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A subgroup $H$ of the group $G$ is normal group in $G$ if and only if $\displaystyle ghg^{-1}\in H$ for all $\displaystyle g\in G$ and $\displaystyle h\in H$. How to use MMA to know the group $H$ is a normal group of group $G$? Such as:

G = SymmetricGroup[6];
H = AlternatingGroup[6];
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  • $\begingroup$ g = GroupElements[SymmetricGroup[6]]; h = GroupElements[AlternatingGroup[6]]; conj[x_] := PermutationProduct[#, x, InversePermutation[#]] & /@ g Brute fprcing: Union[Flatten[conj /@ h]] == Sort[h] yields True $\endgroup$
    – ubpdqn
    May 16, 2022 at 8:43
  • $\begingroup$ @ubpdqn Yes, I note your solution here. But RandomSample[Sort[GroupElements[AlternatingGroup[6]]]]===GroupElements[AlternatingGroup[6]] will return False. You need a parameter Less like my answer in following. And I use the PermutationReplace to get the conjugate is more concise than yours. :) $\endgroup$
    – yode
    May 16, 2022 at 9:10
  • $\begingroup$ Not sure I understand but I'll accept your point $\endgroup$
    – ubpdqn
    May 16, 2022 at 9:12
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    $\begingroup$ @ubpdqn I've just updated the very efficient algorithm. $\endgroup$
    – yode
    May 27, 2022 at 7:49

1 Answer 1

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IsNormalQ[H_, G_] := Sort[DeleteDuplicates[Catenate[
     Outer[PermutationReplace, GroupElements[H], 
      Complement[GroupElements[G], GroupElements[H]]]]], Less] === GroupElements[H]

Or little concise, but sometimes it will be slower:

IsNormalQ[H_, G_] := ContainsExactly[
  Catenate[Outer[PermutationReplace, GroupElements[H], 
    Complement[GroupElements[G], GroupElements[H]]]],GroupElements[H]]

Update(Simplicity and efficiency)

IsNormalQ[H_,G_]:=AllTrue[GroupElements[G],PermutationReplace[H,#]==H&]
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