Magidin May 11 '20 at 1:20
• 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. – verret May 11 '20 at 1:25
• I expect what you want is your condition on each prime part; i.e., each prime part is homocyclic. – Arturo Magidin May 11 '20 at 1:27
• Yes, that's right.Probably I had implicitly done the reduction in my head and forgot to go back to the original problem. – verret May 11 '20 at 1:29