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This question already has an answer here:

How could I show that a normal subgroup of a normal subgroup of a group is not always a normal subgroup of a group?

Here is what I did. I know the Algebra system GAP, so I wrote the following function:

performTheProve:= function()
    local g, gDash, AFour, NormalOfAFour, NormalK;
    AFour:=AlternatingGroup(4);
    NormalOfAFour:=NormalSubgroups(AFour);
    for g in  NormalOfAFour do
        for gDash in NormalSubgroups(g) do
            if(not gDash in NormalOfAFour) then
                return [gDash, g];
            fi;         
        od;
    od;
end; 

This function produces the following result:

[ Group([ (1,2)(3,4) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]) ]

So, now I am stating that the Group([ (1,3)(2,4), (1,2)(3,4) ]) is a normal subgroup of a ${A}_{4}$, Group([ (1,2)(3,4) ]) is a normal subgroup of a Group([ (1,3)(2,4), (1,2)(3,4) ]) and not a normal subgroup of a ${A}_{4}$.

Given this example I am proving that normal subgroup of a normal subgroup of a group is not always a normal subgroup of a group. Am I right?

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marked as duplicate by lulu, Derek Holt group-theory Jun 21 '18 at 20:55

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  • $\begingroup$ Yes, that's a standard example. $\endgroup$ – Lord Shark the Unknown Jun 21 '18 at 19:53
  • $\begingroup$ Just a question for my own amusement: surely if the containing group is abelian, that would be a sufficient, if not necessary condition for transitivity, wouldn't it? $\endgroup$ – Adrian Keister Jun 21 '18 at 20:07