Full invariance is not direct power-closed

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

It is possible to have a group $G$ and a fully invariant subgroup $H$ of $G$ such that in the countable unrestricted direct power of $G$, the corresponding direct power of $H$ is not a fully invariant subgroup.

In the example used here, all groups involved are Abelian group (?)s.

Related facts

• Full invariance is finite direct power-closed
• Characteristicity is not finite direct power-closed

Proof

Further information: quasicyclic group

Let $p$ be a prime number and let $A$ be the quasicyclic group corresponding to the prime $p$. Let $B$ be the countable unrestricted direct power of $A$ and let $C$ be the subgroup of $B$ comprising the elements of finite order. Let $D$ be the quotient group $B/C$, with a natural projection map $\pi :B\to D$. Define $G=A\times D$ and let $H$ be the first direct factor, i.e., the subgroup $A\times \{e\}$.

We note that:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$A$ is periodic, i.e., every element has finite order		$A$ is quasicyclic
2	$C$ is a proper subgroup of $B$, and hence $D$ is nontrivial		$A$ is quasicyclic, definitions of $B,C,D$		$A$ contains elements of order $p^{n}$ for every natural number $n$. Consider an element of $B$ whose $n^{th}$ coordinate is an element of $A$ of order $p^{n}$. This element has infinite order, and is hence not in $C$. This proves that $C$ is proper in $B$.
3	$D$ is torsion-free, i.e., it has no non-identity element of finite order		Definitions of $B,C,D$		If $x\in D$ has finite order, let $y$ be an element of $B$ such that $\pi (y)=x$. Some finite power of $y$ is thus in $C$. But since this finite power itself has finite order, $y$ also has finite order, forcing $y\in B$, and hence forcing $x$ to be trivial.
4	$H$ is precisely the set of elements of finite order in $G$			Steps (1), (3)	From steps (1) and (3), we see that for an element of $G$ to have finite order, its $A$-coordinate can be arbitrary and its $D$-coordinate must be trivial. This precisely characterizes $H$.
5	$H$ is fully invariant in $G$			Step (4)	Under any endomorphism, an element of finite order must go to an element of finite order (in fact, the order of the image element must divide the order of the original element). Thus, under any endomorphism of $G$, the image of $H$ is in $H$.
6	The countable unrestricted direct power $G^{\omega }$ of $G$ can be identified with $A^{\omega }\times D^{\omega }=B\times D^{\omega }$, with $H^{\omega }$ identified with the first direct factor $B\times \{e\}$.
7	There is a nontrivial surjective homomorphism $\varphi :B\times D^{\omega }\to D$ given by $\varphi (b,d)=\pi (b)$				Note that the homomorphism is surjective because $\pi$ is the quotient map from $B$ to $D$. It is nontrivial because, by step (2), $D$ is nontrivial and the map is surjective.
8	There is a nontrivial endomorphism $\alpha :B\times D^{\omega }\to B\times D^{\omega }$ that does not preserve the subgroup $B\times \{e\}$				We obtain $\alpha$ by post-composing $\varphi$ with an inclusion of $D$ in any of the copies of $D$ in the $D^{\omega }$ part of the product. Since $\varphi$ was nontrivial, some element of $B\times \{e\}$ gets sent outside under $\alpha$.
9	The proof is complete			Steps (6), (8)	Step-combination direct