Characteristicity is not finite direct power-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., characteristic subgroup) not satisfying a subgroup metaproperty (i.e., finite direct power-closed subgroup property).
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Statement

Statement with symbols (n = 2 or square case, which implies the general statement)

It is possible to have a group $G$ and a characteristic subgroup $H$ of $G$ such that in the direct square $G\times G$, the corresponding direct square $H\times H$, viewed as a subgroup, is not a characteristic subgroup.

Related facts

• Full invariance is finite direct power-closed, hence, fully invariant implies finite direct power-closed characteristic

Notion of finite direct power-closed characteristic

A subgroup $H$ of a group $G$ is termed a finite direct power-closed characteristic subgroup if for any natural number $n$, $H^{n}$ is characteristic in $G^{n}$.

Proof

Further information: direct product of Z8 and Z2

Suppose $G=\mathbb {Z} _{8}\times \mathbb {Z} _{2}$, and $H$ is the subgroup of order four given by:

$H:=\{(2,1),(4,0),(6,1),(0,0)\}$

Note that this subgroup is characteristic but not fully invariant, and is a standard example of the fact that characteristic not implies fully invariant in finite abelian group.

Then, we claim that $H\times H$ is not a characteristic subgroup inside $G\times G$. To see this, consider the endomorphism of $G$ that projects onto the $\mathbb {Z} _{8}$ coordinate, i.e., the map:

$\pi :(a,b)\mapsto (a,0)$

Now, consider the automorphism of $G\times G$ defined as:

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

This is an endomorphism as it is a sum of endomorphisms; it is an automorphism because it has an inverse:

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

In full, the automorphism is:

$((a,b),(c,d))\mapsto ((a,b),(c+a,d))$

Under this automorphism, the element:

$((2,1),(2,1))\in H\times H$

gets mapped to the element:

$((2,1),(4,1))$

which is not inside $H\times H$.