# The equivalence classes of $N\sim M\Leftrightarrow G/N\cong G/M$.

Let $G$ be a finite group. Given some $N\unlhd G$, define $$\mathfrak{C}_N:=\{M\unlhd G : G/M \cong G/N\}.$$

1. How are the subgroups in $\mathfrak{C}_N$ related? Is there some other description of $\mathfrak{C}_N$?
2. Would there be a more direct way to compute this set than computing $G/N$, then computing $G/M$ for every (appropriate order) $M\unlhd G$ and checking for isomorphism?

For an example of what I mean by "more direct," say I wanted all conjugates of some $g\in G$. I could check each $g^\prime \in G$ to see if there exists an $x$ so that $g^\prime=g^x$, but it would be much easier to just compute $g^x$ for all $x\in G$ (or all $x$ in a transversal of $C_G(g)$ in $G$, if we have that information). If possible, I'd like to do a similar thing to compute $\mathfrak{C}_N$.

Motivation: I am running a computational experiment having to do with this problem (which is becoming somewhat of an obsession) that requires computing the complete partition of the set of normal subgroups of $G$ under $M\sim N \Leftrightarrow M\in \mathfrak{C}_N$. If $M,N\unlhd G$ are related by an outer automorphism, then surely $M\sim N$, but converse is not necessarily true. So, we could start by computing $N^{\operatorname{Out}(G)}$ for each $N\unlhd G$, but we would still have to check whether $N\sim M$ between each of those sets, so this is not much of an improvement.

EDIT: By the way, if this is too tough for general $G$ as @MartinBrandenburg suggests, I would still be interested to hear an answer for any of the following restricted cases: $p$-groups, nilpotent groups, solvable groups.

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I doubt you can relate the groups in the set very easily. Note for example that all maximal subgroups of a $p$-group will be in such a set together, and they need not be related very much, apart from having the same order. –  Tobias Kildetoft Mar 12 '13 at 17:38
Maybe, it would reasonable to consider another problem: when the exact sequences $$0\to M\to G\to G/M$$ and $$0\to N\to G\to G/N$$ are isomorphic (i.e. the corresponding diagram is commutative)? –  Boris Novikov Mar 12 '13 at 17:44
This comment is only remotely related to what you are doing, but I thought maybe you might get some inspiration from Reyes' use of "similar right ideals" in this paper. Right ideals $A$ and $B$ are similar if $R/A$ and $R/B$ are isomorphic as right $R$ modules. It turns out there are several intersting things you can say about such ideals. Good question btw! –  rschwieb Mar 12 '13 at 17:46
This is one of the many questions which are quite easy to ask but which have no good answer (at least for general $G$). My advice (and not more): Don't waste your time ... –  Martin Brandenburg Mar 12 '13 at 18:10
@MartinBrandenburg I understand where you might say that about (1), but don't you find (2) interesting? It seems to me that at least a small improvement ought to be achievable. –  Alexander Gruber Mar 12 '13 at 18:26

## 1 Answer

There is a special case of your question that plays an essential role in the $p$-group generation algorithm. It should be Theorem 2.5 there.

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This is very useful. –  Alexander Gruber Apr 9 '13 at 1:07