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, D_{8}. 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)

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