Normal not implies characteristic

Statement
It is possible to have a group $$G$$ and a normal subgroup $$H$$ of $$G$$ that is not a characteristic subgroup of $$G$$.

Converse
The converse statement is indeed true. That is, every characteristic subgroup is normal.

Stronger formulations using the same examples

 * Normal not implies characteristic in the collection of all groups satisfying a nontrivial finite direct product-closed group property
 * Every nontrivial normal subgroup is potentially normal-and-not-characteristic
 * Conjunction of normality with a nontrivially satisfied group property not implies characteristic

Other related facts

 * Normality is not transitive: A normal subgroup of a normal subgroup need not be normal.
 * Characteristic of normal implies normal: A characteristic subgroup of a normal subgroup is normal.
 * Left transiter of normal is characteristic: If $$H$$ is a subgroup of $$K$$ such that whenever $$K$$ is normal in $$G$$, $$H$$ is normal in $$G$$, then $$H$$ is characteristic in $$K$$.
 * Direct factor not implies characteristic

Related group properties
There are some groups in which every normal subgroup is characteristic.

Example of a direct product
Let $$H$$ be any nontrivial group. Then consider $$G = H \times H$$, viz., the external direct product of $$H$$ with itself. The subgroups $$H_1 := H \times \{ e \}$$ and $$H_2 := \{ e \} \times H$$ are direct factors of $$G$$, and are hence both normal in $$G$$. Note also that they are distinct, since $$H$$ is nontrivial.

However, the exchange automorphism:

$$(x,y) \mapsto (y,x)$$

exchanges the subgroups $$H_1$$ and $$H_2$$. Thus, neither $$H_1$$ nor $$H_2$$ is invariant under all the automorphisms, so neither is characteristic. Thus, $$H_1$$ and $$H_2$$ are both normal subgroups of $$K$$ that are not characteristic.

Note that this example also shows that direct factor does not imply characteristic subgroup.

Particular cases of this example

 * When $$H$$ is the cyclic group of order two, $$G$$ is the particular example::Klein four-group. In particular, this gives a counterexample where the ambient group is an abelian group. More generally, we can start with any nontrivial abelian group $$H$$.

Implementation of the generic example
Before using this generic example, you need to define $$G$$ for GAP, choosing any nontrivial group (double semicolons have been used here to suppress GAP's output for the first three commands, which depends on the specific choice of $$G$$ -- you can use single semicolons instead).

gap> G := DirectProduct(H,H);; gap> H1 := Image(Embedding(G,1));; gap> H2 := Image(Embedding(G,2));; gap> IsSubgroup(G,H1); true gap> IsSubgroup(G,H2); true gap> IsNormal(G,H1); true gap> IsNormal(G,H2); true gap> IsCharacteristicSubgroup(G,H1); false gap> IsCharacteristicSubgroup(G,H2); false

Textbook references

 * , Page 70 (Problem 7(a))
 * , Page 137 (Problem 6)