Here are some silly things I keep forgetting.
Suppose we have a group and two subgroups , we consider three possible hypotheses , , and . How do these properties relate to each other?
In the following table are summarized all the possible implications (except the ones that are trivially true which are indicated by dots)
Let’s start with the second row, it’s pretty easy to see that and that by letting and any non normal subgroup of one finds that .
Now the first row, we want to prove . It suffices to find a group which has a non normal subgroup , then using the property of direct products we find that letting (where is a cyclic group of order ) is a group with the desired property.
The smallest example satisfying the first step appears when the dihedral group of degree and ; indeed .
(technically we could have used a semidirect product but it’s definitely easier this way).
Now let’s focus on , is it true that ? This one is easy to disprove, just take (with ) and a group in which is not normal. For example , we then use the Cayley embedding to find a copy of in the symmetric group on letters which will fill the role of . Since the only nontrivial normal subgroup of are the alternating group and one of its subgroups isomorphic to (thanks GroupNames!) this proves (this is probably not a minimal example but I like the reasoning).
We are almost done, we just need to find an example of a chain of subgroups such that , proving that sadly (?) being a normal subgroup of a subgroup is not a transitive relation.
This was a bit harder to do, for starters it’s reasonable to pick an Abelian group for as all its subgroups will be normal.
Here’s a solution: take for , it’s easy (although rather boring) to prove the desired relations: for starters one should prove , then follows from the properties of the direct product. Then since one has that . Finally to prove it a matter of checking wether and .
Here’s a much more fun way to find the last example that came to my mind after writing the boring one above: the idea is that a normal subgroup is fixed by conjugation, but those are not the only automorphisms a group can have! Indeed if we can find an automorphism such that we can form the semi-direct product thus finding the desired chain .
Here’s an easy example following the above scheme. Take any group , it’s clear that, identifying with the subgroup we have , moreover it’s easy to show that the map is an automorphism of order . By the reasoning above we conclude that is a chain of subgroups with the desired property.
Taking the group has order and is nonabelian meaning it must be isomorphic to one of or the Quaternion group. However all subgroups of the Quaternion group are normal so it must be that .