Characteristicity does not satisfy intermediate subgroup condition

This article gives the statement, and possibly proof, of a subgroup property (i.e., characteristic subgroup) not satisfying a subgroup metaproperty (i.e., intermediate subgroup condition).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about characteristic subgroup|Get more facts about intermediate subgroup condition|

Statement

Statement with symbols

It is possible to have a group $G$ with a characteristic subgroup $H$, and a subgroup $K\le G$ containing $H$, such that $H$ is not characteristic as a subgroup of $K$.

Partial truth

Corresponding statement for normality

Further information: Normality satisfies intermediate subgroup condition

If $H\le K\le G$ are such that $H$ is normal in $G$, then $H$ is normal in $K$. In particular, since every characteristic subgroup is normal, it follows that if $H$ is characteristic in $G$, $H$ is normal in $K$.

Intermediately characteristic subgroup

Further information: Intermediately characteristic subgroup

A subgroup of a group that is characteristic in every intermediate subgroup is termed an intermediately characteristic subgroup. Under certain conditions, we can guarantee a subgroup to be intermediately characteristic. For instance, any normal Sylow subgroup, any more generally, any normal Hall subgroup, is intermediately characteristic. There are certain subgroup-defining functions that yield intermediately characteristic subgroups, for instance, the perfect core of any group is intermediately characteristic. These are, however, extremely rare.

Potentially characteristic subgroup

Further information: Potentially characteristic subgroup

A subgroup $H$ of a group $K$ is termed potentially characteristic if there exists a group $G$ containing $K$ such that $H$ is characteristic in $G$. Clearly, every potentially characteristic subgroup is normal; on the other hand, since characteristicity does not satisfy the intermediate subgroup condition, not every potentially characteristic subgroup is characteristic.

Proof

Note that for any counterexample, $H$ must be a nontrivial subgroup, $K$ must properly contain $H$, and $G$ must properly contain $K$. Thus, the order of $G$ must be at least eight. We give here two counterexamples of size eight: one Abelian, and one non-Abelian.

Example of an Abelian group of order eight

Consider:

• $G=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$.
• $H$ to be the set of squares in $G$, i.e., $H=\{(0,0),(0,2)\}$ -- $H=\stackrel{\mathrm{g}}{\mathrm{A}}(G)$ (the first agemo subgroup) and is thus characteristic.
• $K$ to be the set of elements of order two in $G$, i.e., $K=\{(0,0),(1,0),(0,2),(1,2)\}$ -- $K={\mathrm{\Omega }}_1(G)$ (the first omega subgroup) and is thus characteristic.

Now, $K$ is an internal direct product of $H$ and the subgroup $L=\{(0,0),(1,0)\}$, so it is isomorphic to the Klein-four group, and it admits an isomorphism that exchanges the two direct factors $H$ and $L$. Hence, $H$ is not characteristic in $K$.

Example of the dihedral group

Consider the dihedral group of order eight, with an element $a$ of order $4$, that plays the role of the rotation, and $x$, that plays the role of the reflection. Then, consider:

• $G$ is the dihedral group
• $H=\{e,a^2\}$ is the center of $G$ -- hence it is characteristic in $G$
• $K=\{e,a^2,x,a^{2}x\}$ is a subgroup of $G$ isomorphic to the Klein-four group.

Now, $K$ is the internal direct product of $H$ and the subgroup $L=\{e,x\}$, so it is isomorphic to the Klein-four group, and it admits an automorphism that exchanges the two direct factors $H$ and $L$. Hence, $H$ is not characteristic in $K$.