I was surprised by this question which provides an example of a finite group $G$ where normal isomorphic subgroups $H, K\triangleleft G$ do not imply isomorphic quotient groups, i.e. $G/H\not\cong G/K$.

Question: What condition must be met in order for the quotient groups to be isomorphic?

Trying to sort this out intuitively, I think like this:

  • A finite group is fully described by its multiplication table and isomorphic groups have the same table.
  • Still we evidently can't say "Let $J\triangleleft G$ be a subgroup with table $T$; form the quotient group $G/J$".
  • It breaks down when trying to carry out coset multiplication in $G/J$.
  • Because coset multiplication depends not only on multiplication in $J$ as specified by $T$ but also on multiplication of elements in $J$ with elements in the embedding group $G$.

So an attempt at an answer would be that $\forall g\in G, h\in H:gh=g\psi(h)$ where $\psi$ is an isomorphism between $H$ and $K$. I am trying to express that isomorphic elements of the subgroups interact similarly with elements of the embedding group. Is this the right conclusion?

  • 4
    $\begingroup$ I don't know about necessary assumptions, but a nice sufficient one is that there be an isomorphism of $G$ sending $H$ to $K$; or more generally a surjective morphism $G\to G$ such that $\pi^{-1}(K) = H$. $\endgroup$ – Max Jan 15 at 22:09
  • $\begingroup$ $gh = g\psi(h) \Rightarrow h=\psi(h)$. $\endgroup$ – Derek Holt Jan 15 at 22:30
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    $\begingroup$ $H,K$ don't need to be isomorphic. Given two normal subgroups $H,K$ of same order and a function $f : G \to G$ such that $f(aH) \subset f(a)K$ and $f(ab)\in f(a)f(b)K$ then $F : G/H \to G/K, F(aH) = f(a)K$ is an homomorphism. If $f$ is a permutation of $G$ then $F$ is surjective so it is an isomorphism $G/H \to G/K$. $\endgroup$ – reuns Jan 16 at 5:09
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    $\begingroup$ I suggest that as a preliminary project you ask the same question about finite abelian $p$-groups for a fixed prime $p$. Perhaps explore what can be said if $H=\mathbb{Z}_{p^3}$ and $K=(\mathbb{Z}_p)^3$ (I think smaller cases are trivial). I think you'll find that there's not much that can be said except "$G/H\cong G/K$". $\endgroup$ – ancientmathematician Jan 16 at 8:09
  • $\begingroup$ Thanks to all of you who responded. That set me off in a good direction. $\endgroup$ – Daniel Jan 16 at 17:18

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