Normal not implies characteristic

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal subgroup) need not satisfy the second subgroup property (i.e., characteristic subgroup)
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Statement

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

Related facts

Converse

Further information: Characteristic implies normal

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. Further information: Group in which every normal subgroup is characteristic

Proof

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 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 $H$.

Initial examples

Here are some examples where the ambient group is quite important and easy to understand:

Here are some examples where the ambient group is somewhat more complicated:

Here are some examples where the ambient group is even more complicated:

GAP implementation

Implementation of the generic example

Before using this generic example, you need to define $H$ 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 $H$ -- 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

References

Textbook references

• Topics in Algebra by I. N. Herstein^{More info}, Page 70 (Problem 7(a))
• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347^{More info}, Page 137 (Problem 6)