The property “is a normal subgroup of” is not transitive.
If A is a subgroup of B, and B is a subgroup of C, then A is a subgroup of C. But the corresponding statement about normal subgroups is false. And there’s a simple example that shows it is false.
We need to find a group C with subgroups A and B such that A is normal in B, B is normal in C, but A is not normal in C.
The subgroup A must have at least two elements, otherwise A would just be the group identity and would then be a normal subgroup of C. The order of a subgroup divides the order of the group, so B must have at least twice as many elements as A, and C must have twice as many elements as B. So the smallest possible example would be a group with 8 elements and subgroups of order 2 and 4.
We’re in luck, because there’s a group of order 8 that will work, D8. This is the group of symmetries of a square under flips and rotations. Let A be the subgroup of flips about the vertical axis of symmetry. Let B the symmetries you can find by combinations of such flips and 180 degree rotations. You can show that A is normal in B, and B is normal in C.
Now let c be a 90 degree clockwise turn and let a be a flip. You can show that cac-1 is not a flip or the identity, so A is not a normal subgroup of C.
Related post: A 3,000 page proof (classification of finite simple groups)
2 thoughts on “Normal subgroups are not transitive”
Now suppose S is the only normal subgroup of H and H is a normal subgroup of G, is S a normal subgroup of G?
Very helpful post! Now mentioned on stackexchange.