Can $G≅H$ and $G≇H$ in two different views?

We have two isomorphic groups $G$ and $H$, so $G≅H$ as groups and suppose that they act on a same finite set, say $\Omega$. Can we see $G≇H$ as permutation groups. Honestly, I am intrested in this point in the following link. It is started by:

Notice that different permutation groups may well be isomorphic as ....

See here

  • $\begingroup$ I like this question! +1 $\endgroup$
    – amWhy
    Mar 13, 2013 at 0:57

2 Answers 2


The definition of "isomorphic as permutation groups" given in the wikipedia page you refer to is equivalent to the images of $G$ and $H$ in their actions on $\Omega$ being conjugate in the symmetric group on $\Omega$.

You should be able to think of examples of subgroups $G$ and $H$ of $S_n$ for some $n$ such that $G$ and $H$ are isomorphic as groups, but not conjugate in $S_n$. Try groups of order 2 in $S_4$, for example.

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    $\begingroup$ Dear Prof.,here are two subgroups of $S_4$ of order two; Group([ (), (1,2) ]) and Group([ (), (1,2)(3,4) ]). $\endgroup$
    – Mikasa
    Jun 17, 2012 at 10:29
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    $\begingroup$ There you go, @Babak! Those two groups you mention are isomorphic as groups yet they act differently on $\,\{1,2,3,4\}\,$ and thus they are not isomorphic as permutation groups, according to Derek's definition, as they clearly are not conjugate subgroups of $\,S_4\,$ $\endgroup$
    – DonAntonio
    Jun 17, 2012 at 10:32
  • $\begingroup$ @DonAntonio: I got it from 2 answers here. Thanks for your comment. You always come suddenly and hint me nicely. :-) $\endgroup$
    – Mikasa
    Jun 17, 2012 at 10:35
  • $\begingroup$ Nice to know that, @Babak. +1 for yet another interesting question that, again, makes me think. $\endgroup$
    – DonAntonio
    Jun 17, 2012 at 10:37
  • $\begingroup$ To clarify what is meant, it's not that the two groups act differently on $\{1,2,3,4\}$ that's important, but that the two actions are not isomorphic as $\mathbf{Z}/2\mathbf{Z}$-sets. $\endgroup$
    – user14972
    Jun 17, 2012 at 12:31

Let $G$ and $H$ act faithfully on a finite set $\Omega$. This is equivalent to say that $G$ and $H$ are subgroups of the total permutation group $\mathfrak S_\Omega$.

Here are two notion of isomorphism which are stronger than a simple group isomorphism :

  1. The action of $G$ and $H$ are isomorphic if there exists a group morphism $\phi : G \to H$ and a bijection $\sigma : \Omega \to \Omega$ such that for all $g\in G$ and $x\in \Omega$ $$ g\cdot \sigma x = \sigma(\phi g \cdot x). $$ In particular, the morphism $\phi$ is an isomorphism, because if $\phi g = 1$ then for all $x\in\Omega$ $g\cdot \sigma x = \sigma x$, and since the action of $G$ is faithful and $\sigma$ surjective, this implies that $g = 1$.

  2. $G$ and $H$ are said to be conjugate in $\mathfrak S_\Omega$ is there exist a permutation $\sigma$ of $\Omega$ such that $G = \sigma H \sigma^{-1}$. In particular the map $g\in G \mapsto \sigma^{-1} g \sigma\in H$ is an isomorphism.

In fact, both notion are the same (easy exercise for you !), and its what Wikipedia call permutation group isomorphism.

This notion is strictly stronger that the notion of group isomorphism. For example, take $\Omega = \{1,2,3,4\}$, $G$ the group of order $2$ generated by the transposition $(1 2)$ and $H$ the group of order $2$ generated by the double transposition $(1 2)(3 4)$. As groups, $G$ and $H$ are isomorphic, because they are both isomorphic to $\Bbb Z/2\Bbb Z$. However, they are not isomorphic as permutation group. Indeed, the conjugate of a transposition is always a transposition, it cannot be a double transposition. More precisely, the conjugate $\sigma (1 2) \sigma^{-1}$ is the transposition $(\sigma 1, \sigma 2)$.

When classifying the subgroups of a given group, it is often important to classify them up to isomorphism but also up to conjugation, because isomorphism class can split into several conjugation classes.

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    $\begingroup$ Before I could see your nice answer, I was taking groups $H$ and $G$ as you noted kindly. May I deduce that since $G$ is transitive on $\Omega$ and the other group is not, so they are not isomorphic as permutaion groups? I got more than I had expected. Thanks. $\endgroup$
    – Mikasa
    Jun 17, 2012 at 10:34
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    $\begingroup$ @BabakSorouh — That would have been a good reason to say that G and H are not isomorphic actions, however H is not transitive since 2 can only be send on itself or on 1. But it is true that H fixes no points, whereas G fixes 2 points (3 and 4), so they cannot be isomorphic. $\endgroup$
    – Lierre
    Jun 17, 2012 at 10:49
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    $\begingroup$ You probably want to use the word "transposition" instead of "permutation". $\endgroup$ Jun 22, 2012 at 0:16
  • $\begingroup$ @GenericHuman — I do, thank you for pointing it out. $\endgroup$
    – Lierre
    Jun 22, 2012 at 5:13

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